HILBERT SPACES AND INNER PRODUCT SPACES -
FOUNDATIONS OF FUNCTIONAL ANALYSIS
1 INTRODUCTION TO INNER PRODUCT SPACES AND HILBERT SPACES
Inner product spaces and Hilbert spaces are fundamental concepts in functional analysis,
providing a framework for studying infinite-dimensional vector spaces with additional structure.
Through exercises and discussions, we’ll explore their properties, relationships, and
applications.
2 EXERCISE 1: INNER PRODUCT SPACES
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
1.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
2.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
3.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
1. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
2. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
3. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
3 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
4 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
5 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
6 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
4.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
5.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
6.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
7. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
8. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
9. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
7 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
8 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
9 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
10 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
10.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
11.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
12.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
13. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
14. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
15. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
11 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
12 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
13 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
14 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
16.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
17.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
18.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
19. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
20. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
21. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
15 EXERCISE 6: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
16 EXERCISE 7: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
17 EXERCISE 8: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
18 EXERCISE 9: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
22.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
23.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
24.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
25. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
26. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
27. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
19 EXERCISE 10: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
20 EXERCISE 11: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
21 EXERCISE 12: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
22 EXERCISE 13: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
28.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
29.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
30.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
31. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
32. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
33. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
23 EXERCISE 14: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
24 EXERCISE 15: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
25 EXERCISE 16: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
26 EXERCISE 17: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
34.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
35.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
36.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
37. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
38. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
39. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
27 EXERCISE 18: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
28 EXERCISE 19: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
29 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
30 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
40.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
41.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
42.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
43. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
44. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
45. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
31 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
32 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
33 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
34 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
46.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
47.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
48.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
49. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
50. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
51. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
35 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
36 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
37 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
38 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
52.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
53.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
54.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
55. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
56. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
57. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
39 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
40 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
41 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
42 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
58.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
59.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
60.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
61. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
62. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
63. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
43 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
44 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
45 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
46 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
64.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
65.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
66.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
67. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
68. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
69. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
47 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
48 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
49 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
50 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
70.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
71.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
72.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
73. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
74. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
75. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
51 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
52 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
53 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
54 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
76.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
77.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
78.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
79. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
80. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
81. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
55 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
56 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
57 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
58 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
82.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
83.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
84.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
85. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
86. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
87. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
59 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
60 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
61 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
62 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
88.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
89.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
90.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
91. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
92. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
93. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
63 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
64 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
65 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
66 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
94.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
95.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
96.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
97. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
98. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
99. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
67 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
68 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
69 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
70 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
Definition 1.
An inner product space is a vector space
𝑉
over a field
𝔽
(usually
ℝ
or
ℂ
) with
an inner product
⟨⋅,⋅⟩:𝑉 × 𝑉 → 𝔽
satisfying:
100.
Conjugate symmetry:
⟨𝑥,𝑦⟩ = ⟨𝑦,𝑥⟩
101.
Linearity in the first argument:
⟨𝑎𝑥 +𝑏𝑦,𝑧⟩ = 𝑎⟨𝑥,𝑧⟩+𝑏⟨𝑦,𝑧⟩
102.
Positive definiteness:
⟨𝑥,𝑥⟩ ≥ 0
, and
⟨𝑥,𝑥⟩ = 0
if and only if
𝑥 = 0
Prove that the space 𝐶[0,1] of continuous functions on [0,1] with the inner product ⟨𝑓,𝑔⟩ =
∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 is an inner product space.
Solution: We need to verify the three properties of an inner product:
103. Conjugate symmetry: ⟨𝑓,𝑔⟩ = ∫𝑓
1
0(𝑥)𝑔(𝑥)𝑑𝑥 =∫𝑔
1
0(𝑥)𝑓(𝑥)𝑑𝑥 = ⟨𝑔,𝑓⟩ (Note: For real-
valued functions, conjugation has no effect)
104. Linearity in the first argument: ⟨𝑎𝑓 +𝑏𝑔,ℎ⟩ = ∫(𝑎𝑓(𝑥)+𝑏𝑔(𝑥))
1
0ℎ(𝑥)𝑑𝑥 =
𝑎∫𝑓
1
0(𝑥)ℎ(𝑥)𝑑𝑥 +𝑏∫𝑔
1
0(𝑥)ℎ(𝑥)𝑑𝑥 = 𝑎⟨𝑓,ℎ⟩+𝑏⟨𝑔,ℎ⟩
105. Positive definiteness: ⟨𝑓,𝑓⟩ = ∫𝑓
1
0(𝑥)2𝑑𝑥 ≥ 0 for all 𝑓 ∈ 𝐶[0,1] If ⟨𝑓,𝑓⟩ = 0, then
∫𝑓
1
0(𝑥)2𝑑𝑥 = 0, which implies 𝑓(𝑥)= 0 for all 𝑥 ∈ [0,1]
Therefore, 𝐶[0,1] with this inner product is indeed an inner product space.
