Abstract Nonsense

Crushing one theorem at a time

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 nth 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}.


					
Advertisements

March 28, 2010 - Posted by | General Topology, Topology, Uncategorized | , , ,

3 Comments »

  1. […] 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 […]

    Pingback by Interesting Theorem Regarding Polynomials « Abstract Nonsense | February 9, 2011 | Reply

  2. […] isn’t the extent of one’s exposure to rings of continuous functions. Indeed, given a compact Hausdorff space one considers the ring  of continuous functions as a Banach algebra in the general […]

    Pingback by Rings of Functions (Pt. I) « Abstract Nonsense | August 1, 2011 | Reply

  3. […] it’s fairly easy to see that the map given by is a continuous map when equipping with the infinity norm and since we’d have that is a closed subset of and so the same analysis as above would […]

    Pingback by Closed Curves « Abstract Nonsense | September 23, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: