📂ヒルベルト空間

ヒルベルト空間における一般化されたフーリエ係数、フーリエ級数

説明

$x=f$、$\left\{ u_{\alpha} \right\}=\left\{ e^{-i n x} \right\}$とすると、$\hat{x}(\alpha)$はよく知られている複素フーリエ係数になる。

$$\hat{x}(\alpha)=c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$

また、有限次元ヒルベルト空間について以下の定理が成立する。ここで、定理(a)の$(2)$は有限次元ヒルベルト空間におけるピタゴラスの定理、$(4)$は有限次元ヒルベルト空間におけるベッセルの不等式だ。

定理(b)で定義される有限和$s_{F}(x)$を$x$のフーリエ級数と呼ぶ。

$(3)$は、$s_{F}(x)$が$M_{F}$で$x$に対する唯一の最良近似であることを示している。

Theorem

$\hat{x}$ is a linear functional of $H$, i.e., an element of the dual of $H$.

Let $\left\{ u_{\alpha} \right\}_{\alpha \in A}$ be a normal orthogonal set in the Hilbert space $H$, and let $F$ be a finite subset of $A$. And let’s say $M_{F}=\text{span} \left\{ u_{\alpha} : \alpha \in F \right\}$.

(a) There exists the following $y\in M_{F}$.

$$\begin{equation}\begin{aligned} y&=\sum \limits_{\alpha \in F}\hat{y}(\alpha)u_{\alpha} \\ &=\sum \limits_{\alpha \in F}\left\langle y,u_{\alpha} \right\rangle u_{\alpha} \end{aligned} \end{equation}$$

Also, the following holds:

$$\begin{equation} \left\| y \right\|^{2} = \sum \limits_{\alpha \in F} \left| \hat{y}(\alpha) \right|^{2} \end{equation}$$

(b) If $x\in H$ and $s_{F}(x)=\sum \limits_{\alpha \in F}\hat{x}(\alpha)u_{\alpha}$, then the following holds for all $s\in M_{F}$ except for $s_{F}$.

$$\begin{equation} \left\| x-s_{F}(x) \right\| < \left\| x-s \right\| \end{equation}$$

And

$$\begin{equation} \sum \limits_{\alpha \in F} \left| \hat{x}(\alpha) \right|^{2} \le \left\| x \right\|^{2} \end{equation}$$

Proof

(a)

$(1)$ is obvious by the definition of $M_{F}$, and $(2)$ can easily be known by orthogonality.

$$\begin{align*} \left\| y \right\|^{2} =&\ \left\langle y,y \right\rangle \\ =&\ \left\langle \sum \limits_{\alpha \in F}\hat{y}(\alpha) u_{\alpha},\sum \limits_{\alpha \in F}\hat{y}(\alpha) u_{\alpha} \right\rangle \\ =&\ \left\langle \hat{y}(\alpha_{1}) u_{\alpha_{1}}+ \cdots + \hat{y}(\alpha_{k}) u_{\alpha_{k}} , \hat{y}(\alpha_{1}) u_{\alpha_{1}}+ \cdots + \hat{y}(\alpha_{k}) u_{\alpha_{k}} \right\rangle \\ =&\ \left| \hat{y}(\alpha_{1}) \right|^{2}\left\langle u_{\alpha_{1}},u_{\alpha_{1}} \right\rangle + \cdots + \left| \hat{y}(\alpha_{k}) \right|^{2}\left\langle u_{\alpha_{k}},u_{\alpha_{k}} \right\rangle \\ =&\ \sum \limits_{\alpha \in F} \left| \hat{y}(\alpha) \right|^{2} \end{align*}$$

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(b)

Firstly, let’s simply denote $s_{F}(x)$ as $s_{F}$. Then, since $\left\{ u_{\alpha} \right\}$ is a normal orthogonal set,

$$\hat{s_{F}}(\alpha)=\left\langle \sum \limits_{\alpha \in F}\hat{x}(\alpha)u_{\alpha},u_{\alpha} \right\rangle=\hat{x}(\alpha)\quad \forall \alpha \in F$$

holds. This means that for $\alpha\in F$, $(x-s_{F})\perp u_{\alpha}$ happens. Then since $s,s_{F}\in M_{F}$, $(x-s_{F})\perp (s_{F}-s)$ holds. Therefore,

$$\left\| x-s \right\| ^{2} = \left\| (x-s_{F})+(s_{F}-s) \right\| ^{2}=\left\| x-s_{F} \right\| ^{2}+\left\| s_{F}-s \right\| ^{2}$$

holds and from this, $(3)$ holds. Also, substituting $s=0$ into the above equation,

$$\left\| x \right\| ^{2} = \left\| x-s_{F} \right\| ^{2}+\left\| s_{F} \right\| ^{2}$$

gives us

$$\left\| s_{F} \right\| ^{2}\le \left\| x \right\| ^{2}$$

and thus

$$\sum \limits_{\alpha \in F} \left| \hat{x}(\alpha) \right|^{2}=\left\| s_{F} \right\|^{2} \le\left\| x \right\|$$

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