User:TakuyaMurata/Continuous functions on a compact space

In this section, we will undergo a throughout study of $C(K)$, the space of all real-valued or complex-valued functions on a compact space. The most important result in the line of this study is Ascoli's theorem and Stone-Weierstrass theorem. For simplicity, we assume functions in $C(K)$ are real-valued. (A discussion will be given later as to why this does not diminish the generality.)

As usual, we topologizes $C(K)$ by the norm $\|\cdot \|=\sup |\cdot |$. To say that $C(K)$ is complete is precisely:
2. Lemma The limit of a uniformly convergent sequence of continuous functions is continuous.
Proof: Suppose $f_{n}\in C(K)$ is a sequence such that $\sup _{K}|f_{n}-f|\to 0$ for some function $f$ defined on $K$. For any $x\in K$, by the iterated limit theorem,

$\lim _{y\to x}f(y)=\lim _{n\to 0}\lim _{y\to x}f_{n}(y)=\lim _{n\to 0}f_{n}(x)=f(x).$ $\square$

Hence, $C(K)$ is a Banach space. (For the definition and basic results of Banach spaces, see Functional Analysis.)

2 Theorem (Ascoli) Let $\Gamma \subset C(K)$. Then $\Gamma$ is relatively compact if and only if

• (i) Given an $\epsilon >0$ and $x\in K$, we can find a neighborhood $G$ of $x$ such that

  $|f(y)-f(x)|<\epsilon$ for every $y\in G$ and every $f\in \Gamma$

• (ii) $\sup _{\Gamma }|\cdot (x)|<\infty$ for every $x\in K$

Proof: First, assume that $\Gamma$ is relatively compact. (ii) is then obvious. For (i), let $\epsilon >0$ and $x\in K$ be given. For each $f\in \Gamma$, by continuity, we can find a neighborhood $G_{f}$ of $x$ such that:

$|f(y)-f(x)|<\epsilon$ for every $y\in G_{f}$.

Since $\Gamma$ is relatively compact, $\Gamma$ contains a finite subset $\gamma$ such that $\Gamma$ is the union of the sets of the form

$\{f;f\in \Gamma ,\sup _{K}|f-g|<\epsilon /3\}$

over $g\in \gamma$. Let $G$ be the intersection of $G_{g}$ over $g\in \gamma$. Then for every $f\in \Gamma$, there is $g\in \gamma$ with $\sup _{K}|f-g|<\epsilon /3$, and so:

$|f(y)-f(x)|\leq |f(y)-g(y)|+|g(y)-g(x)|+|g(x)-f(x)|<\epsilon$ for any $y\in G$.

This proves (i). Next, suppose $E$ satisfies (i) and (ii). To show that $\Gamma$ is totally bounded, let $\epsilon >0$ be given. For each $x\in K$, by (i), we can find a neighborhood $G_{x}$ of $x$ such that:

$|f(y)-f(x)|<\epsilon /3$ for every $y\in G_{x}$ and $f\in \Gamma$.

Since $K$ is compact, we can find $z_{1},...z_{n}\in K$ such that $K$ is the union of $G_{z_{j}}$ over $j=1,2,...n$. Let

$A=\{(f(z_{1}),f(z_{2}),...f(z_{n}));f\in \Gamma \}$.

By (ii), $A$ is a bounded (thus totally bounded) subset of $\mathbf {R} ^{n}$. That means that $\Gamma$ contains a finite subset $\gamma$ such that:

$A\subset \bigcup _{g\in \gamma }\{(t_{1},...t_{n});t_{j}\in \mathbf {R} ,\max _{j}|t_{j}-g(z_{j})|<\epsilon /3\}$

It now follows: given $f\in \Gamma$, we can find $g\in \gamma$ such that:

$\max _{j}|f(z_{j})-g(z_{j})|<\epsilon /3$.

Then, for each $x\in K$, since $x\in G_{z_{k}}$ for some $z_{k}$,

$|f(x)-g(x)|\leq |f(x)-f(z_{k})|+|f(z_{k})-g(z_{k})|+|g(z_{k})-g(x)|<\epsilon$

In other words, $\sup _{K}|f-g|<\epsilon$. Hence, $\Gamma$ is totally bounded, or equivalently, relatively compact. $\square$

2 Corollary Let $f_{n}\in C(K)$. $f_{n}$ is uniformly convergent if and only if it is pointwise convergent and equicontinuous.

2 Theorem Let $f_{n}$ converge pointwise to $f\in C(K)$. If ${f_{n}}'$ exists and converges uniformly to $g$, then $f_{n}$ converges uniformly to $f$. Moreover, $f$ is differentiable and its derivative is $g$. Proof: Let $M=\sup\{|f_{n}'(x)||n\geq 1,x\in K\}$. $M$ is finite by uniform convergence. By the mean value theorem,

$|f_{n}(y)-f_{n}(x)|\leq M|y-x|$

Thus, $f_{n}$ is equicontinuous and converges uniformly by Ascoli's thoerem (or one of its corollaries.) $\square$

2 Theorem Let $\Gamma$ be an equicontinous set of real-valued functions on $\mathbf {R}$. If $\sup _{f\in \Gamma }|f(0)|=b<\infty$, then there exist an $a$ such that:

$\sup _{f\in \Gamma }|f(x)|\leq a|x|+b+1$

Proof: Let $\delta >0$ be such that:

$|f(x)-f(y)|<1$ for all $f\in \Gamma$ and $|x-y|<2\delta$

($2\delta$ isn't a typo; it is meant to simplify the computation.) Let $x>0$ be fixed. Then, for any $f\in \Gamma$,

$|f(x)|\leq |f(x)-f(0)|+|f(0)|$,

and we estimate:

$|f(0)-f(x)|\leq \sum _{k=0}^{n-1}|f(k\delta )-f((k+1)\delta )|+|f(n\delta )-f(x)|\leq n+1$

where $n$ is such that $n\delta <x\leq (n+1)\delta$. Thus,

$|f(x)|\leq \delta ^{-1}|x|+1+b$

Since we can get the same estimate for $x<0$, the proof is complete.$\square$

2 Corollary (Dini's theorem) Let $f_{n}\in {\mathcal {C}}(K)$ be a sequence such that $f_{n}(x)\to f(x)$ for every $x\in K$. If $f_{n}$ is increasing, then $f_{n}\to f$.
Proof: Set $g_{n}=f-f_{n}$. Then $g_{n}$ is decreasing and thus satisfies the hypothesis of Ascoli's theorem. Hence, $g_{n}$ admits a convergent subsequence, which converges to 0 since the subsequence converges pointwise to 0. Since $g_{n}$ is decreasing, $g_{n}$ converges as well. $\square$

2 Theorem Suppose $K$ is a metric space. Then $\Gamma \subset C(K)$ is equicontinuous if and only if for every $\epsilon >0$ there exists $\delta >0$ such that

$|f(x)-f(y)|<\epsilon$

for every $f\in \Gamma$ and $x,y\in K$ with $|x-y|<\delta$.
Proof: $(\Leftarrow )$ holds vacuously. For the converse, let $\epsilon >0$ be given. Then for each $x\in K$, we can find $\delta _{x}$ such that

$|x-y|<\delta _{x}$ implies $|f(x)-f(y)|<\epsilon /2$ for every $f\in \Gamma$

By compactness, we find $x_{1},x_{2},...x_{n}\in K$ such that:

$K\subset \bigcup _{j}B(x_{j},\delta _{x_{j}}/2)$

Let $\delta =\min\{\delta _{x_{1}},\delta _{x_{2}},...\delta _{x_{n}}\}/2$, and then suppose we are given $x,y\in K$ with $|x-y|<\delta$. It follows: there is a $j$ with $|x-x_{j}|<\delta _{x_{j}}/2$. Since

$|y-x_{j}|\leq |y-x|+|x-x_{j}|<\delta +\delta _{x_{j}}/2\leq \delta _{x_{j}}$,

we have:

$|f(x)-f(y)|\leq |f(x)-f(x_{j})|+|f(x_{j})-f(y)|<\epsilon$

for every $f\in \Gamma$. $\square$

The Stone-Weierstrass theorem states that polynomials are dense in C(K, R). It is however not the case that the space of polynomials in z is a dense in C(K, C). If it were, we have the equality in the below

${\overline {P(K)}}\subset A(K)\subset C(K)$

But $A(K)\neq C(K)$ if K has nonempty interior.

2 Theorem (intermediate value theorem) A function $f:[a,b]\to \mathbf {R}$ is continuous if and only if

• (i) If $f(a)<c<f(b)$, then c is in $f((a,b))$.
• (ii) If $f^{-1}(\{c\})$ is closed for every real c.

Proof: ($\Rightarrow$) Obvious. ($\Leftarrow$) Suppose $f(x)<c$. Since the complement of $f^{-1}(\{c\})$, which contains x, is open, we have: $f\neq c$ in some interval U in $[a,b]$ containing x. We actually have: $f<c$ in U. In fact, if $y\in U$ and $c<f(y)$, then $f(x)<c<f(y)$, which implies U contains a point z such that $f(z)=c$, a contradiction. Hence, f is upper semicontinous at x. The same argument applied to $-f$ shows that f is also lower semicontinous at x. $\square$

2 Theorem Let $f$ be a real-valued continuous function on an open interval. Then the following are equivalent.

• (i) f is injective.
• (ii) f is strictly monotonic.
• (iii) f is an open mapping.

