[Fourier Analysis] The Weierstrass approximation theorem / 바이어슈트라스 근사정리

참고문헌 : Stein, Elias M., Shakarchi, Rami 2003. Fourier Analysis. Princeton University Press.

 

Stein 해석학 시리즈 1권 [Fourier Analysis] 5단원 Fourier Transfrom on R을 공부한 내용을 TeX으로 적어 정리하는 김에 블로그에도 글을 옮겨 적는다. 책의 내용 중 보충설명이 필요한 부분 그리고 증명을 생략하고 연습문제로 넘긴 명제들 중 일부를 함께 설명하였다. 본문은 PC에 최적화되어있다.

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1.8 The Weierstrass approximation theorem

We have proved the Weierstrass approximation theorem for the continuous function on the circle in Chapter 2. The outline is as follow : Since the Fejér kernel is a good kernel, $(f*F_N)(x)$ converges uniformly to $f$ if $f$ is continuous. As a result, if $f$ is continuous function on the circle then the Fourier series of $f$ is uniformly Césaro summable to $f$. Hence a continuous function on the circle can be uniformly approximated by trigonometric polynomials. Then this can be extended to the fact that any continuous function on the circle can be uniformly approximated by polynomials, considering the relation between  trigonometric polynomials and polynomials. Here we extend the result to the functions on the real line.

Theorem 1. Let $f$ be a continuous function on the closed and bounded interval $[a, b]\subset\mathbb{R}$. Then for any $\epsilon>0$, there exists a polynomial $P$ such that

    $$\sup_{x\in[a, b]} |f(x)-P(x)|<\epsilon$$

In other words, $f$ can be uniformly approximated by polynomials.

proof. Let $[-M, M]$ be any interval containing $[a, b]$ in its interior and $g$ be any continuous function on $\mathbb{R}$ that equals $0$ outside $[-M, M]$ and equals $f$ in $[a, b]$. For example, we can construct $g$ as

    $$g(x)=\begin{cases}
        f(x) & \text{if}\;\;a\le x\le b \\
        l_1(x) & \text{if}\;\;-M\le x<a \\
        l_2(x) & \text{if}\;\; b<x\le M \\
        0 & \text{otherwise}
    \end{cases}$$

where $l_1$ is a line segment connecting $(-M, 0)$ and $(a, f(a))$, $l_2$ is a line segment connecting $(M, 0)$ and $(b, f(b))$. Since $f$ is bounded in $[a, b]$ and $g$ vanishes at $|x|>M$, $g$ is bounded in $\mathbb{R}$. Say, $|g(x)|\le B$. Recall the proof of \textbf{Proposition 2.9}. It states that $(f*K_\delta)(x)$ uniformly converges to $f(x)$ as $\delta\to 0$ if $f\in\mathcal{S}$. The key property of $f$ that establishes this proposition was that $f$ is uniformly continuous. Since $g$ is uniformly continuous, we have analogous proposition that $(g*K_\delta)(x)$ uniformly converges to $g(x)$ as $\delta\to 0$. Then we have $\delta_0>0$ such that

    $$|g(x)-(g*K_{\delta_0})(x)|<\epsilon/2$$

for all $x\in\mathbb{R}$. Now recall that

    $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$

then since $e^x$ is analytic, the partial sum $S_N$ converges to $e^x$ uniformly on any compact subset of $\mathbb{R}$ as $N\to\infty$. Therefore for $K_{\delta_0}(x)=\delta_0^{-1/2} e^{-\pi x^2/\delta_0}$ we have $N'$ such that $N>N'$ implies

    $$|K_{\delta_0}(x)-R_N(x)|<\epsilon/4MB\;\;\;\;\text{where}\;\; R_N(x)=\delta_0^{-1/2}\sum_{n=0}^N\frac{(-\pi x^2/\delta_0)^n}{n!}$$

for all $x\in[-2M, 2M]$ from uniform convergence. Since $g$ vanishes outside of $[-M, M]$, we have

    \begin{align*}
        |(g*K_{\delta_0})(x)-(g*R_N)(x)|&=\Bigg|\int_{-M}^M g(t)\cdot\left(K_{\delta_0}(x-t)-R_N(x-t)\right) dt\Bigg|\\
        &\le \int_{-M}^{M} |g(t)|\cdot |K_{\delta_0}(x-t)-R_N(x-t)| dt\\
        &\le \frac{\epsilon}{4MB}\int_{-M}^{M} |g(t)| dt < \frac{\epsilon}{2}
    \end{align*}

for all $x\in[-M, M]$. Then

    $$|g(x)-(g*R_N)(x)|\le |g(x)-(g*K_{\delta_0})(x)|+|(g*K_{\delta_0})(x)-(g*R_N)(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}$$

for all $x\in[-M, M]$. Hence

    $$\sup_{x\in[a,b]}|f(x)-(g*R_N)(x)|<\epsilon$$

holds. Finally, it is clear that $(g*R_N)(x)$ is a polynomial in $x$ since

    $$(g*R_N)(x)=\int_{-M}^{M} g(t)R_N(x-t) dt$$

and

    $$R_N(x-t)=\delta_0^{-1/2}\sum_{n=0}^N\frac{(-\pi (x-t)^2/\delta_0)^n}{n!}=\sum_{n=0}^{2N} a_n(t) x^n$$

 

This completes the theorem.

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2학년 1학기 때 해석개론 및 연습1에서 같은 정리를 증명했었다. 그때는 해석학이 익숙하지 않아 이 정리가 무슨 의미인지, 심지어는 uniform convergence가 무슨 '함의'를 가지고 있는지도 잘 이해하지 못했는데 이제는 많이 익숙해진 것 같다. 그렇지만 공부를 하면 할수록 여전히 가야할 길이 너무 멀게만 느껴진다.