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Textbook by: Lawrence Baggett. E-mail the author

# Deeper Analytic Properties of Continuous Functions

Module by: Lawrence Baggett. E-mail the author

Summary: We collect here some theorems that show some of the consequences of continuity. Some of the theorems apply to functions either of a real variable or of a complex variable, while others apply only to functions of a real variable. We begin with what may be the most famous such result, and this one is about functions of a real variable.

We collect here some theorems that show some of the consequences of continuity. Some of the theorems apply to functions either of a real variable or of a complex variable, while others apply only to functions of a real variable. We begin with what may be the most famous such result, and this one is about functions of a real variable.

## Theorem 1: Intermediate Value Theorem

If f:[a,b]Rf:[a,b]R is a real-valued function that is continuous at each point of the closed interval [a,b],[a,b], and if vv is a number (value) between the numbers f(a)f(a) and f(b),f(b), then there exists a point cc between aa and bb such that f(c)=v.f(c)=v.

### Proof

If v=f(a)v=f(a) or f(b),f(b), we are done. Suppose then, without loss of generality, that f(a)<v<f(b).f(a)<v<f(b). Let SS be the set of all x[a,b]x[a,b] such that f(x)v,f(x)v, and note that SS is nonempty and bounded above. (aS,aS, and bb is an upper bound for S.S.) Let c=supS.c=supS. Then there exists a sequence {xn}{xn} of elements of SS that converges to c.c. (See (Reference).) So, f(c)=limf(xn)f(c)=limf(xn) by (Reference). Hence, f(c)v.f(c)v. (Why?)

Now, arguing by contradiction, if f(c)<v,f(c)<v, let ϵϵ be the positive number v-f(c).v-f(c). Because ff is continuous at c,c, there must exist a δ>0δ>0 such that |f(y)-f(c)|<ϵ|f(y)-f(c)|<ϵ whenever |y-c|<δ|y-c|<δ and y[a,b].y[a,b]. Since any smaller δδ satisfies the same condition, we may also assume that δ<b-c.δ<b-c. Consider y=c+δ/2.y=c+δ/2. Then y[a,b],|y-c|<δ,y[a,b],|y-c|<δ, and so |f(y)-f(c)|<ϵ.|f(y)-f(c)|<ϵ. Hence f(y)<f(c)+ϵ=v,f(y)<f(c)+ϵ=v, which implies that yS.yS. But, since c=supS,c=supS,cc must satisfy cy=c+δ/2.cy=c+δ/2. This is a contradiction, so f(c)=v,f(c)=v, and the theorem is proved.

The Intermediate Value Theorem tells us something qualitative about the range of a continuous function on an interval [a,b].[a,b]. It tells us that the range is “connected;” i.e., if the range contains two points cc and d,d, then the range contains all the points between cc and d.d. It is difficult to think what the analogous assertion would be for functions of a complex variable, since “between” doesn't mean anything for complex numbers. We will eventually prove something called the Open Mapping Theorem in (Reference) that could be regarded as the complex analog of the Intermediate Value Theorem.

The next theorem is about functions of either a real or a complex variable.

## Theorem 2

Let f:SCf:SC be a continuous function, and let CC be a compact (closed and bounded) subset of S.S. Then the image f(C)f(C) of CC is also compact. That is, the continuous image of a compact set is compact.

### Proof

First, suppose f(C)f(C) is not bounded. Thus, let {xn}{xn} be a sequence of elements of CC such that, for each n,n,|f(xn)|>n.|f(xn)|>n. By the Bolzano-Weierstrass Theorem, the sequence {xn}{xn} has a convergent subsequence {xnk}.{xnk}. Let x=limxnk.x=limxnk. Then xCxC because CCis a closed subset of C.C. Co, f(x)=limf(xnk)f(x)=limf(xnk) by (Reference). But since |f(xnk)|>nk,|f(xnk)|>nk, the sequence {f(xnk)}{f(xnk)} is not bounded, so cannot be convergent. Hence, we have arrived at a contradiction, and the set f(C)f(C) must be bounded.

Now, we must show that the image f(C)f(C) is closed. Thus, let yy be a limit point of the image f(C)f(C) of C,C, and let y=limyny=limyn where each ynf(C).ynf(C). For each n,n, let xnCxnC satisfy f(xn)=yn.f(xn)=yn. Again, using the Bolzano-Weierstrass Theorem, let {xnk}{xnk} be a convergent subsequence of the bounded sequence {xn},{xn}, and write x=limxnk.x=limxnk. Then xC,xC, since CC is closed, and from (Reference)

y = lim f ( x n ) = lim f ( x n k ) = f ( x ) , y = lim f ( x n ) = lim f ( x n k ) = f ( x ) ,
(1)

showing that yf(C),yf(C), implying that f(C)f(C) is closed.

