Ascoli-Azrela and Stone-Weirstrass Theorem

Theorem (Ascoli-Azrela)

Let $\{f_n\}$ be a sequence of functions defined on a set $K$. If $K$ is compact, $\{f_n\}$ is pointwise bounded on $K$, $\{f_n\}$ is equicontinuous then

(a) $\{f_n\}$ is uniformly bounded on $K$

(b) $\{f_n\}$ has a uniformly convergence subsequence.

<Proof>

(a) Fix an $\epsilon>0$. By equicontinuity, there is a $\delta>0$ such that

\[\begin{align*} d(x,y) < \delta \Rightarrow d(f_n(x), f_n(y)) <\epsilon \end{align*}\]

for all $x,y \in K$ and for all $n\in\mathbb{N}$. Now cover the set $K$ with $\{N_\delta (x): x\in K\}$. Since $K$ is compact, there is a finite sub-cover $\{N_\delta(x_1), \ldots, N_\delta(x_n)\}$. Since $\{f_n\}$ is pointwise bounded, for each $x_i$, there is $M_i>0$ such that

\[\begin{align*} \lvert f_n(x_i) \rvert \leq M_i \end{align*}\]

for all $n\in\mathbb{N}$. Now let $M :=\max\{M_1, \ldots, M_n\}+\epsilon$. For any $x\in K$, choose $x_i$ such that $x\in N_\delta(x_i)$, i.e., $d(x, x_i) <\delta$. Then

\[\begin{align*} \lvert f_n(x) \rvert &= \lvert f_n(x) - f_n(x_i)+ f_n(x_i) \rvert \\ &\leq \lvert f_n(x) - f_n(x_i)\rvert + \lvert f_n(x_i) \rvert \\ &< \epsilon + M_i \\ &\leq M \end{align*}\]

(b) Exercise) If $K$ is compact metric space, then there is countable dense subset $E \subset K$, i.e., $K=E\cup E^\prime$. By previous Theorem, there exists a subsequence $\{f_{n_k}\}$ pointwise converges on $E$.

We want to show that $\{f_{n_k}\}$ is uniformly convergent subsequence.

Let $\epsilon >0$ be given. By equicontinuity, there exists a $\delta>0$ such that

\[\begin{align*} d(x,y) < \delta \Rightarrow d(f_{n_k}(x) - f_{n_k}(x)) < \epsilon \end{align*}\]

for all $k\in\mathbb{N}$. Cover $E$ with $\{N_\delta (e): e\in E\}$. Since $E$ is dense, it also covers $K$. By compactness of $K$, there exists a finite sub-cover $\{N_\delta (e_1), \ldots, N_\delta (e_p)\}$ of $K$. Since $\{f_{n_k}\}$ converges pointwise on $E$, for each $e_i$, there exists $N\in\mathbb{N}$ such that

\[\begin{align*} k,l > N_i \Rightarrow \lvert f_{n_k}(e_i) - f_{n_l}(e_i)\rvert <\epsilon. \end{align*}\]

Let $N=\max\{N_1, \ldots, N_p\}$ and let $x\in K$ be given. Choose $e_i$ such that $x\in N_\delta(e_i)$. Then, if $k,l >N$

\[\begin{align*} \lvert f_{n_k}(x) - f_{n_l}(x) \rvert &\leq \lvert f_{n_k}(x) - f_{n_k}(e_i)\rvert + \lvert f_{n_k}(e_i) - f_{n_l}(e_i)\rvert + \lvert f_{n_l}(e_i)-f_{n_l}(x) \rvert \\ &<3\epsilon. \end{align*}\]

$\therefore \{f_{n_k}\}$ converges uniformly on $K$ by Cauchy criterion.

\[\tag*{$\square$}\]

Theorem (Stone-Weierstrass)

Given any function $f\in\mathscr{C}[a,b]$, there exists a sequence $\{P_n\}$ of polynomials that converges uniformly to $f$ on $[a,b]$. In other words, the space polynomial of $[a,b]$ is dense in $\mathscr{C}[a,b]$ with respect to supnorm metric.

<Proof>

Without loss of generality, assume $[a,b]$ is $[0,1]$ , $f(0)=f(1)=0$, and think $f(x)=0$ for all $x\in \mathbb{R}\setminus [a,b]$. Since $[a,b]$ is closed and bounded, $[a,b]$ is compact. Thus $f$ is uniformly continuous on $[a,b]$ and also on $\mathbb{R}$.

