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

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 65

Lecture 16 - 11/7 - Continuity of sums, products and compositions. Open set characterization. Intermediate Value Theorem.

Using (13), the algebraic limit rules immediately help us prove that products and sums of functions are continuous, and ratios of continuous functions are continuous wherever they are defined. Let us first introduce some notation:

Given f,g : I → R, we can define the sum f + g : I → R, and product f · g : I → R, by the formula, for any x ∈ I, (f + g)(x) := f(x) + g(x) and f ·g(x) := f(x) ·g(x).

Theorem 65. Let I an interval and let f : I → R and g : I → R. (i) If f and g are continuous at a ∈ I, then f + g and f ·g are continuous at a. (ii) If in addition g(a) 6= 0, then there exists a neighborhood (a − δ,a + δ) where g does not

vanish, and f g

: (a−δ,a + δ) → R is continuous at a.

Proof. We use the sequential characterization of continuity. To prove (i), suppose {xn} converges to a. Since f and g are both continuous at a, this implies that the sequences {f(xn)} and {g(xn)} both converge to f(a) and g(a) respectively. By algebraic limit rules, this implies that

lim n→∞

(f + g)(xn) = lim n→∞

(f(xn) + g(xn)) = lim n→∞

f(xn) + lim n→∞

g(xn) = f(a) + g(a) = (f + g)(a),

so f + g is continuous at a. The proof that f ·g is continuous at a is similar. To prove (ii), since g(a) 6= 0, using ε = |g(a)|

2 > 0 and applying the ε−δ definition of continuity,

there exists δ > 0 such that for every x ∈ (a − δ,a + δ), we have |g(x) − g(a)| < ε = |g(a)| 2

. By triangle inequality, this implies

|g(a)|− |g(x)| ≤ |g(a) −g(x)| < |g(a)|

2 ,

which implies |g(x)| > |g(a)| 2

> 0, for every x ∈ (a− δ,a + δ). On to showing that f g

is continuous at a, let {xn} a sequence converging to a. Then there exists N ∈ N such that n ≥ N implies |xn −a| < δ, and thus for n ≥ N, g(xn) 6= 0 and

f(xn) g(xn)

is well-defined. By the algebraic limit rules,

we then have limn→∞ f(xn) g(xn)

= limn→∞f(xn) limn→∞g(xn)

= f(a) g(a)

. Hence the result.

If f : I → R and g : J → R are such that f(I) ⊂ J, then we define the composition g ◦f : I → R by g ◦f(x) := g(f(x)).

Theorem 66. In the setting just defined, suppose f is continuous at a ∈ I and g is continuous at f(a), then g ◦f is continuous at a.

Proof. We again use the sequential characterization of continuity. Namely, suppose {xn} converges to x0 (and let us prove that g◦f(xn) converges to g◦f(a)). Since f is continuous at a, the sequence yn = f(xn) converges to b = f(a). The sequence {yn} is a sequence in J where g is defined, and since g is continuous at b = f(a), the sequence {g(yn)} converges to g(b). Untwining the definitions, this exactly means that the sequence g ◦f(xn) = g(f(xn)) converges to g ◦f(a) = g(f(a)).

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 66

Example 30. The function f(x) = 1 1+x2

is continuous on R. One way to view this is as a ratio N(x) D(x)

where N(x) = 1 is continuous on R, and D(x) = 1 + x2 is continuous on R and is never zero. By the result on ratios of continuous functions, f is continuous on R.

A second proof is to write f as the composition h◦g, where g(x) = 1 + x2 and h(x) = 1 x

. Notice that g(x) ≥ 1 for every x ∈ R. Then we have g : R → R is continuous, h : (0,∞) → (0,∞) is continuous AND g(R) ⊂ (0,∞), therefore the composite h◦g is continuous on R.

Continuity and open sets. When the domain of definition is an open interval, we have yet another way to describe continuity, namely, in terms of open sets.

Theorem 67. Suppose f : I → R with I an open interval. Then f is continuous on I if and only if for every open set O ⊂ R, the preimage set f−1(O) = {x ∈ I : f(x) ∈O} is open.

Proof. ( =⇒ ) Suppose f is continous, let O ⊂ R an open set and x ∈ f−1(O). Then f(x) ∈ O thus there exists ε > 0 such that (f(x) − ε,f(x) + ε) ⊂ O. By continuity of f at x, there exists δ > 0 such that if |x′−x| < δ, then |f(x′)−f(x)| < ε. This exactly means that f((x−δ,x + δ)) ⊂ (f(x) −ε,f(x) + ε) ⊂O, and therefore (x−δ,x + δ) ⊂ f−1(O). Therefore f−1(O) is open.