This theorem tells us something about the range of a continuous function of a real or complex variable. It says that if a subset of the domain is closed and bounded, so is the image of that subset.

The next theorem is about continuous real-valued functions of a complex variable, and it is one of the theorems to remember.

## Theorem 3

Let ff be a continuous real-valued function on a compact subset SS of C.C. Then ff attains both a maximum and a minimum value on S.S. That is, there exist points z1z1 and z2z2 in SS such that f(z1)f(z)f(z2)f(z1)f(z)f(z2) for all zS.zS.

### Proof

We prove that ff attains a maximum value, leaving the fact that ff attains a minimum value to the exercise that follows. Let M0M0 be the supremum of the set of all numbers f(x)f(x) for xS.xS. (How do we know that this supremum exists?) We will show that there exists an z2Sz2S such that f(z2)=M0.f(z2)=M0. This will finish the proof, since we would then have f(z2)=M0f(z)f(z2)=M0f(z) for all zS.zS. Thus, let {yn}{yn} be a sequence of elements in the range of ff for which the sequence {yn}{yn} converges to M0.M0. (This is Exercise 2.20 again.) For each n,n, let xnxn be an element of SS such that yn=f(xn).yn=f(xn). Then the sequence {f(xn)}{f(xn)} converges to M0.M0. Let {xnk}{xnk} be a convergent subsequence of {xn}.{xn}. (How?) Let z2=limxnk.z2=limxnk. Then z2S,z2S, because SS is closed, and f(z2)=limf(xnk),f(z2)=limf(xnk), because ff is continuous. Hence, f(z2)=M0,f(z2)=M0, as desired.

## Exercise 1

1. Prove that the ff of the preceding theorem attains a minimum value on S.S.
2. Give an alternate proof of Theorem 3 by using Theorem 2, and then proving that a closed and bounded subset of RR contains both its supremum and its infimum.
3. Let SS be a compact subset of C,C, and let cc be a point of CC that is not in S.S. Prove that there is a closest point to cc in S.S. That is, show that there exists a point wSwS such that |w-c||z-c||w-c||z-c| for all points zS.zS. HINT: The function z|z-c|z|z-c| is continuous on the set S.S.

## Exercise 2

Let f:[a,b]Rf:[a,b]R be a real-valued function that is continuous at each point of [a,b].[a,b].

1. Prove that the range of ff is a closed interval [a',b'].[a',b']. Show by example that the four numbers f(a),f(b),f(a),f(b),a'a' and b'b' can be distinct.
2. Suppose ff is 1-1. Show that, if cc is in the open interval (a,b),(a,b), then f(c)f(c) is in the open interval (a',b').(a',b').

We introduce next a different kind of continuity called uniform continuity. The difference between regular continuity and uniform continuity is a bit subtle, and well worth some thought.

Definition 1:

A function f:SCf:SC is called uniformly continuous on SS if for each positive number ϵ,ϵ, there exists a positive number δδ such that |f(x)-f(y)|<ϵ|f(x)-f(y)|<ϵ for all x,ySx,yS satisfying |x-y|<δ.|x-y|<δ.

Basically, the difference between regular continuity and uniform conintuity is that the same δδ works for all points in S.S.

Here is another theorem worth remembering.

## Theorem 4

A continuous complex-valued function on a compact subset SS of CC is uniformly continuous.

### Proof

We argue by contradiction. Thus, suppose ff is continuous on SS but not uniformly continuous. Then, there exists an ϵ>0ϵ>0 for which no positive number δδ satisfies the uniform continuity definition. Therefore, thinking of the δδ's as ranging through the numbers 1/n,1/n, we know that for each positive integer n,n, there exist two points xnxn and ynyn in SS so that

1.  |yn-xn|<1/n,|yn-xn|<1/n, and
2.  |f(yn)-f(xn)|ϵ.|f(yn)-f(xn)|ϵ.

Otherwise, some 1/n1/n would suffice for a δ.δ. Let {xnk}{xnk} be a convergent subsequence of {xn}{xn} with limit x.x. By (1) and the triangle inequality, we deduce that xx is also the limit of the corresponding subsequence {ynk}{ynk} of {yn}.{yn}. But then f(x)=limf(xnk)=limf(ynk),f(x)=limf(xnk)=limf(ynk), implying that 0=lim|f(ynk)-f(xnk)|,0=lim|f(ynk)-f(xnk)|, which implies that |f(ynk)-f(xnk)|<ϵ|f(ynk)-f(xnk)|<ϵ for all large enough k.k. But that contradicts (2), and this completes the proof.