We will create an approximation of the identity. Let

\(\begin{align*} Q_n(x) = c_n(1-x^2)^n \end{align*}\) on $[-1,1]$, where

\[\begin{align*} c_n = \frac{1}{\int_{-1}^1 Q_n(x)dx}. \end{align*}\]

Observe that $\int_{-1}^1 Q_n(x)dx=1$ and for any $\delta\in (0,1)$ $Q_n$ converges uniformly to $0$ on $\delta \leq \lvert x \rvert \leq 1$.

$f(x)= (1-x^2)^n, g(x) = 1-nx^2$ on $[0,1]$. $f(0)=g(0)=1,\text{and } f(1)=g(1)=0$. Since

\[\begin{align*} f^\prime(x) = n(1-x^2)^n(-2x) \geq -2nx = g^\prime(x), \forall x \in [0,1] \end{align*}\]

$f(x) \geq g(x)$ for all $x\in [0,1]$.

Since

\[\begin{align*} \int_{-1}^1 (1-x^2)^n dx &= 2\int_0^1(1-x^2)^n dx\\ &\geq 2\int_0^{1/\sqrt{n}}(1-x^2)^ndx \\ &\geq 2\int_0^{1/\sqrt{n}}(1-nx^2)dx \quad (\because (1-x^2)^n \geq 1-nx^2 \text{ on } [0,1]) \\ &=\frac{4}{3\sqrt{n}} \\ &> \frac{1}{\sqrt{n}} \end{align*}\]

Thus $c_n <\sqrt{n}$. Then

\[\begin{align*} Q_n(x) < \sqrt{n}(1-x^2)^n \leq \sqrt{n}(1-\delta^2)^n. \end{align*}\]

By L’Hôpital’s rule

\[\begin{align*} \lim_{n\to\infty}\sqrt{n}(1-\delta^2)^n = \lim_{n\to\infty} \frac{\sqrt{n}}{r^n} = \lim_{n\to\infty}\frac{1}{2\sqrt{n}r^{n-1}}=0, \end{align*}\]

where $r=(1-\delta^2)^{-n}>1$.

Thus, $Q_n$ converges uniformly to $0$ on $\delta \leq \lvert x \rvert \leq 1$. Define

\[\begin{align*} P_n(x) = \int_{-1}^1 f(x+t)Q_n(t) dt. \end{align*}\]

Intuitively, if $n\to\infty$ $Q_n(t)$ becomes dirac-delta, i.e., $\int_{-1}^1Q_n(t) dt =1$ but $Q_n(t)=\mathbf{1}\{t=0\}$. Since

\[\begin{align*} P_n(x) &= \int_{-1}^1 f(x+t)Q_n(t)dt \\ &= \int_{-x}^{1-x}f(x+t)Q_n(t) dt \quad (\because 0\leq x+t \leq 1 \Rightarrow -x \leq t \leq 1-x) \\ &=\int_0^1 f(t) Q_n(t-x)dt \quad (\because \text{change of variable})\\ &=\int_0^1 f(t) c_n(1-(t-x)^2)^n dt \end{align*}\]

Now we want to show that $P_n \to f$ uniformly. Let $\epsilon >0$ be given. Since $f \in \mathscr{C}[0,1]$, it is bounded, i.e., there is $M>0$ such that $\lvert f(x) \rvert < 0$ for all $x\in [0,1]$. Moreover, since $f$ is uniformly continuous on $[0,1]$, we can choose $\delta >0$ such that

\[\begin{align*} \lvert t \rvert <\delta \Rightarrow \lvert f(x+t) -f(x) \rvert <\frac{\epsilon}{2}. \end{align*}\]

Lastly, $Q_n\to 0$ uniformly on $\delta \leq \lvert x \rvert \leq 1$,

\[\begin{align*} \lim_{n\to\infty}\int_{\delta \leq \lvert t\rvert \leq 1}Q_n(t) &= \int_{\delta \leq \lvert t \rvert \leq 1}\lim_{n\to\infty}Q_n(t)dt \\ &=\int_{\delta \leq \lvert t \rvert \leq 1} 0dt =0. \end{align*}\]