71 EXERCISE 2: HILBERT SPACES
Definition 2.
A Hilbert space is a complete inner product space, i.e., an inner product space in
which every Cauchy sequence converges to an element in the space.
Prove that the space ℓ2 of square-summable sequences, defined as
ℓ2= {(𝑥𝑛)𝑛=1
∞:∑|𝑥𝑛|2
∞
𝑛=1 < ∞}
with inner product ⟨𝑥,𝑦⟩ = ∑𝑥𝑛
∞
𝑛=1 𝑦𝑛, is a Hilbert space.
Solution: We need to show that ℓ2 is an inner product space and that it is complete.
1. Inner product space properties: The inner product properties can be verified similarly to
Exercise 1.
2. Completeness: Let (𝑥(𝑘))𝑘=1
∞ be a Cauchy sequence in ℓ2. We need to show it converges to
an element in ℓ2.
For each fixed 𝑛, (𝑥𝑛
(𝑘))𝑘=1
∞ is a Cauchy sequence in ℂ. Since ℂ is complete, this sequence
converges to some 𝑥𝑛.
We need to show that (𝑥𝑛)𝑛=1
∞∈ ℓ2 and that 𝑥(𝑘)→ 𝑥 in ℓ2.
Given 𝜖 > 0, there exists 𝑁 such that for all 𝑘,𝑚 ≥ 𝑁:
∑|𝑥𝑛
(𝑘)−𝑥𝑛
(𝑚)|2
∞
𝑛=1 < 𝜖2
Fixing 𝑘 and letting 𝑚 → ∞:
∑|𝑥𝑛
(𝑘)−𝑥𝑛|2
∞
𝑛=1 ≤ 𝜖2
This shows that 𝑥(𝑘)→ 𝑥 in ℓ2.
To show (𝑥𝑛)𝑛=1
∞∈ ℓ2, note that:
∑|𝑥𝑛|2
𝑀
𝑛=1 = lim
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
𝑀
𝑛=1 ≤ limsup
𝑘→∞ ∑|𝑥𝑛
(𝑘)|2
∞
𝑛=1 < ∞
Therefore, ℓ2 is complete and thus a Hilbert space.
72 EXERCISE 3: ORTHOGONAL COMPLEMENTS
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace of 𝐻. Prove that 𝐻 = 𝑀 ⊕𝑀⊥, where 𝑀⊥ is
the orthogonal complement of 𝑀.
Solution: We need to show that every 𝑥 ∈ 𝐻 can be uniquely written as 𝑥 = 𝑦 +𝑧 where 𝑦 ∈ 𝑀
and 𝑧 ∈ 𝑀⊥.
1. Existence: Let 𝑥 ∈ 𝐻. Consider the optimization problem:
min
𝑦∈𝑀 ∥ 𝑥 −𝑦 ∥2
Since 𝑀 is closed, a minimizer 𝑦 exists. Let 𝑧 = 𝑥 − 𝑦.
For any 𝑣 ∈ 𝑀, consider 𝑓(𝑡)=∥ 𝑥 −(𝑦+𝑡𝑣)∥2. Since 𝑦 is a minimizer, 𝑓′(0)= 0.
Calculating 𝑓′(0):
𝑓′(0)= −2Re⟨𝑥 − 𝑦,𝑣⟩ = −2Re⟨𝑧,𝑣⟩ = 0
This implies 𝑧 ⊥ 𝑀, so 𝑧 ∈ 𝑀⊥.
2. Uniqueness: Suppose 𝑥 = 𝑦1+𝑧1= 𝑦2+𝑧2 with 𝑦1,𝑦2∈ 𝑀 and 𝑧1,𝑧2∈ 𝑀⊥. Then 𝑦1−𝑦2=
𝑧2−𝑧1∈ 𝑀 ∩𝑀⊥= {0}, so 𝑦1= 𝑦2 and 𝑧1= 𝑧2.
