フーリエ級数展開でのベッセルの不等式
フーリエ級数展開でのベッセルの不等式
周期\(2\pi\)の区分的に滑らかな関数\(f\left(x\right)\)をフーリエ級数展開の第\(n\)部分和を
\[ F_{n}\left(x\right)=\sum_{k=-n}^{n}C_{k}e^{ikx} \] \[ C_{k}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx \] とする。
このとき、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] が成り立つ。
周期\(2\pi\)の区分的に滑らかな関数\(f\left(x\right)\)をフーリエ級数展開の第\(n\)部分和を
\[ F_{n}\left(x\right)=\sum_{k=-n}^{n}C_{k}e^{ikx} \] \[ C_{k}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx \] とする。
このとき、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] が成り立つ。
内積空間でのベッセルの不等式より、
\begin{align*} \sum_{k=-n}^{n}\left|\left\langle x,e_{k}\right\rangle \right|^{2} & \leq\sum_{k=-\infty}^{\infty}\left|\left\langle x,e_{k}\right\rangle \right|^{2}\\ & \leq\left\langle x,x\right\rangle \end{align*} となるので、\(f\left(x\right)\in L^{2}\left[-\pi,\pi\right]\)として内積を\(\left\langle x,y\right\rangle =\frac{1}{2\pi}\int_{-\pi}^{\pi}x\overline{y}dx\)、正規直交列を\(\left\{ e^{inx}\right\} _{n\in\mathbb{Z}}\)とすると、
\[ \sum_{k=-n}^{n}\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] が成り立ち、左辺は
\[ \sum_{k=-n}^{n}\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx\right|^{2}=\sum_{k=-n}^{n}\left|C_{k}\right|^{2} \] となるので、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] となり、フーリエ級数展開でのベッセルの不等式となる。
\begin{align*} \sum_{k=-n}^{n}\left|\left\langle x,e_{k}\right\rangle \right|^{2} & \leq\sum_{k=-\infty}^{\infty}\left|\left\langle x,e_{k}\right\rangle \right|^{2}\\ & \leq\left\langle x,x\right\rangle \end{align*} となるので、\(f\left(x\right)\in L^{2}\left[-\pi,\pi\right]\)として内積を\(\left\langle x,y\right\rangle =\frac{1}{2\pi}\int_{-\pi}^{\pi}x\overline{y}dx\)、正規直交列を\(\left\{ e^{inx}\right\} _{n\in\mathbb{Z}}\)とすると、
\[ \sum_{k=-n}^{n}\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] が成り立ち、左辺は
\[ \sum_{k=-n}^{n}\left|\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx\right|^{2}=\sum_{k=-n}^{n}\left|C_{k}\right|^{2} \] となるので、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] となり、フーリエ級数展開でのベッセルの不等式となる。
元の関数\(f\left(x\right)\)とフーリエ級数展開の第\(n\)部分和\(F_{n}\left(x\right)\)との2乗誤差は、
\begin{align*} 0 & \leq\int_{-\pi}^{\pi}\left|f\left(x\right)-F_{n}\left(x\right)\right|^{2}dx\\ & =\int_{-\pi}^{\pi}\left(\left|f\left(x\right)\right|^{2}-f\left(x\right)\overline{F_{n}\left(x\right)}-\overline{f\left(x\right)}F_{n}\left(x\right)+\left|F_{n}\left(x\right)\right|^{2}\right)dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\int_{-\pi}^{\pi}f\left(x\right)\sum_{k=-n}^{n}\overline{C_{k}e^{ikx}}dx-\int_{-\pi}^{\pi}\overline{f\left(x\right)}\sum_{k=-n}^{n}C_{k}e^{ikx}dx+\int_{-\pi}^{\pi}\left|\sum_{k=-n}^{n}C_{k}e^{ikx}\right|^{2}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\int_{-\pi}^{\pi}f\left(x\right)\sum_{k=-n}^{n}\overline{C_{k}}e^{-ikx}dx-\int_{-\pi}^{\pi}\overline{f\left(x\right)}\sum_{k=-n}^{n}C_{k}e^{ikx}dx+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}C_{k}e^{ikx}\overline{\sum_{j=-n}^{n}C_{j}e^{ijx}}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\sum_{k=-n}^{n}\overline{C_{k}}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx-\sum_{k=-n}^{n}C_{k}\overline{\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx}+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}C_{k}e^{ikx}\sum_{j=-n}^{n}\overline{C_{j}}e^{-ijx}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\sum_{k=-n}^{n}\overline{C_{k}}\left(2\pi