( ⇐= ) Suppose that f−1(O) is open for every open set O. Let x ∈ I and ε > 0. Then O := (f(x) − ε,f(x) + ε) is an open set and as such satisfies that f−1(O) is open. Since x ∈ f−1(O), then there exists δ > 0 such that (x − δ,x + δ) ⊂ f−1(O). But then this means that f((x− δ,x + δ)) ⊂ O = (f(x) − ε,f(x) + ε). Since this is true for every ε > 0, f is continuous at x. Theorem 67 is proved.

Example 31. To prove that a function is not continuous, one may find an open set O such that f−1(O) is not open. As an example, the Heaviside function (defined by H(x) = 1 if x > 0 and H(x) = 0 if x ≤ 0) is not continuous because, with O = (−1

2 , 1 2 ), we have H−1(O) = (−∞, 0],

which is not open.

An interesting consequence of Theorem 67 is that a continuous function cannot oscillate more than countably many times in the following sense: we say that a function oscillate around zero on I if there exist x1 < x2 < x3 in I such that f(x) > 0 for all x ∈ (x1,x2) and f(x) ≤ 0 for all x ∈ (x2,x3). Suppose f is continuous on R, then with O = (0,∞), f−1(O) is open, and can therefore be written as a countable union of disjoint intervals f−1(O) =

⋃ k∈Z(ak,bk) (the union

could also be finite but we treat the most general case) and where bk ≤ ak+1 for every k. Then we see that an oscillation scenario can happen at most on any setting of the form (ak,bk,ak+1), which is at most countably many times (around zero, or around any other value after proper definition of what that means).

The Intermediate Value Theorem.

Theorem 68 (Intermediate Value Theorem, IVT). Suppose f : D → R continuous, let a < b such that [a,b] ⊂ D, and suppose f(a) < f(b). Then for every y ∈ (f(a),f(b)), there exists c ∈ (a,b) such that f(c) = y

The conclusion still holds if f(a) > f(b), in which case one must change (f(a),f(b)) into (f(b),f(a)) in the statement above.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 67

Proof. Set a1 = a, b1 = b, and construct two sequences satisfying, for every n ∈ N, (i) an ≤ an+1 ≤ bn+1 ≤ bn, (ii) |bn+1 − an+1| = 12|bn − an| and (iii) f(an) ≤ y ≤ f(bn). The construction goes as follows.

Suppose the sequences have been constructed up to index n, satisfying (i)-(ii)-(iii). Consider the midpoint m = an+bn

2 . If f(m) = y we are done; if f(m) < y, set an+1 = m and bn+1 = bn; if

f(m) > y, set an+1 = an and bn+1 = m.

Since {an} is increasing and bounded above, it converges to some c by Theorem 31 amd (ii) implies that {bn}n also converges to c. Since f is continuous at x, then limn→∞f(an) = f(x) = limn→∞f(bn). Since for every n ∈ N, we have f(an) ≤ y ≤ f(bn), in the limit we have limn→∞f(an) ≤ y ≤ limn→∞f(bn), but this means f(x) ≤ y ≤ f(x), in other words, f(x) = y.

Here are a couple of important applications of the intermediate value theorem:

1. The intermediate value theorem gives us another immediate construction of the p-th roots, for instance for any x ∈ [0, 1], as follows: fix p ∈ N. The function f(x) = xp is continous on [0, 1], with f(0) = 0 and f(1) = 1. By the intermediate value theorem, for every y ∈ (0, 1) = (f(0),f(1)), there exists x ∈ (0, 1) such that f(x) = y, that is to say xp = y.

2. More generally, every continuous function which is monotonically strictly increasing or de- creasing on an interval admits an inverse function, which is itself continuous.

3. Every polynomial of odd degree has at least one real root.

Exercises for Lectures 16:

1. In this problem, use the ε− δ definition of continuity.

(a) Show that x 7→ √ x is continuous on (0,∞).

(b) Show that x 7→ x 1 3 is continuous on (0,∞).

(c) Generalize the previous argument to x 7→ x1/p for any p ∈ N.

2. Let f1, . . . ,fn, n functions which are continuous on the real line. Show that the set

A = {x ∈ R : fj(x) > 0, ∀ j = 1, . . . ,n},

is open.

3. Show that f−1(A ∪ B) = f−1(A) ∪ f−1(B) and f−1(A ∩ B) = f−1(A) ∩ f−1(B) for any function f. Is the same true of images (as opposed to inverse images) ?

4. (a) Show that f(x) = x5 + x3 + x− 1 has a real root between 0 and 1. (b) Show that the graphs of f(x) = 1

1+x2 and g(x) = 5x5 + 4x4 must intersect.

5. Another proof of Theorem 68 (IVT) goes as follows. Define S := {x ∈ [a,b], f(x) ≤ y}.

(a) Show that c = sup S exists.

(b) Show that, by using continuity of f at c, the options f(c) < y and f(c) > y both contradict that c = sup S. (and thus we must have f(c) = y).