L2 Recovery of Integrable Functions and Consequences
Corollary 1.2 (Parseval’s Identity)
Let $f$ be integrable function, and $a_n= \hat{f}(n)$. Then $\lim_{N\to\infty}\sum_{n=-N}^N\lvert a_n\rvert^2$ converges to $\lVert f \rVert^2$.
<Proof>
By Pythagorean theorem,
\[\begin{align*} \lVert f \rVert^2 & = \lVert f-S_N(f)\rVert^2 + \lVert S_N(f)\rVert^2 \\ & = \lVert f-S_N(f)\rVert^2 + \frac{1}{2\pi} \int_{-\pi}^{\pi}\left( \sum_{n=-N}^N a_n e^{inx}\right)\left(\overline{\sum_{m=-N}^N a_m e^{imx}}\right) dx \\ &=\lVert f-S_N(f)\rVert^2 + \frac{1}{2\pi} \int_{-\pi}^\pi \sum_{n=-N}^N \lvert a_n\rvert^2 dx \\ &=\lVert f-S_N(f)\rVert^2 + \sum_{n=-N}^N \lvert a_n\rvert^2 \end{align*}\]We know that $\lim_{N\to\infty}\lVert f-S_N(f)\rVert^2=0$. Thus,
\[\begin{align*} \lim_{N\to\infty}\sum_{n=-N}^N \lvert a_n\rvert^2&=\lVert f\rVert^2 - \lim_{N\to\infty}\lVert f-S_N(f)\rVert^2 \\ &= \lVert f\rVert^2. \end{align*}\]$\therefore \lVert a_n\rVert_{\ell^2}= \lVert f\rVert_{L^2}$.
\[\tag*{$\square$}\]Theorem (Polarized Parseval’s identity)
Let $f$ and $g$ be integrable on the circle, with Fourier coefficients $\{a_n\}$ and $\{b_n\}$, respectively. Then
\[\begin{align*} \frac{1}{2\pi} \int_0^{2\pi} f(\theta)\overline{g(\theta)}d\theta = \sum_{n=-\infty}^\infty a_n \overline{b_n}. \end{align*}\]<Proof>
In any Hermitian inner product space, one has the polarization identity,
\[\begin{align*} \langle f, g \rangle = \frac{1}{4}\left(\lVert f+g\rVert^2 - \lVert f-g\rVert^2 + i(\lVert f+ig \rVert^2-\lVert f-ig\rVert^2)\right). \end{align*}\]Using this in $\mathcal{R}$ and Parseval’s indentity, we get
\[\begin{align*} \langle f, g\rangle_{L^2} &= \frac{1}{4}\left( \sum \lvert a_n +b_n\rvert^2 - \sum \lvert a_n-b_n\rvert^2 + i(\sum\lvert a_n +ib_n\rvert^2 -\sum\lvert a_n-ib_n\rvert^2) \right) \\ &=\frac{1}{4}\left(\lVert \{a_n + b_n\}\rVert^2-\lVert \{a_n-b_n\}\rVert^2 +i (\lVert \{a_n+ib_n\}\rVert^2-\lVert \{a_n-ib_n\}\rVert^2) \right) \\ &=\langle \{a_n\}, \{b_n\}\rangle_{\ell^2}. \end{align*}\] \[\tag*{$\square$}\]Theorem 1.3 (Riemann-Lebesgue Lemma)
If $f$ is integrable on the circle, then $\hat{f}(n) \to 0$ as $\lvert n\rvert\to\infty$.
