math upper homework-Real analysis due tomorrow 9：00 am

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 68

Lecture 17 - 11/9 - Continuity on compact domains. Infinite limits.

Uniform continuity

Definition 17. Let a function f : I → R. We say that f is continuous on I if f is continuous at every point of I, that is to say,

∀x ∈ I, ∀ε > 0, ∃δ > 0, ∀y ∈ I, |x−y| < δ =⇒ |f(x) −f(y)| < ε.

We say that f is uniformly continuous on I if

∀ε > 0, ∃δ > 0, ∀x ∈ I, ∀y ∈ I, |x−y| < δ =⇒ |f(x) −f(y)| < ε.

Judging by the order of the quantifiers, the main difference between “continuity on I” and “uniform continuity on I” is that in the former, δ depends on ε and x, while in the latter, δ only depends on ε and not on x. In particular, if f is uniformly continuous on I, then it is continuous at every x ∈ I, and the same δ(ε) applies to all x’s. The converse is false in general, as the following examples show.

Example 32. 1. The function f(x) = 1 x

is continuous on (0, 1] but not uniformly continuous on (0, 1]. To show this, we must show

∃ε > 0, ∀δ > 0, ∃ x,y ∈ (0, 1], |x−y| < δ and |f(x) −f(y)| > ε.

Notice that if x > 0, then f(x/2) − f(x) = 1 x

. Set ε = 1 and pick 0 < δ < 1 arbitrary14.

Choose x = δ and y = δ 2 . Then |x−y| < δ and |f(x)−f(y)| = 1/δ ≥ 1 = ε. Hence the claim

is proved.

2. The function f(x) = x2 is continuous on R but not uniformly continuous on R.

On the other hand, compactness comes here handy to prove that uniform continuity happens every time we consider a continuous function on a compact set. Note how the proof does not give us the δ(ε) explicitely, which is a feature of compactness-based proofs (they prove existence of objects without showing us what they look like).

Theorem 69. Suppose f : D → R is continuous on D and K ⊂ D is a compact set. Then f is uniformly continuous on K.

Proof. By contradiction, suppose f is not uniformly continuous on K. This means

∃ε > 0, ∀δ > 0, ∃x,y ∈ K, |x−y| < δ and |f(x) −f(y)| > ε.

In particular, for δ = 1 n

, we can find xn,yn in K such that |xn −yn| < 1n and |f(xn) −f(yn)| > ε. By compactness, xn has a convergent subsequence in K, xnk , converging to some x ∈ K. Since |xnk −ynk| <

1 nk

for every k, we get that ynk also converges to x. Then by continuity of f, we get that

lim k→∞

f(xnk ) −f(ynk ) = f(x) −f(x) = 0,

which contradicts the fact that |f(xnk )−f(ynk )| > ε for every k. Hence, f is uniformly continuous on K.

14the case δ ≥ 1 does not matter because any two numbers in (0, 1] are less than a distance 1 apart

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 69

Continuous functions on compact domains. Extreme Values Theorem

We say that a function f : D → R is bounded above on D′ ⊂ D if the set f(D′) = {f(x), x ∈ D′} is bounded above. Equivalently, f is bounded above on D′ if there exists M ∈ R such that

f(x) ≤ M, ∀x ∈ D′.

Similarly, f is bounded below on D′ if f(D′) is bounded below. Finally, f is bounded on D′ if it is both bounded above and below on D′.

When f is bounded above on D′, f(D′) has a supremum, and one may ask whether this supremum is achieved on D′, that is, whether there is x ∈ D′ such that f(x) = sup{f(y), y ∈ D′}. That this is possible requires topological restrictions on the set D′.

Example 33. 1. The function f(x) = 1 1+x2

does not achieve its infimum on R.

2. The function f(x) = 1 −x does not achieve its supremum on (0, 1).

We see that if the domain of definition is not bounded, or not closed, a function need not achieve its supremum or infimum. We now show however that it always does so on compact (i.e., closed and bounded) sets, as Theorem 71 will show.

Let us recall that a compact set K ⊂ R is a closed and bounded set (by Theorem 61) and is such that every sequence in K has a convergent subsequence, converging to a point in K.

Theorem 70. Suppose f : D → R is continuous on D. Then for any compact set K ⊂ D, f(K) is compact.

Proof. Suppose K is compact and consider a sequence {yn} in f(K). This means that for every n, there exists xn ∈ K such that f(xn) = yn. Since K is compact, there exists a subsequence {xnk}k of {xn} converging to some x ∈ K. By continuity of f at x,

lim k→∞

ynk = lim k→∞

f(xnk ) = f( lim k→∞

xnk ) = f(x).

Since x ∈ K, f(x) ∈ f(K) and the subsequnce {ynk} converges in f(k) to f(x). Since any sequence in f(K) has a subsequence converging in f(K), f(K) is compact.

Theorem 71. Let f : D → R continuous on D. Then for any compact set K ⊂ D, f is bounded on K and there exists y,z ∈ K such that f(y) = sup{f(x), x ∈ K} and f(z) = inf{f(x), x ∈ K}.

