# Abstract Nonsense

## Stone-Weierstrass Theorems

This post is the first in a line of posts which deal with theorems of approximation due to Stone and/or Weierstrass. The simplest and first to be discussed is that the set of all polynomials defined on $[a,b]$ (denoted $\mathbb{P}[a,b]$) is dense in $\mathcal{C}[a,b]$. From there we will make a generalization to $\mathcal{C}\left[X,\mathbb{R}\right]$ where $X$ is compact Hausdorff. Finally, after a brief discussion of locally compact Hausdorff spaces we shall make our final statement regarding $\mathcal{C}[X,\mathbb{R}]$ on such spaces. But, we begin with baby steps.

Theorem: The subspace $\mathbb{P}[a,b]$ is dense in $\mathcal{C}[a,b]$.

Proof: To do this we must merely show that given any $\varepsilon>0$ and $f\in\mathcal{C}[a,b]$  there exists some $p\in\mathbb{P}[a,b]$ such  that $\|f-p\|_{\infty}<\varepsilon$. We take a constructionist approach.

We first note that it suffices to prove the case when $a=0,b=1$.

So, define $B_n(x)$ to be the $n$th Bernstein Polynomial given by

$\displaystyle B_n(x)=\sum_{j=0}^{n}{ n\choose j}x^j (1-x)^{n-j}f\left(\frac{j}{n}\right)$

We begin by noting some identities. By the binomial theorem

$\displaystyle \sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}=(x-(1-x))^n=1$

Differentiating with respect to $x$ and doing some rearrangement gives

$\displaystyle \sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}(j-nx)=0$

Differentiating with respect to $x$ again gives us

$\displaystyle \sum_{j=0}^{n}{n\choose j}\left\{-nx^j(1-x)^{n-j}+x^{j-1}(1-x)^{n-j-1}(j-nx)^2\right\}=0$

Splitting the sum over the addition and apply the first identity to the first sum gives

$\displaystyle \sum_{j=0}^{n}{n\choose j}x^{j-1}(1-x)^{n-j-1}(j-nx)^2=n$

Lastly, multiplying by $\displaystyle \frac{x(1-x)}{n^2}$ gives

$\displaystyle \sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}\left(x-\frac{j}{n}\right)^2=\frac{x(1-x)}{n}$

Now, using the first identity we see that

$\displaystyle f(x)=f(x)\sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}=\sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}f(x)$

And so

$\displaystyle f(x)-B_n(x)=\sum_{j=0}^{n}{n\choose j}x^j(1-x)^{n-j}\left[f(x)-f\left(\frac{j}{n}\right)\right]$

And so

$\displaystyle \left|f(x)-B_n(x)\right|\leqslant \sum_{j=0}^{n}{n\choose j}x^j (1-x)^{n-j}\left|f(x)-f\left(\frac{j}{n}\right)\right|$

Since $f$ is a continuous map on a compact metric space it follows that $f$ is uniformly continuous and so we may find a $\delta>0$ such that

$\displaystyle \left|x-\frac{j}{n}\right|<\delta\implies\left|f(x)-f\left(\frac{j}{n}\right)\right|<\frac{\varepsilon}{2}$

From here we break our sum into two terms

$\displaystyle \sum_{j,\text{ }|x-\frac{j}{n}|<\delta}+\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}$

Clearly the first sum is less than $\displaystyle \frac{\varepsilon}{2}$ (just replace the term with $f$ by $\displaystyle \frac{\varepsilon}{2}$, yank it out, and apply the first identity). So, we must merely show the second sum can be made less than $\displaystyle \frac{\varepsilon}{2}$ independent of $x$. We know appealing to $f$‘s continuity again that it is bounded and so $|f(x)|\leqslant K$ for all $x\in[0,1]$. It clearly follows that

$\displaystyle \sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}\left|f(x)-f\left(\frac{j}{n}\right)\right|\leqslant 2 K\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}$

Thus, we merely need to make the right hand sum less than $\displaystyle \frac{\varepsilon}{2}$. Note that the last identity in the above shows that

$\displaystyle 2\delta^2K\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}\leqslant\frac{x(1-x)}{n}$

and thus

$\displaystyle 2K\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}\leqslant \frac{x(1-x)}{\delta^2 n}$

But, it is easily verified that $x(1-x)\leqslant \frac{1}{4}$ for $x\in[0,1]$ and so

$\displaystyle 2K\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}\leqslant\frac{1}{4\delta^2 n}$

Taking $n$ larger than $\displaystyle \frac{K}{\delta^2\varepsilon}$ implies that

$\displaystyle 2K\sum_{j,\text{ }|x-\frac{j}{n}|\geqslant\delta}{n\choose j}x^j (1-x)^{n-j}<\frac{\varepsilon}{4K}$

and so

$\displaystyle \sum_{j,\text{ }|x-\frac{j}{n}|\geqslant \delta}{n\choose j}x^j (1-x)^{n-j}\left|f(x)-f\left(\frac{j}{n}\right)\right|<\frac{\varepsilon}{2}$.