An equivalent reformulation of this proposition is that if $f$ is integrable on $[0,2\pi]$, then
\[\begin{align*} &\lim_{\lvert N \rvert\to0}\int_0^{2\pi} f(\theta)\sin(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty \\ &\lim_{\lvert N \rvert\to0}\int_0^{2\pi} f(\theta)\cos(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty \end{align*}\]<Proof> Let $\epsilon >0$ be given. Since $f$ is integrable, there is a partition $\mathcal{P}=\{ 0=x_0, \ldots, 2\pi=x_l\}$ of $[0,2\pi]$ such that
\[\begin{align*} \left\lvert \int_0^{2\pi}f(x)dx - L(f, \mathcal{P}) \right\rvert < \frac{\epsilon}{2}. \end{align*}\]Now define a step function $\varphi: [0,2\pi]\to\mathbb{R}$ such that
\[\begin{align*} \varphi(x)= \begin{cases} c_i=\inf_{[x_{i-1}, x_i]}f(x) \quad &\text{if } x \in (x_{i-1}, x_i) \\ f(x_i) \quad &\text{if } x \in \{x_0, \ldots, x_l\} \end{cases} \end{align*}\]Then $\int_0^{2\pi}\varphi(x)dx = \sum_{i=1}^l c_i(x_i - x_{i-1}) = L(\mathcal{P},f)$, i.e.
\[\begin{align*} \left\lvert \int_0^{2\pi} f - \int_0^{2\pi}\varphi\right\rvert < \frac{\epsilon}{2}. \end{align*}\]By triangular inequality,
\[\begin{align*} \left\lvert \int_0^{2\pi} f(x)\cos(nx) dx\right\rvert & \leq \left\lvert \int_0^{2\pi}(f(x)-\varphi(x))\cos(nx) \right\rvert + \left\lvert \int_0^{2\pi} \varphi(x)\cos(nx) dx\right\rvert \\ &\leq \int_0^{2\pi} f(x)-\varphi(x) dx + \sum_{i=1}^l \frac{\lvert c_i\rvert}{\lvert n\rvert} \lvert \sin(nx_{i})-\sin(nx_{i-1}) \rvert \\ &=\left\lvert \int_0^{2\pi}f - \int_0^{2\pi}\varphi \right\rvert + \sum_{i=1}^l \frac{\lvert c_i\rvert}{\lvert n \rvert} \lvert \sin(nx_{i})-\sin(nx_{i-1}) \rvert \\ &<\frac{\epsilon}{2} + \frac{\epsilon}{2} \end{align*}\]$\therefore \lim_{\lvert n \rvert \to \infty} \int_0^{2\pi}f(x)\cos(nx)dx=0$. Similarly we can show that
\[\begin{align*} &\lim_{\lvert n \rvert\to0}\int_0^{2\pi} f(\theta)\cos(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty \\ &\lim_{\lvert n \rvert\to0}\int_0^{2\pi} f(x)e^{inx}dx \to 0 \quad \text{as } \lvert n\rvert\to\infty \end{align*}\] \[\tag*{$\square$}\]Lemma 1.4
Let $f$ be a bounded function on the compact interval $[a,b]$. If $c\in (a,b)$ and if for small $\delta>0$ the function $f$ is integrable on the intervals $[a,c-\delta]$ and $[c+\delta, b]$, then $f$ is integrable on $[a,b]$.
<Proof>
Suppose that $\lvert f(x) \rvert \leq M$ for all $x\in [a,b]$. Let $\epsilon>0$ be given and take $\delta>0$ such that $4\delta M < \epsilon/3$. Then there is a partition $\mathcal{P}_1$ of $[a,c-\delta]$ and partition $\mathcal{P}_2$ of $[c+\delta, b]$ such that
\[\begin{align*} &U(f, \mathcal{P}_1) -L(f, \mathcal{P}_1) < \frac{\epsilon}{3} \\ &U(f, \mathcal{P}_2)- L(f, \mathcal{P}_2) < \frac{\epsilon}{3} \end{align*}\]since $f$ is integrable on $[a, c-\delta]$ and $[c+\delta, b]$, respectively. Take a partition $\mathcal{P}_3=\{c-\delta, c+\delta \}$ of $[c-\delta, c+\delta]$. Then
\[\begin{align*} \sup_{x\in [c-\delta, c+\delta]} f(x) &\leq M \\ \inf_{x\in [c-\delta, c+\delta]}f(x) &\geq -M. \end{align*}\]Thus,
\[\begin{align*} U(f, \mathcal{P}_3) - L(f, \mathcal{P}_3) \leq 4 \delta M < \frac{\epsilon}{3}. \end{align*}\]Let $\mathcal{P} = \bigcup_{i=1}^3 \mathcal{P}_i$. Then $U(f, \mathcal{P})- L(f,\mathcal{P}) < \epsilon$.