Proof. Suppose K compact. Then by the previous theorem, f(K) is compact, hence closed and bounded. In particular f(K) is bounded, therefore M := sup{f(x),x ∈ K} and L := inf{f(x),x ∈ K} exist and are finite. Now, for every n ∈ N, M − 1

n is not an upper bound for f(K), so there

exists xn ∈ K such that M − 1n < f(xn) ≤ M. Since {xn} is a sequence in a compact set, it has a convergent subsequence xnk , which converges to some y ∈ K. By continuity of f, f(xnk ) converges to f(y) but also by the estimate M − 1

nk < f(xnk ) ≤ M, we have by squeezing that

f(y) = limk→∞f(xnk ) = M. Hence f achieves its supremum on K. Similarly, one can prove that f achieves its infimum on K.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 70

Infinite Limits

So far, we have not discussed about infinite limits, as these are a special case of limits for which cer- tain rules must be modified. The main difference is in how we express that something “approaches infinity”. A sequence which approaches a real number x0 eventually falls into any neighborhood of the form (x0−δ,x0 +δ) for any δ > 0. However, (∞−δ,∞+δ) makes no sense and we must redefine what a “neighborhood of infinity” is. We may think of a sequence escaping to ∞ if it eventually falls into the interval (M,∞) no matter how large M is. So a way to view a neighborhood of ∞ is as a set which contains an open interval of the form (M,∞). • Infinite limits for sequences: Given a sequence {xn}n, we say limn→∞xn = ∞ if

∀M ∈ R, ∃N ∈ N, ∀n ∈ N, n ≥ N =⇒ xn ≥ M.

One could replace “M ∈ R” by “M ≥ M0” because the whole point is that the statement be true for “M large enough”. Given a sequence {xn}n, we say limn→∞xn = −∞ if

∀M ∈ R, ∃N ∈ N, ∀n ∈ N, n ≥ N =⇒ xn ≤ M.

Example 34. For any p ∈ N, the sequence xn = np satisfies limn→∞xn = ∞. For p = 1, this is an immediate consequence of the Archimedean property. For p ≥ 1, then np ≥ n, so np should diverge to ∞ as well.

A few comments as to why such limits do not behave as finite limits:

• A sequence whose limit is ±∞ is not bounded. This is why it is sometimes said that the sequence diverges to ∞ (and not “converges” to ∞).

• The algebraic limit rules do not always apply as in the case of finite limits.

◦ If xn → ∞ and yn → −∞, xn + yn could converge to anything or not converge at all. For example, if xn = n and yn = −n + a with a ∈ R fixed, then xn + yn converges to a. If xn = n

2 and yn = −n, then limn→∞xn + yn = limn→∞n(n− 1) = ∞. ◦ If xn →∞ and yn →∞, xnyn could converge to anything or not converge at all. ◦ If xn →∞ and yn → 0, then xn ·yn could converge to anything or not converge at all.

Some properties which one may prove:

• If xn is bounded and yn →∞, then xn + yn →∞.

• If xn is bounded and yn →−∞, then xn + yn →−∞.

• If xn →∞ and yn →∞, then xn ·yn →∞ and xn + yn →∞ .

• Infinite limits of functions at finite points. On to the case of functions, suppose f is defined in a neighborhood (x0 −δ0,x0 + δ0) of x0 ∈ R, except possibly at x0. We say that limx→x−0 f(x) = +∞ if

∀M ∈ R, ∃δ ∈ (0,δ0), x0 −δ < x < x0 =⇒ f(x) ≥ M.

You may define similarly lim x→x−0

f(x) = −∞, lim x→x+0

f(x) = ±∞, and limx→x0 f(x) = ±∞.

Similar to the sequential characterization for finite limits, we can show that lim x→x−0

f(x) = +∞

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 71

• Limits of functions as x →±∞. As x becomes arbitrarily large positive or negative, a function may approach a limit. If the limit is a finite number, we can define limx→∞f(x) = L if

∀ε > 0, ∃M ∈ R, ∀x ∈ R, x ≥ M =⇒ |f(x) −L| < ε.

A similar definition can be given to define limx→−∞f(x) = L.

Example 35. 1. limx→∞ 1 x

= 0. Indeed, let ε > 0, set M = 1 ε . If x > M, then 0 < 1

x < 1

M = ε.

2. limx→∞ 2x+1 x+3

= 2. Indeed for x ≥ 0, setting f(x) = 2x+1 x+3

, we have the preliminary bound

|f(x) − 2| = ∣∣∣∣2x + 1x + 3 − 2

∣∣∣∣ = 5|x + 3| ≤ 5x. So if x →∞, 5

x → 0 and therefore f(x) → 2.

If the limit is infinite when x becomes infinite, we have both x and f(x) become large, which means that ε-neighbourhoods are completely gone from the definition. Namely, we say that limx→∞f(x) = ∞ if

∀M ≥ 0, ∃L ≥ 0, ∀x ∈ R, x ≥ L =⇒ f(x) ≥ M.

