# User:TakuyaMurata/Continuous functions on a compact space

In this section, we will undergo a throughout study of ${\displaystyle C(K)}$, 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 ${\displaystyle C(K)}$ are real-valued. (A discussion will be given later as to why this does not diminish the generality.)

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

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

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

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

• (i) Given an ${\displaystyle \epsilon >0}$ and ${\displaystyle x\in K}$, we can find a neighborhood ${\displaystyle G}$ of ${\displaystyle x}$ such that
${\displaystyle |f(y)-f(x)|<\epsilon }$ for every ${\displaystyle y\in G}$ and every ${\displaystyle f\in \Gamma }$
• (ii) ${\displaystyle \sup _{\Gamma }|\cdot (x)|<\infty }$ for every ${\displaystyle x\in K}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof: Let ${\displaystyle \delta >0}$ be such that:

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

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

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

and we estimate:

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

where ${\displaystyle n}$ is such that ${\displaystyle n\delta . Thus,

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

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

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

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

${\displaystyle |f(x)-f(y)|<\epsilon }$

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

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

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

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

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

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

we have:

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

for every ${\displaystyle f\in \Gamma }$. ${\displaystyle \square }$

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

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

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

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

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

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

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

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

${\displaystyle f(b)-f(a)=f'(c)(b-a)}$

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

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

Then since x is a maximum,

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

If ${\displaystyle y>x}$,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus, if we let

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

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

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

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

is said to be homogeneous of degree ${\displaystyle k}$.

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

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

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

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

${\displaystyle h(y)=f(y)g(x)-f(x)g(y)}$

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

${\displaystyle 0=h'(c)=f'(c)g(x)-f(x)g'(c)}$

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

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

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

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 ${\displaystyle f\in C([0,1])}$, and define

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

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

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

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

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

It thus follows: for any ${\displaystyle \delta >0}$

${\displaystyle |f(x)-f_{n}(x)|\leq \sum _{k:|nx-k|

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

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

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

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

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

Then there is an ${\displaystyle R>0}$ such that

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

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

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

${\displaystyle \square }$

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

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

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

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

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

and so:

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

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

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

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

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

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