$\therefore f$ is integrable on $[a,b]$.
\[\tag*{$\square$}\]Remark
Convergence in $L^2$ does not guarantee that the Fourier series converges for any $\theta$.
Theorem 2.1
Let $f$ be an integrable function on the circle which is differentiable at $\theta_0$. Then $\lim_{N\to\infty}S_N(f)(\theta_0) = f(\theta_0)$.
<Proof>
Define
\[\begin{align*} F(t) = \begin{cases} \frac{f(\theta_0-t)-f(\theta_0)}{t} \quad &\text{ if } t \neq 0 \text{ and } \lvert t \rvert < \pi \\ -f^\prime(\theta_0) & \text{ if } t= 0 \end{cases} \end{align*}\]$F$ is bounded near $0$ since $f$ is differentiable there. For a small $\delta >0$, the function $F$ is integrable on $[-\pi, -\delta] \cup [\delta, \pi]$ since $f$ is integrable. By lemma 1.4, $F$ is integrable on $[-\pi, \pi]$.
Let $D_N(x)= \sum_{n=-N}^N e^{inx}$ be Dirichlet kernel and recall that $S_N(f)(x) = (f*D_N)(x)$. Then
\[\begin{align*} S_N(f)(\theta_0) - f(\theta_0) &= \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta_0-t)D_N(t) dt - f(\theta_0) \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi (f(\theta_0-t)-f(\theta_0))D_N(t) dt \quad (\because \frac{1}{2\pi}\int_{-\pi}^\pi D_N(t)dt=1) \\ &=\frac{1}{2\pi} \int_{-\pi}^\pi F(t)tD_N(t)dt \end{align*}\]Recall that
\[\begin{align*} tD_N(t) = \frac{t}{\sin(t/2)}\sin((N+1/2)t) \end{align*}\]and
\[\begin{align*} \sin((N+1/2)t)=\sin(Nt)\cos(t/2) + \cos(Nt)\sin(t/2). \end{align*}\]Thus,
\[\begin{align*} S_N(f)(\theta_0) - f(\theta_0) &= \frac{1}{2\pi} \int_{-\pi}^\pi F(t) \frac{t}{\sin(t/2)}\left( \sin(Nt)\cos(t/2) + cos(Nt)\sin(t/2)\right) dt \\ &= \frac{1}{2\pi} \int_{-\pi}^\pi F(t)\frac{t\cos(t/2)}{\sin(t/2)} \sin(Nt) dt + \frac{1}{2\pi}\int_{-\pi}^\pi F(t)t\cos(Nt)dt. \end{align*}\]By Riemann-Lebesgue lemma, $S_N(f)(\theta_0) \to 0$ as $N\to \infty$.
$\therefore \lim_{N\to\infty}S_N(f)(\theta_0)=f(\theta_0)$.
\[\tag*{$\square$}\]Corollary 2.2 (Localization principle of Riemann)
Suppose $f$ and $g$ are two integrable functions defined on the circle, and for some $\theta_0$ there exists and open interval $I$ containing $\theta_0$ such that $f(\theta)=g(\theta)$ for all $\theta\in I$. Then $S_N(f)(\theta_0)- S_N(g)(\theta_0) \to 0$ as $N\to \infty$.
The function $f-g$ is 0 on $I$, so it is differentiable at $\theta_0$. By the previous theorem, $S_N(f-g)(\theta_0)\to (f-g)(\theta_0)$ as $N\to\infty$.
$\therefore \lim_{N\to\infty}S_N(f)(\theta_0) - S_N(g)(\theta_0) =0$.
\[\tag*{$\square$}\]Reference
- Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
- Math 139 Fourier Analysis Notes