Example 36. If f(x) = √ x2 + 1, then limx→∞f(x) = ∞. Indeed, given M large enough (M ≥ 1,

say), one may set L = √ M2 − 1. If x ≥ L =

√ M2 − 1, then x2 ≥ M2−1, then x2 + 1 ≥ M2, then

f(x) ≥ M. Hence the result.

Similarly, one can define limx→∞f(x) = −∞, limx→−∞f(x) = ∞ and limx→−∞f(x) = −∞.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 72

Exercises for Lecture 17:

1. We say that f is Lipschitz-continuous with constant M on I if

∀x,y ∈ I, |f(x) −f(y)| ≤ M|x−y|.

Suppose that f and g are both Lipschitz continuous on R.

(a) Show that f + g is Lipschitz continuous on R. (b) Show that if in addition f and g are bounded, then f ·g is Lipschitz continuous. (c) Give a counterexample to show that it is necessary to assume boundedness in (b).

2. On Lipschitz-continuity and absolute continuity.

(a) Show that if f : I → R is Lipschitz-continuous on an interval I, then f is uniformly continuous on I.

(b) Show that the function f(x) = √ x is absolutely continuous on [0, 1] but not Lipschitz-

continuous on [0, 1]. (for the latter, you need to show that ratios of the form f(x)−f(y)

x−y with x,y ∈ [0, 1] and x 6= y are unbounded)

3. Suppose that f and g are both uniformly continuous on R.

(a) Show that f + g is uniformly continuous on R. (b) Show that if in addition f and g are bounded, then f ·g is uniformly continuous. (c) Give a counterexample to show that it is necessary to assume boundedness in (b).

4. Show that the function x 7→ x2 is not uniformly continuous on R.

5. Show that the function x → sin(1/x) is not uniformly continuous on (0, 1].

6. Show that the function f(x) = 1 x2−4 is uniformly continuous on [0, 1]. (use a theorem)

7. Show that if f is uniformly continuous on R, there exists positive constants M1,M2 such that |f(x)| ≤ M1 + |x|M2 for all x ∈ R.

8. (a) Show that if xn is bounded and yn →∞, then xn + yn →∞. (b) Show that if xn →∞ and yn →∞, then xn ·yn →∞.

9. Write down the definition of limx→1− f(x) = −∞ and use this definition to show that limx→1− f(x) = +∞ where f(x) := 11−x2 for x 6= ±1.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 73

Extra: Zoology of dicontinuities.

Which are the possible ways that a function may fail to be continuous ?

In what follows, suppose x0 ∈ R and let f : I → R where I is some open interval containing x0. Then f is continuous at x0 if and only if

lim x→x−0

f(x) = lim x→x+0

f(x) = f(x0).

Let us explore the various cases for which the above condition fails, by increasing order of “badness”.

• Removable discontinuities. The left-limit and the right-limit exist, are finite and equal, but do not equal f(x0) (or f has not yet been defined at x0). In that case, the function f can be defined at x0 in such a way that the modified function is continuous at x0. Namely, define f̃ : I → R by f̃(x) = f(x) if x 6= x0, and f̃(x0) := limx→x0 f(x). Then f̃ is continuous at x0. Example 37. 1. Set f(x) = sinx

x for any x 6= 0. One may see that limx→0 f(x) = 1, and

therefore, one may define f̃ : R → R by f̃(x) = sinx x

for x 6= 0, and f̃(0) = 1. See Fig. 5.

2. For x 6= 2, define f(x) := x 3−8 x−2 , continuous for all x 6= 2. One may notice that for any

x 6= 2, f(x) = x2 + 2x + 4, and as such, has a well-defined limit as x approaches 2, namely limx→2 f(x) = 12. Therefore we can extend f in such a way that f is continuous on R.

Figure 5: The sinc function f(x) = sinx x

: an example of removable discontinuity at x = 0

• Discontinuities of the first kind = jump discontinuities. The left-limit and the right-limit exist and are finite, but are not equal.

Example 38. 1. The Heaviside function H(x) = 1 if x > 0 and H(x) = 0 if x ≤ 0.

lim x→0−

H(x) = 0 6= 1 = lim x→0+

H(x).

2. The signum function f(x) = x|x| for x 6= 0, and f(0) = 0 (see Fig. 6). In this case, we have

lim x→0−

f(x) = −1 6= 1 = lim x→0+

f(x).

• Discontinuities of the second kind. Either left or right-limit does not exist or is infinite. Example 39. 1. If f = χQ, for any x0 ∈ R, limx→x−0 f(x) does not exist, neither does limx→x+0 f(x).

2. If f(x) = sin(x−1) for x 6= 0, limx→0− f(x) or limx→0+ f(x) do not exist. (see Fig. 7, left).

3. If f(x) = 1 x

for x 6= 0, limx→0+ f(x) = ∞ and limx→0− f(x) = −∞.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 74

Figure 6: The signum function f(x) = x|x|: an example of jump discontinuity at x = 0

Figure 7: Examples of discontinuities of the second kind. Left: the function f(x) = sin(x−1). Right: the function f(x) = 1

x for x > 0 and f(x) = 1 for x ≤ 0.