C_{k}\right)-\sum_{k=-n}^{n}C_{k}\overline{\left(2\pi C_{k}\right)}+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}\sum_{j=-n}^{n}C_{k}\overline{C_{j}}e^{i\left(k-j\right)x}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+\sum_{k=-n}^{n}\sum_{k=-n}^{n}C_{k}\overline{C_{j}}\int_{-\pi}^{\pi}e^{i\left(k-j\right)x}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+\sum_{k=-n}^{n}\sum_{j=-n}^{n}C_{k}\overline{C_{j}}\left(2\pi\delta_{k,j}\right)\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+2\pi\sum_{k=-n}^{n}C_{k}\overline{C_{k}}\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2} \end{align*} となるので、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] となる。
従って与式は成り立つ。
\begin{align*} 0 & \leq\int_{-\pi}^{\pi}\left|f\left(x\right)-F_{n}\left(x\right)\right|^{2}dx\\ & =\int_{-\pi}^{\pi}\left(\left|f\left(x\right)\right|^{2}-f\left(x\right)\overline{F_{n}\left(x\right)}-\overline{f\left(x\right)}F_{n}\left(x\right)+\left|F_{n}\left(x\right)\right|^{2}\right)dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\int_{-\pi}^{\pi}f\left(x\right)\sum_{k=-n}^{n}\overline{C_{k}e^{ikx}}dx-\int_{-\pi}^{\pi}\overline{f\left(x\right)}\sum_{k=-n}^{n}C_{k}e^{ikx}dx+\int_{-\pi}^{\pi}\left|\sum_{k=-n}^{n}C_{k}e^{ikx}\right|^{2}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\int_{-\pi}^{\pi}f\left(x\right)\sum_{k=-n}^{n}\overline{C_{k}}e^{-ikx}dx-\int_{-\pi}^{\pi}\overline{f\left(x\right)}\sum_{k=-n}^{n}C_{k}e^{ikx}dx+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}C_{k}e^{ikx}\overline{\sum_{j=-n}^{n}C_{j}e^{ijx}}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\sum_{k=-n}^{n}\overline{C_{k}}\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx-\sum_{k=-n}^{n}C_{k}\overline{\int_{-\pi}^{\pi}f\left(x\right)e^{-ikx}dx}+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}C_{k}e^{ikx}\sum_{j=-n}^{n}\overline{C_{j}}e^{-ijx}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-\sum_{k=-n}^{n}\overline{C_{k}}\left(2\pi C_{k}\right)-\sum_{k=-n}^{n}C_{k}\overline{\left(2\pi C_{k}\right)}+\int_{-\pi}^{\pi}\sum_{k=-n}^{n}\sum_{j=-n}^{n}C_{k}\overline{C_{j}}e^{i\left(k-j\right)x}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+\sum_{k=-n}^{n}\sum_{k=-n}^{n}C_{k}\overline{C_{j}}\int_{-\pi}^{\pi}e^{i\left(k-j\right)x}dx\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+\sum_{k=-n}^{n}\sum_{j=-n}^{n}C_{k}\overline{C_{j}}\left(2\pi\delta_{k,j}\right)\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+2\pi\sum_{k=-n}^{n}C_{k}\overline{C_{k}}\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-4\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}+2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2}\\ & =\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx-2\pi\sum_{k=-n}^{n}\left|C_{k}\right|^{2} \end{align*} となるので、
\[ \sum_{k=-n}^{n}\left|C_{k}\right|^{2}\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f\left(x\right)\right|^{2}dx \] となる。
従って与式は成り立つ。
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複素フーリエ級数
\[
c_{m}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f\left(x\right)e^{-imx}dx
\]
ディリクレの収束定理
\[
F\left(x\right)=\lim_{\epsilon\rightarrow0}\frac{f\left(x+\epsilon\right)+f\left(x-\epsilon\right)}{2}
\]
簡単な関数のフーリエ級数展開
\[
F\left(x\right)=\sum_{k=1}^{\infty}\frac{4}{\pi\left(2k-1\right)}\sin\left(\left(2k-1\right)x\right)
\]
複素フーリエ係数の関係
\[
c_{-n}=\overline{c_{n}}
\]