3.5 Deeper analytic properties of continuous functions  (Page 2/2)

showing that $y\in f\left(C\right),$ implying that $f\left(C\right)$ 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.

Let $f$ be a continuous real-valued function on a compact subset $S$ of $C.$ Then $f$ attains both a maximum and a minimum value on $S.$ That is, there exist points $z_1$ and $z_2$ in $S$ such that $f\left(z_1\right)\le f\left(z\right)\le f\left(z_2\right)$ for all $z\in S.$

We prove that $f$ attains a maximum value, leaving the fact that $f$ attains a minimum value to the exercise that follows. Let $M_0$ be the supremum of the set of all numbers $f\left(x\right)$ for $x\in S.$ (How do we know that this supremum exists?) We will show that there exists an $z_2\in S$ such that $f\left(z_2\right)=M_0.$ This will finish the proof, since we would then have $f\left(z_2\right)=M_0\ge f\left(z\right)$ for all $z\in S.$ Thus, let $\left\{y_n\right\}$ be a sequence of elements in the range of $f$ for which the sequence $\left\{y_n\right\}$ converges to $M_0.$ (This is Exercise 2.20 again.) For each $n,$ let $x_n$ be an element of $S$ such that $y_n=f\left(x_n\right).$ Then the sequence $\left\{f\left(x_n\right)\right\}$ converges to $M_0.$ Let $\left\{x_{n_k}\right\}$ be a convergent subsequence of $\left\{x_n\right\}.$ (How?) Let $z_2=lim{x}_{n_k}.$ Then $z_2\in S,$ because $S$ is closed, and $f\left(z_2\right)=limf\left(x_{n_k}\right),$ because $f$ is continuous. Hence, $f\left(z_2\right)=M_0,$ as desired.

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.

A function $f:S\to C$ is called uniformly continuous on $S$ if for each positive number $ϵ,$ there exists a positive number $\delta$ such that $\left|f\right(x)-f(y\left)\right|<ϵ$ for all $x,y\in S$ satisfying $|x-y|<\delta .$

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

Here is another theorem worth remembering.

A continuous complex-valued function on a compact subset $S$ of $C$ is uniformly continuous.

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

1.   $|y_n-x_n|<1/n,$ and
2.   $|f\left(y_n\right)-f\left(x_n\right)|\ge ϵ.$

Otherwise, some $1/n$ would suffice for a $\delta .$ Let $\left\{x_{n_k}\right\}$ be a convergent subsequence of $\left\{x_n\right\}$ with limit $x.$ By (1) and the triangle inequality, we deduce that $x$ is also the limit of the corresponding subsequence $\left\{y_{n_k}\right\}$ of $\left\{y_n\right\}.$ But then $f\left(x\right)=limf\left(x_{n_k}\right)=limf\left(y_{n_k}\right),$ implying that $0=lim|f\left(y_{n_k}\right)-f\left(x_{n_k}\right)|,$ which implies that $|f\left(y_{n_k}\right)-f\left(x_{n_k}\right)|<ϵ$ for all large enough $k.$ 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.

Let $f:S\to T$ be a continuous 1-1 function from a compact (closed and bounded) subset of $C$ onto the (compact) set $T.$ Let $g:T\to S$ denote the inverse function $f^{-1}$ of $f.$ Then $g$ is continuous. The inverse of a continuous function, that has a compact domain, is also continuous.

We prove that $g$ is continuous by using [link] ; i.e., we will show that $g^{-1}\left(A\right)$ is closed whenever $A$ is a closed subset of $C.$ But this is easy, since $g^{-1}\left(A\right)=g^{-1}(A\cap S)=f(A\cap S),$ and this is a closed set by [link] , because $A\cap S$ is compact. See part (e) of [link] .

REMARK Using the preceding theorem, and the exercise below, we will show that taking $n$ th roots is a continuous function. that is, the function $f$ defined by $f\left(x\right)=x^{1/n}$ is continuous.

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

Since $f$ is 1-1, we clearly have that $f\left(a\right)\ne f\left(b\right),$ and, without loss of generality, let us assume that $c=f\left(a\right)<f\left(b\right)=d.$ It will suffice to show that if $\alpha$ and $\beta$ belong to the open interval $(a,b),$ and $\alpha <\beta ,$ then $f\left(\alpha \right)\le f\left(\beta \right).$ (Why will this suffice?) Suppose by way of contradiction that there exists $\alpha <\beta$ in $(a,b)$ for which $f\left(\alpha \right)>f\left(\beta \right).$ We use the intermediate value theorem to derive a contradiction. Consider the four points $a<\alpha <\beta <b.$ Either $f\left(a\right)<f\left(\alpha \right)$ or $f\left(\beta \right)<f\left(b\right).$ (Why?) In the first case ( $f\left(a\right)<f\left(\alpha \right)$ ), $f\left(\right[a,\alpha \left]\right)$ contains every value between $f\left(a\right)$ and $f\left(\alpha \right).$ And, $f\left(\right[\alpha ,\beta \left]\right)$ contains every value between $f\left(\alpha \right)$ and $f\left(\beta \right).$ So, let $v$ be a number such that $f\left(a\right)<v,$ $f\left(\beta \right)<v,$ and $v<f\left(\alpha \right)$ (why does such a number $v$ exist?). By the Intermediate Value Theorem, there exists $x_1\in (a,\alpha )$ such that $v=f\left(x_1\right),$ and there exists an $x_2\in (\alpha ,\beta )$ such that $v=f\left(x_2\right).$ But this contradicts the hypothesis that $f$ is 1-1, since $x_1\ne x_2.$ A similar argument leads to a contradiction in the second case $f\left(\beta \right)<f\left(b\right).$ (See the following exercise.) Hence, there can exist no such $\alpha$ and $\beta ,$ implying that $f$ is strictly increasing on $[a,b].$