Therefore, 𝐻 = 𝑀 ⊕𝑀⊥.
73 EXERCISE 4: RIESZ REPRESENTATION THEOREM
State and prove the Riesz Representation Theorem for Hilbert spaces.
Theorem 1 (Riesz Representation Theorem).
Let
𝐻
be a Hilbert space and
𝜙:𝐻 → 𝔽
a
bounded linear functional on
𝐻
. Then there exists a unique vector
𝑦 ∈ 𝐻
such that
𝜙(𝑥)= ⟨𝑥,𝑦⟩
for all
𝑥 ∈ 𝐻
.
Proof: 1. If 𝜙 = 0, then 𝑦 = 0 satisfies the theorem. Assume 𝜙 ≠ 0.
2. Let 𝑁 = ker𝜙 = {𝑥 ∈ 𝐻:𝜙(𝑥)= 0}. 𝑁 is a closed subspace of 𝐻.
3. Since 𝜙 ≠ 0, 𝑁 ≠ 𝐻. Choose 𝑣 ∈ 𝑁⊥, 𝑣 ≠ 0.
4. Define 𝑦 = 𝜙(𝑣)
∥𝑣∥2𝑣.
5. For any 𝑥 ∈ 𝐻, let 𝑧 = 𝑥 − 𝜙(𝑥)
𝜙(𝑣)𝑣. Then 𝑧 ∈ 𝑁.
6. Since 𝑣 ⊥ 𝑁, we have ⟨𝑧,𝑣⟩ = 0.
7. Expanding this: ⟨𝑥,𝑣⟩−𝜙(𝑥)
𝜙(𝑣)⟨𝑣,𝑣⟩ = 0
8. Rearranging: 𝜙(𝑥)=𝜙(𝑣)
∥𝑣∥2⟨𝑥,𝑣⟩ = ⟨𝑥,𝑦⟩
9. Uniqueness follows from the fact that if ⟨𝑥,𝑦1⟩ = ⟨𝑥,𝑦2⟩ for all 𝑥 ∈ 𝐻, then 𝑦1= 𝑦2.
This theorem establishes a one-to-one correspondence between bounded linear functionals and
vectors in a Hilbert space, which is crucial in many areas of functional analysis.
74 EXERCISE 5: PROJECTIONS IN HILBERT SPACES
Let 𝐻 be a Hilbert space and 𝑀 a closed subspace. Define the orthogonal projection 𝑃𝑀:𝐻 → 𝑀
by 𝑃𝑀(𝑥)= 𝑦 where 𝑥 = 𝑦 +𝑧, 𝑦 ∈ 𝑀, 𝑧 ∈ 𝑀⊥.
Prove that 𝑃𝑀 is a bounded linear operator with ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}).
Solution: 1. Linearity: For 𝑥1,𝑥2∈ 𝐻 and scalars 𝑎,𝑏: 𝑃𝑀(𝑎𝑥1+𝑏𝑥2)= 𝑎𝑃𝑀(𝑥1)+𝑏𝑃𝑀(𝑥2)
(follows from uniqueness of decomposition)
2. Boundedness: For any 𝑥 ∈ 𝐻, 𝑥 = 𝑃𝑀(𝑥)+(𝑥 −𝑃𝑀(𝑥)) where 𝑃𝑀(𝑥)⊥(𝑥 −𝑃𝑀(𝑥)) By the
Pythagorean theorem: ∥ 𝑥 ∥2=∥ 𝑃𝑀(𝑥)∥2+∥ 𝑥 − 𝑃𝑀(𝑥)∥2 Therefore, ∥ 𝑃𝑀(𝑥)∥2≤∥ 𝑥 ∥2, so ∥
𝑃𝑀∥≤ 1
3. ∥ 𝑃𝑀∥= 1 (unless 𝑀 = {0}): If 𝑀 ≠ {0}, choose any unit vector 𝑦 ∈ 𝑀 Then 𝑃𝑀(𝑦)= 𝑦, so ∥
𝑃𝑀(𝑦)∥=∥ 𝑦 ∥= 1 Therefore, ∥ 𝑃𝑀∥= 1
This result shows that orthogonal projections are well-behaved operators in Hilbert spaces,
which is crucial in many applications.
