From eec5ed3ab67f4a0763821a66a5650404f4fbe78a Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 13 Oct 2026 16:45:34 -0400 Subject: [PATCH] Weierstrass approximation via Korovkin's Theorem, first draft. --- src/topology/functions/sw.tex | 104 ++++++++++++++++++++++++++++++++++ 1 file changed, 104 insertions(+) diff --git a/src/topology/functions/sw.tex b/src/topology/functions/sw.tex index 9134d3a..4dfcc9c 100644 --- a/src/topology/functions/sw.tex +++ b/src/topology/functions/sw.tex @@ -1,6 +1,110 @@ \section{The Stone-Weierstrass Theorem} \label{section:stone-weierstrass} +\begin{theorem}[Korovkin] +\label{theorem:korovkin-convergence} + For each $x \in [0, 1]$, let $g_x(t) = (t - x)^2$, then + \begin{enumerate} + \item For each $f \in C([0, 1]; \complex)$ and $\eps > 0$, there exists $C > 0$ such that + \[ + |f(t) - f(x)| \le \eps + cg_x(t) + \] + + for all $x, t \in [0, 1]$. + \item Let $f \in C([0, 1]; \complex)$ and $\eps > 0$, then there exists $C > 0$ such that for any positive map $\phi: C([0, 1]; \complex) \to C([0, 1]; \complex)$ and $x \in [0, 1]$, + \[ + -\eps\norm{\phi(1)}_u - C\phi(g_x)(x) \le \phi(f)(x) - f(x) \le \eps\norm{\phi(1)}_u + C\phi(g_x)(x) + \] + + and + + \[ + \norm{\phi(f) - f}_u \le \eps \norm{\phi(1)}_u + C\sup_{x \in [0, 1]}|\phi(g_x)(x)| + \] + \item For each $n \in \natp$, let $\phi_n: C([0, 1]; \complex) \to C([0, 1]; \complex)$ be a positive map. If $\phi_n(1) \to 1$, $\phi_n(t) \to t$, and $\phi_n(t^2) \to t^2$ uniformly as $n \to \infty$, then $\phi_n(f) \to f$ uniformly as $n \to \infty$ for all $f \in C([0, 1]; \complex)$. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[Exercise 2.20]{Paulsen}}}. ] + (1): By \autoref{proposition:uniform-continuous-compact}, there exists $\delta > 0$ such that $|f(t) - f(x)| < \eps$ for all $x, t \in [0, 1]$ with $|x - t| < \delta$. Thus for each $x, t \in [0, 1]$, + \[ + |f(t) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil + \] + + Let $C = \eps/\delta^2$, then for each $s \ge 0$, + \[ + \eps + Cs^2 = \eps + \eps\paren{\frac{s}{\delta}}^2 \ge \eps + \eps\left\lfloor \abs{\frac{s}{\delta}} \right\rfloor \ge \eps\left\lceil \abs{\frac{s}{\delta}} \right\rceil + \] + + so + \[ + |f(x) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil \le \eps + C(s - t)^2 = \eps + Cg_x(t) + \] + + (3): Let $f \in C([0, 1]; \complex)$ and $\eps > 0$. By (2), there exists $C > 0$ such that + \[ + \norm{\phi_n(f) - f}_u \le \eps \norm{\phi_n(1)}_u + C\sup_{x \in [0, 1]}|\phi_n(g_x)(x)| + \] + + for all $n \in \natp$, where + \begin{align*} + \sup_{x \in [0, 1]}|\phi_n(g_x)(x)| &\le \sup_{x \in [0, 1]}|\phi_n(t^2 - 2tx + x^2)(x)| \\ + &\le \sup_{x \in [0, 1]}|\phi_n(t^2)(x) - \phi_n(tx)(x) + x^2| \\ + &\le \normn{\phi_n(t^2) - t^2}_u + 2\sup_{x \in [0, 1]}\normn{\phi_n(xt) - xt}_u + \end{align*} + + which tends to $0$ as $n \to \infty$. Thus + \[ + \limsup_{n \to \infty}\norm{\phi_n(f) - f}_u \le \eps + \] + + As the above holds for all $\eps > 0$, $\phi_n(f) \to f$ uniformly. + +\end{proof} + +\begin{theorem}[Weierstrass Approximation Theorem] +\label{theorem:weierstrass-bernstein-polynomial} + For each $n \in \natz$, let + \[ + \phi_n: C([0, 1]; \real) \to C([0, 1]; \real) \quad \phi_n(f)(t) = \sum_{k = 0}^n {n \choose k}f\paren{\frac{k}{n}}t^k(1-t)^{n-k} + \] + + then: + \begin{enumerate} + \item For every $n \in \natp$, $\phi_n$ is a positive map. + \item For every $n \in \natp$, + \[ + \phi_n(1) = 1 \quad \phi_n(t) = t \quad \phi_n(t^2) = t^2 + \frac{t - t^2}{n} + \] + \item For every $f \in C([0, 1]; \real)$, $\phi_n(f) \to f$ uniformly as $n \to \infty$. + \item $\real[t]$ is dense in $C([0, 1]; \real)$. + \end{enumerate} +\end{theorem} +\begin{proof} + (1): For each $n \in \natz$ and $0 \le k \le n$, $t^k(1 - t)^{n-k} \ge 0$ on $[0, 1]$. + + (2): Let $n \in \natp$, then by the Binomial formula, + \[ + \phi_n(1) = \sum_{k = 0}^n{n \choose k}t^k(1 - t)^{n-k} = (t + (1 - t))^n = 1 + \] + + and + \begin{align*} + \phi_n(t) &= \sum_{k = 1}^n\frac{n!}{k!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\ + &= \sum_{k = 1}^n\frac{(n-1)!}{(k-1)!(n-k)!}t^k(1 - t)^{n-k} \\ + &= t\sum_{k = 0}^{n-1}\frac{(n-1)!}{k!(n-k-1)!}t^{k}(1 - t)^{n-k-1} = t + \end{align*} + + Finally, + \begin{align*} + \phi_n(t^2) &= \sum_{k = 1}^n \frac{n!}{k!(n-k)!}\frac{k^2}{n^2}t^k(1 - t)^{n-k} \\ + &= \sum_{k = 1}^n \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\ + &= t \sum_{k = 0}^{n-1} \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k+1}{n}t^k(1 - t)^{n-k} \\ + &= t \phi_{n-1}\paren{\frac{(n-1)t + 1}{n}} = t^2 + \frac{t - t^2}{n} + \end{align*} + + (3): Since $\seq{\phi_n}$ are positive with $\phi_n(1) \to 1$, $\phi_n(t) \to t$, and $\phi_n(t^2) \to t^2$ uniformly as $n \to \infty$, $\phi_n \to \text{Id}$ pointwise by \hyperref[Korovkin's Theorem]{theorem:korovkin-convergence}. +\end{proof} + \begin{lemma} \label{lemma:sqrt-polynomial} There exists a sequence of polynomials $\seq{p_n} \subset \real[x]$ such that $p_n(x) \upto \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$.