It follows that

$\|f-B_n(x)\|_{\infty}<\varepsilon$

The conclusion follows. $\blacksquare$.

With this theorem we may prove some very cool stuff. For example:

Theorem: $\mathcal{C}[a,b]$ is separable.

Proof: Let $\mathcal{Q}[a,b]\subseteq\mathbb{P}[a,b]$ be the set of all polynomials with rational coefficients. We prove that $\mathcal{Q}[a,b]$ is dense in $\mathcal{C}[a,b]$. To do this let $\varphi\in\mathcal{C}[a,b]$ and $\varepsilon>0$ be arbitrary. Since $\mathbb{P}[a,b]$ is dense in $\mathcal{C}[a,b]$ there exists some $a_0+\cdots+a_n x^n=p(x)\in\mathbb{P}[a,b]$ such that $\|\varphi-p\|_{\infty}<\frac{\varepsilon}{2}$.

Now, the mapping $\eta:[a,b]\to\mathbb{R}$ given by $\displaystyle x\mapsto\sum_{j=0}^{n}|x|^j$ is continuous and so it attains a maximum $M>1$ on $[a,b]$. So, choose numbers $q_0,\cdots,q_n\in\mathbb{Q}$ such that $\displaystyle |a_k-q_k|<\frac{\varepsilon}{2M}$. Then, clearly $q_0+\cdots+q_n x^n=q(x)\in\mathcal{Q}[a,b]$ and

$\left|p(x)-q(x)\right|\leqslant |a_0-q_0|+\cdots+|a_n-q_n||x|^n<\frac{\varepsilon}{2M}\left(1+\cdots+|x|^n\right)\leqslant \frac{\varepsilon}{2}$

It follows that $\|q-p\|_{\infty}\leqslant\frac{\varepsilon}{2}$ and so

$\displaystyle \|\varphi-q\|_{\infty}\leqslant \|\varphi-p\|_{\infty}+\|p-q\|_{\infty}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

Thus, $\mathcal{Q}[a,b]$ is dense in $\mathcal{C}[a,b]$. It remains to show that it is countable. To see this define $\displaystyle \psi:\bigcup_{n=1}^{\infty}\mathbb{Q}^n\to\mathcal{Q}[a,b]$ by $(q_0,\cdots,q_{n-1})\mapsto q_0+\cdots+q_{n-1}x^{n-1}$. Clearly $\psi$ is surjective and since $\displaystyle \bigcup_{n=1}^{\infty}\mathbb{Q}^n$ is the countable union of countable sets it’s countable. The conclusion follows. $\blacksquare$.

The next problem is a particular favorite of mine, for no particular reason.

Theorem: Define a moment of $f\in\mathcal{C}[0,1]$ by $\displaystyle \int_0^1 f(x)x^n dx$. If $f,g\in\mathcal{C}[0,1]$ have the same moments for $n=0,1,2,\cdots$ then $f=g$.

Proof: We prove (mainly for fun) a lemma from real analysis.

Lemma: Let $f:[a,b]\to [0,\infty)$ be continuous. Then, $\displaystyle \int_a^b f(x)dx=0$ only if $f=0$.

Proof: Suppose that $f\ne 0$ then $f(x)>0$ for some $x\in [a,b]$, call that point $x_0$. Since $f(x_0)\in(0,\infty)$ we clearly have $B_{\varepsilon}(f(x_0))\subseteq (0,\infty)$ for some $\varepsilon>0$. So, by $f$‘s continuity we see that there exists some $B_{2\delta }(x_0)$ such that $f(B_{2\delta}(x_0))\subseteq B_{\varepsilon}(f(x_0))\subseteq (0,\infty)$. Taking the concentric closed ball of radius $\delta$ $B_{\delta}[x_0]$ we see that $f(x)>0$ for every $x\in B_{\delta}[x_0]$. Thus, by $f$‘s continuity and the compactness of $B_{\delta}[x_0]$ we have that $f(x)\geqslant \xi>0$ for some $\xi$. It follows that

$\displaystyle \int_a^b f(x) dx=\int_a^{x_0-\delta}+\int_{B_{\delta}[x_0]}f(x)+\int_{x+\delta}^{b}f(x)dx\geqslant 0+2\xi \delta+0>0$

Which of course contradicts the assumption that the integral is zero. The conclusion follows. $\blacksquare$.

Using this lemma we may assume that $\displaystyle \int_0^1 |f(x)-g(x)|dx\ne 0$ otherwise we would have that

$|f(x)-g(x)|=0\implies f(x)-g(x)=0\implies f(x)=g(x)$

and so we’d be done.

So, under this assumption define the functional $\varphi:\mathcal{C}[0,1]\to\mathbb{R}$ by

$\displaystyle h\mapsto\int_0^1 (f(x)-g(x))h(x) dx$

We claim that $\varphi$ is continuous. To see this $h_0\in\mathcal{C}[0,1]$ and $\varepsilon>0$ be given. Then, choosing $h\in\mathcal{C}[0,1]$ such that

$\displaystyle \|h_0-h\|_{\infty}<\varepsilon\left(\int_0^1 |f(x)-g(x)|dx\right)^{-1}$

we see that

$\displaystyle \left|\varphi[h_0]-\varphi[h]\right|=\left|\int_0^1 (f(x)-g(x))(h_0(x)-h(x))dx\right|$

$\displaystyle \leqslant \int_0^1|f(x)-g(x)||h_0(x)-h(x)|dx$

$\displaystyle <\varepsilon\left(\int_0^1 |f(x)-g(x)|dx\right)^{-1}\int_0^1|f(x)-g(x)|dx=\varepsilon$

$\varphi$‘s continuity follows. We next note that that $\varphi$ is linear and so

$\displaystyle \varphi\left[\sum_{j=0}^{n}a_j x^j\right]=\sum_{j=0}^{n}a_j\varphi\left[x^j\right]=\sum_{j=0}^{n}0=0$

The last part gotten by noticing that

$\displaystyle \varphi\left[x^j\right]=\int_0^1 (f(x)-g(x))x^jdx=\int_0^1 f(x)x^jdx-\int_0^1 g(x)x^jdx=0$

It follows that $\mathbb{P}[a,b]\subseteq\varphi^{-1}[\{0\}]$ and since the right hand side is closed we see that

$\mathcal{C}[0,1]=\overline{\mathbb{P}[a,b]}\subseteq\overline{\varphi^{-1}[\{0\}]}=\varphi^{-1}[\{0\}]$

It follows that $\varphi[h]=0$ for every $h\in\mathcal{C}[0,1]$. In particular

$\displaystyle \varphi[f-g]=\int_0^1(f(x)-g(x))^2dx=0\implies (f(x)-g(x))^2=0\implies f(x)=g(x)$

The conclusion follows. $\blacksquare$.

We now prove a nice little extension of the Weierstrass theorem.

Theorem: Let $X\subseteq\mathbb{R}$ be closed and bounded. Prove that $\mathbb{P}[X]$ is dense in $\mathcal{C}[X,\mathbb{R}]$.

Proof: Note that since $X$ is closed and bounded it is a subset of $[a,b]$ for some $a,b\in\mathbb{R}$. Now, given any $\varphi\in\mathcal{C}[X,\mathbb{R}]$ there exists by Tietze’s extension theorem an extension $\varphi^*\in\mathcal{C}[a,b]$ such that $\varphi^*\mid X=\varphi$. Now, for every $\varepsilon>0$ there exists some $p_{\varepsilon}\in\mathbb{P}[a,b]$ such that $\|\varphi^*-p\|_{\infty}<\varepsilon$. Let $\Omega_\varphi=\left\{p_{\varepsilon}:\varepsilon>0\right\}$. Furthermore, let

$\displaystyle \Omega^*=\bigcup_{\varphi\in\mathcal{C}[X,\mathbb{R}]}\Omega_\varphi$

and finally let $\Omega=\left\{p\mid X:p\in\Omega^*\right\}$. Clearly, $\Omega\subseteq\mathbb{P}[X]$. So, it remains to show that given any $\varepsilon>0$ and $\varphi\in\mathcal{C}[X,\mathbb{R}]$ there exists some $p\in\Omega$ such that $\displaystyle \sup_{x\in X}|\varphi(x)-p(x)|<\varepsilon$. To see this let $\varphi^*$ and $p_{\varepsilon}$ be as above and let $p=p_{\varepsilon}\mid X$. Then, clearly $p\in\Omega$ and

$\displaystyle \sup_{x\in X}|\varphi(x)-p(x)|=\sup_{x\in X}|\varphi^*(x)-p_{\varepsilon}(x)|\leqslant\sup_{x\in[a,b]}|\varphi(x)-p_{\varepsilon}(x)|<\varepsilon$

The conclusion follows. $\blacksquare$

Corollary: Using the exact same idea except taking $q\in\mathcal{Q}[a,b]$ one may prove that $\mathcal{C}[X,\mathbb{R}]$ is separable for every closed and bounded $X\subseteq \mathbb{R}$.

March 28, 2010 -