Continuous functions whose domains are not compact sets may or may not be uniformly continuous, as the next exercise shows.

## Exercise 3

1. Let f:(0,1)Rf:(0,1)R be defined by f(x)=1/x.f(x)=1/x. Prove that ff is continuous at each xx in its domain but that ff is not uniformly continuous there. HINT: Set ϵ=1,ϵ=1, and consider the pairs of points xn=1/nxn=1/n and yn=1/(n+1).yn=1/(n+1).
2. Let f:[1,)[1,)f:[1,)[1,) be defined by f(x)=x.f(x)=x. Prove that ff is not bounded, but is nevertheless uniformly continuous on its domain. HINT: Take δ=ϵ.δ=ϵ.

## Theorem 5

Let f:STf:ST be a continuous 1-1 function from a compact (closed and bounded) subset of CC onto the (compact) set T.T. Let g:TSg:TS denote the inverse function f-1f-1 of f.f. Then gg is continuous. The inverse of a continuous function, that has a compact domain, is also continuous.

### Proof

We prove that gg is continuous by using (Reference); i.e., we will show that g-1(A)g-1(A) is closed whenever AA is a closed subset of C.C. But this is easy, since g-1(A)=g-1(AS)=f(AS),g-1(A)=g-1(AS)=f(AS), and this is a closed set by Theorem 2, because ASAS is compact. See part (e) of (Reference).

REMARK Using the preceding theorem, and the exercise below, we will show that taking nnth roots is a continuous function. that is, the function ff defined by f(x)=x1/nf(x)=x1/n is continuous.

## Exercise 4

Use the preceding theorem to show the continuity of the following functions.

1. Show that if nn is an odd positive integer, then there exists a continuous function gg defined on all of RR such that g(x)g(x) is an nnth root of xx for all real numbers x.x. That is, (g(x))n=x(g(x))n=x for all real x.x. (The function f(x)=xnf(x)=xn is 1-1 and continuous.)
2. Show that if nn is any positive integer then there exists a unique continuous function gg defined on [0,)[0,) such that g(x)g(x) is an nnth root of xx for all nonnegative x.x.
3. Let r=p/qr=p/q be a rational number. Prove that there exists a continuous function g:[0,)[0,)g:[0,)[0,) such that g(x)q=xpg(x)q=xp for all x0;x0; i.e., g(x)=xrg(x)=xr for all x0.x0.

## Theorem 6

Let ff be a continuous 1-1 function from the interval [a,b][a,b] onto the interval [c,d].[c,d]. Then ff must be strictly monotonic, i.e., strictly increasing everywhere or strictly decreasing everywhere.

### Proof

Since ff is 1-1, we clearly have that f(a)f(b),f(a)f(b), and, without loss of generality, let us assume that c=f(a)<f(b)=d.c=f(a)<f(b)=d. It will suffice to show that if αα and ββ belong to the open interval (a,b),(a,b), and α<β,α<β, then f(α)f(β).f(α)f(β). (Why will this suffice?) Suppose by way of contradiction that there exists α<βα<β in (a,b)(a,b) for which f(α)>f(β).f(α)>f(β). We use the intermediate value theorem to derive a contradiction. Consider the four points a<α<β<b.a<α<β<b. Either f(a)<f(α)f(a)<f(α) or f(β)<f(b).f(β)<f(b). (Why?) In the first case (f(a)<f(α)f(a)<f(α)), f([a,α])f([a,α]) contains every value between f(a)f(a) and f(α).f(α). And, f([α,β])f([α,β]) contains every value between f(α)f(α) and f(β).f(β). So, let vv be a number such that f(a)<v,f(a)<v,f(β)<v,f(β)<v, and v<f(α)v<f(α) (why does such a number vv exist?). By the Intermediate Value Theorem, there exists x1(a,α)x1(a,α) such that v=f(x1),v=f(x1), and there exists an x2(α,β)x2(α,β) such that v=f(x2).v=f(x2). But this contradicts the hypothesis that ff is 1-1, since x1x2.x1x2. A similar argument leads to a contradiction in the second case f(β)<f(b).f(β)<f(b). (See the following exercise.) Hence, there can exist no such αα and β,β, implying that ff is strictly increasing on [a,b].[a,b].

## Exercise 5

Derive a contradiction from the assumption that f(β)<f(b)f(β)<f(b) in the preceding proof.

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