Proof: (ii) $\Rightarrow$ (i) is obvious. (iii) $\Rightarrow$ (ii): If (ii) is false, then we can assume there exists $a<c<b$ such that $f(a)<f(c)$ and $f(c)>f(b)$. By continuity and compactness, f attains a maximum in some point x in $[a,b]$ but by hypothesis $x\in (a,b)$ and so $f(x)$ is a non-interior point of $f((a,b))$, falsifying (iii). If (iii) is false, then $(a,b)$ contains a x such that $f(x)$ is not an interior point of $f((a,b))$. Since $f((a,b))$ is an interval, we may assume that $\sup _{(a,b)}f=f(x)$. It then follows from the intermediate value theorem that f is not injective. $\square$

2 Theorem (mean value theorem) Suppose $f\in C([a,b],\mathbf {R} )$ is differentiable on the open interval $(a,b)$. Then

$f(b)-f(a)=f'(c)(b-a)$

for some $c\in (a,b)$
Proof: FIrst assume $0=f(a)=f(b)$. By the theorem preceding this one, $f$ attains a maximum or minimum at $x\in (a,b)$; say, a maximum. By definition, we can write:

$f(y)=f(x)+f'(x)(y-x)+o(|y-x|)$ as $y\to x$

Then since x is a maximum,

$0\geq f(y)-f(x)=f'(x)(y-x)+o(|y-x|)$.

If $y>x$,

$0\geq f'(x)+O(|y-x|)$

and letting $x\to 0$ gives that $f'(x)\leq 0$. If $y<x$, then by the same argument, we find that $f'(x)\geq 0$. Thus, $f'(x)=0$. For the general case, let

$g(x)=f(x)-rx$ where $r={f(b)-f(a) \over b-a}$.

Then $g(a)=g(b)=0$. Hence, applying the first part of the proof gives: $g'(c)=0$ for some c. Since $0=g'(c)=f'(c)-r$, c is a solution of the equation. $\square$

2 Corollary Let $f:\mathbf {R} \to \mathbf {R}$ be a differentiable function. If $f'(a)<c<f'(b)$, then c is in $f'((a,b))$.
Proof:Let $g(x)=f(x)-cx$. Then $g'(a)<0$ and $g'(b)<0$. In other words, $g$ is increasing at a and decreasing at b. By the theorem above, g is not injective on $(a,b)$; i.e., $g(x)=g(x')$ for some $x\neq x'$ in $(a,b)$. It follows:

$0=g(x)-g(x')=g'(y)(x-x')$ for some $y\in (a,b)$.

and $g'(y)=0$. $\square$

2 Corollary A strictly monotonic continuous function with closed range is a homeomorphism.

A real-valued function $f$ on $\mathbf {R} ^{n}$ is said to be homogeneous of degree $k$ ($k$ could be any real number) if

$f(tx)=t^{k}f(x)$

for all $x\in \mathbf {R} ^{n}$ and all $t>0$.

2 Theorem (Euler's relation) Let $f:\mathbf {R} ^{n}\to \mathbf {R}$ be a function differentiable on $\mathbf {R} ^{n}\backslash \{0\}$. Then $f$ is homogeneous of degree $k$ if and only if:

$\left(\sum _{j}x_{j}\partial _{j}-k\right)f(x)=0$

for all $x\in \mathbf {R} ^{n}$.
Proof: ($\Rightarrow$) Differentiate with respect to $t$ both sides of $f(tx)=t^{k}f(x)$ and then put $t=1$. ($\Leftarrow$) Note

$(f(tx))'=\sum _{j}x_{j}{\partial _{j}f}(tx)={k \over t}f(tx)$.

Thus, if we let

$g(t)=\log \left|{f(tx) \over t^{k}f(x)}\right|=\log |f(tx)|-k\log t-\log |f(x)|$,

then the derivative of $g$ vanishes identically. Since $g(1)=0$, $g$ is identically zero. $\square$

The theorem permits a generalization. By definition, any (distribution) solution of the equation

$\left(\sum _{j}x_{j}\partial _{j}-k\right)f=0$

is said to be homogeneous of degree $k$.

A homogeneous distribution is tempered and its Fourier transform is homogeneous.

2 Theorem (l'Hôpital's rule) Let $0\leq a\leq \infty$. If $f(x),g(x)\to 0$ as $x\to a$ or if $g(x)\to \infty$ as $x\to a$, then

$\lim _{x\to a}{f(x) \over g(x)}=\lim _{x\to a}{f'(x) \over g'(x)}$

provided the limit in the right-hand side exists.
Proof: First assume $a$ is finite. We may redefine $f(a)=g(a)=0$ (since the values of functions at $a$ are immaterial when we compute the limit.) Fix $x$, and define

$h(y)=f(y)g(x)-f(x)g(y)$

Since $h(x)=0=h(a)$, by the mean value theorem, we can find a $c$ between $x$ and $a$ such that

$0=h'(c)=f'(c)g(x)-f(x)g'(c)$

Since when $x$ is close to $a$ we may assume $g'$ never vanishes,

${f(x) \over g(x)}={f'(c) \over g'(c)}$

Since $c\to a$ as $x\to a$, the proof of this case is complete. (TODO: handle other cases.) $\square$

The next theorem can be skipped without the loss of continuity, for more general results will later be obtained.

