# Abstract Nonsense

## Polynomial Rings (Pt. III)

Point of Post: This post is a continuation of this one.

July 19, 2011

## Polynomial Rings (Pt. II)

Point of Post: This post is a continuation of this one.

$\text{ }$

July 19, 2011

## Interesting Theorem Regarding Polynomials

Point of post: This post proves that the uniform limit of polynomials of a fixed degree is a polynomial of degree less than or equal to that fixed degree.

Motivation

Surely most of us are aware of the Stone-Weierstrass theorem (see here for a more generalized version for compact Hausdorff spaces) which states that for any $[a,b]\subseteq\mathbb{R}$ one has that the polynomials $\mathbb{P}[a,b]$ are dense in the space $C[a,b]$ of continuous functions with the infinity norm. Another interpretation of this is that every continuous function $f$ can be arbitrarily and uniformly approximated by a sequence of polynomials $\{p_n\}_{n\in\mathbb{N}}$ is the sense that $\|p_n-f\|_{\infty}\to 0$. One may begin to wonder “what kinds of polynomials do I need to approximate these functions?” in the sense that could I approximate $f$ with polynomials of only even degree? What about odd degree? What bout prime degree? This seems like a valid question since, either the Stone-Weierstrass theorem is done by constructing the polynomials (see the first link and the notion of the Bernstein polynomials) or it’s proven in full generality (as in the second link). The first way leaves one wondering if there isn’t a ‘better’ choice of polynomials to approximate $f$, I mean, it’s conceivable that using Bernstein polynomials while a valid way to prove the theorem is overly complicated. The second proof gives the reader absolutely zero idea what the polynomials used to approximate the functions look like. So with this in mind, one may ask an an example of such a thought experiment (although a naive one) whether continuous functions can be approximate by polynomials which all have a fixed degree $N$. Well, in this post we will show that not only can you not approximate all the continuous functions with such sequences of polynomials but you can really only approximate other polynomials of degree less than or equal to $N$! In other words, we’ll prove that the space of all polynomials of degree less than or equal to $N$ is closed in $C[a,b]$ with the infinity norm. In fact, we’ll prove something MUCH stronger

February 9, 2011

## Halmos Sections 34 and 35:Products and Polynomials (Pt. II)

Point of post: This post is a continuation of

November 29, 2010

## Permutations (Pt. VI Alternate Definitions of the Sign of a Permutation cont.)

Point of post: This is a literal continuation of this post. I just hate when that dark grey to light grey thing happens. Just pretend that there was no lapse in the posts

And so we continue…

November 8, 2010

## 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