Then, we can choose $N\in\mathbb{N}$ such that

\[\begin{align*} n\geq N\Rightarrow \int_{\delta \leq \lvert t \rvert \leq 1} Q_n(t) dt < \frac{\epsilon}{4M}. \end{align*}\] \[\begin{align*} \lvert P_n(x) - f(x) \rvert &=\left\lvert \int_{-1}^1f(x+t)Q_n(t) dt - f(x) \right\rvert \\ &=\left\lvert \int_{-1}^1f(x+t)Q_n(t) dt - f(x)\int_{-1}^1Q_n(t)dt \right\rvert \\ &=\left\lvert \int_{-1}^1(f(x+t)-f(x)Q_n(t) dt \right\rvert \\ &\leq \int_{-1}^1\left\lvert f(x+t)-f(x)\right\rvert Q_n(t)dt \\ &\leq \int_{\lvert t\rvert \leq \delta}\left\lvert f(x+t)-f(x)\right\rvert Q_n(t)dt + \int_{\delta \leq \lvert t\rvert \leq 1}\left\lvert f(x+t)-f(x)\right\rvert Q_n(t)dt \\ &\leq \frac{\epsilon}{2}\int_{\lvert t\rvert \leq \delta}Q_n(t)dt + 2M\int_{\delta \lvert t \rvert \leq 1}Q_n(t) dt \\ &\leq \frac{\epsilon}{2}\int_{-1 \leq t \leq 1}Q_n(t)dt + 2M\int_{\delta \leq \lvert t \rvert \leq 1}Q_n(t) dt \\ &< \frac{\epsilon}{2} + 2M\frac{\epsilon}{4M}\\ &=\epsilon \end{align*}\] \[\tag*{$\square$}\]

Definition (Analytic Function)

Any function $f$ that has a power series expansion, i.e., can be expressed in the form of

\[\begin{align*} f(x) = \sum_{n=0}^\infty c_n x^n \end{align*}\]

or

\[\begin{align*} f(x) = \sum_{n=0}^\infty c_n(x-a)^n, \quad a \in \mathbb{R} \end{align*}\]

is called analytic function.

Remark

Any such power series has a radius of convergence $R$, obtained via root test,

\[\begin{align*} \limsup_{n\to\infty}\sqrt[n]{\lvert c_n x^n \rvert} <1 \Rightarrow \sum_{n=0}^\infty c_n x^n \text{ converges.} \end{align*}\]

In other words, convergence radius $R=1/\sqrt[n]{c_n}$.

Lemma

$\lim_{n\to\infty}\sqrt[n]{n}=1$

<Proof>

Let $\sqrt[n]{n}:=1+\delta$. By binomial theorem,

\[\begin{align*} n&=(1+\delta_n)^n\\ &=1 + n\delta_n + {n \choose 2 } \delta^2_n + \cdots + \delta^n_n.\\ \end{align*}\]

Thus $n > n(n-1)/2 \delta^2_n$, i.e.,

\[\begin{align*} 0 < \delta_n < \sqrt{\frac{2}{n-1}}. \end{align*}\]

$\therefore \lim_{n\to\infty}\sqrt[n]{n}=1$.

\(\tag*{$\square$}\)

Theorem

Suppose $f(x) = \sum c_n x^n$ converges in $(-R, R)$. Then

i) $\sum c_n x^n$ converges uniformly in $[-R+\epsilon, R-\epsilon]$ for any $\epsilon >0$.

ii) $f$ is continuous and differentiable on $(-R,R)$ and $f^\prime(x) = \sum_{n=1}^\infty n c_n x^{n-1}$.

<Proof>

Use Weierstrass M-test. If $x\in [-R+\epsilon, R-\epsilon]$, then $\lvert x\rvert \leq R-\epsilon$, which implies that $\lvert c_n x^n \rvert \leq \lvert c_n (R-\epsilon)^n\rvert$. Since every power series converges absolutely in the interior of convergence by the root test, $\sum c_n x^n$ converges uniformly by Weierstrass M-test.

Now consider $\sum_{n=1}^\infty n c_n x^{n-1}$. By Lemma,

\[\begin{align*} \limsup_{n\to\infty}\sqrt[n]{\lvert n c_n x^{n-1}\rvert}= \limsup_{n\to\infty}\sqrt[n]{\lvert c_n\rvert}\lvert x_n\rvert <1. \end{align*}\]

Thus, $\sum_{n=1}^\infty n c_n x^{n-1}$ has the same radius of convergence, which implies that $\sigma_N=\sum_{n=1}^N n c_n x^{n-1}$ converges uniformly on $[-R+\epsilon, R-\epsilon]$. So by the previous theorem,

\[\begin{align*} \frac{d}{dx}\lim_{N\to\infty}\sum_{n=0}^N c_n x^n=\lim_{N\to\infty}\frac{d}{dx}\left(\sum_{n=0}^Nc_nx^n\right)=\lim_{N\to\infty}\sigma_N. \end{align*}\] \[\tag*{$\square$}\]

Reference

• Walter Rudin, 『Principles of Mathematical Analysis』