2 Theorem (The Weierstrass approximation theorem) Let $f\in C([0,1])$, and define

$f_{n}(x)=\sum _{k=0}^{n}f(k/n)p_{k}^{n}(x)$ with $p_{k}^{n}(x)={n \choose k}x^{k}(1-x)^{n-k}$ (w:Bernstein polynomial).

Then $f_{n}\to f$ uniformly on $[0,1]$
Proof [1]: First note that

$1=\sum _{k=0}^{n}p_{k}^{n}(x)$

is a partition of unity, by the binomial theorem applied to $(x+1-x)^{n}$. Moreover, a simple computation gives the identity:

$\sum (nx-k)^{2}p_{k}^{n}(x)=nx(x-1)$

It thus follows: for any $\delta >0$

$|f(x)-f_{n}(x)|\leq \sum _{k:|nx-k|<n\delta }|f(x)-f(k/n)|+{nx(x-1) \over n^{2}\delta ^{2}}$

Since $f$ is uniformly continuous on $[0,1]$ by compactness, the theorem now follows. $\square$

2 Corollary Any continuous function vanishing at infinity is uniformly approximated by Hermite function. (TODO: made the statement more precise and give a proof.)

Example: Let $f\in L^{2}([0,1])$. Let $M$ be the linear span of polynomials. Then $M$ is a dense subspace of $L^{2}([0,1])$ by the above theorem since

$\int _{0}^{1}|f-p_{n}|^{2}dx\leq \sup |f-p_{n}|^{2}\to 0$

2 Theorem If $f\in C(\mathbf {R} )$ is uniformly continuous and integrable, then $f\in C_{0}(\mathbf {R} )$.
Proof: Define $F(x)=\int _{-\infty }^{x}f(x)dx$. That $f$ is integrable means that $M=\lim _{x\to \infty }F(x)$ exists and is finite. Let $\epsilon >0$ be given. By uniform continuity, there is a $\delta >0$ such that

$|f(y)-f(x)|<\epsilon$ whenever $|y-x|<\delta$.

Then there is an $R>0$ such that

$|F(x+\delta )-F(x)|<\delta \epsilon$ whenever $x>R$

Now, let $x>R$ be given. By the mean value theorem, we find $y$ such that $\delta f(y)=F(x+\delta )-F(x)$ and $|x-y|<\delta$. Thus,

$|f(x)|\leq |f(x)-f(y)|+\delta ^{-1}|F(x+\delta )-M|+\delta ^{-1}|M-F(x)|<3\epsilon$

$\square$

Example No function is continuous only on rational points. To see this, let $f:\mathbf {R} \to \mathbf {R}$ be a function, and let $E$ be the set of all points at which $f$ is continuous. It follows immediately from the definition of continuity that $E$ is $G_{\delta }$; i.e., it is an intersection of countably many open sets. On the other hand, $\mathbf {Q}$ is not $G_{\delta }$.

2 Theorem Let $U\subset \mathbf {R}$ be a nonempty open subset.

$\sup _{U}|f|\leq 2{\sqrt {\sup _{U}|f'|\sup _{U}|f''|}}$

Proof: Let $A=\sup _{U}|f'|$ and $B=\sup _{U}|f'|$. We assume $A$ and $B$ are finite; otherwise the inequality is trivial. Given $x\in U$, we can find $h>0$ so that an interval $[x,x+h]\subset U$. By Taylor's formula,

$f(x+h)=f(x)+f'(x)h+f''(x+\theta h)h^{2}$ (where $0<\theta <1$)

and so:

$|f'(x)|\leq {A \over h}+Bh$

Now, take $h={{\sqrt {A}} \over {\sqrt {B}}}$. $\square$

2 Theorem (Whitney extension theorem) Every real-valued L-Lipschitz function on a subset of $\mathbf {R} ^{n}$ is the restriction of a L-Lipschitz function on $\mathbf {R} ^{n}$.
Proof ([2] pg. 5): Let $f$ be a L-Lipschitz function on a subset $A$. Define

$F(x)=\inf _{y\in A}(f(y)+L|x-y|)\qquad (x\in \mathbf {R} ^{n})$

It is clear that $F$ is L-Lipschitz continuous. $\square$

Exercise: Every closed set is a zero set of a $C^{\infty }$ function.