Theorem 0.5 (Parseval for Continuous Functions)
Let \(f\) be a continuous \(2\pi\)-periodic function. Then \[ \lim_{N\to\infty} \int_0^{2\pi} |f(x)-S_Nf(x)|^2\,dx = 0, \] and \[ \sum_{n=-\infty}^{\infty} |\widehat{f}(n)|^2 = \|f\|^2. \]

Proof. We use the normalized norm

$$ \begin{align*} \|g\|_2^2=\frac{1}{2\pi}\int_0^{2\pi}|g(x)|^2\,dx. \end{align*} $$

Recall from chapter two that the Fejér kernels \(\{ F_N \}\) form a family of good kernel. If we write \(\sigma_N f=F_N*f\). If we are additionally given a periodic continuous function, then Fejér’s theorem gives uniform convergence \(\sigma_N f\to f\). Moreover Corollary 5.4 says

Corollary 5.4: Continuous functions on the circle can be uniformly approximated by trigonometric polynomials.

This is true because \(\sigma_N f\) is a trigonometric polynomial itself since it is a finite sum:

$$ \begin{align*} \sigma_N f(x)=\sum_{n=-N}^{N} \left(1-\frac{|n|}{N}\right)\widehat f(n)e^{inx}. \end{align*} $$

Uniform convergence means that for every \(\epsilon>0\), choosing \(N\) sufficiently large gives

$$ \begin{align*} \sup_x|f(x)-\sigma_N f(x)|<\epsilon. \end{align*} $$

So \(\sigma_N f\) itself is the trigonometric polynomial promised by Corollary 5.4. Since \(\sigma_N f\) contains frequencies with \(|n|\le N\), it is a trigonometric polynomial of degree at most \(N\). Since \(\sigma_N f\) has degree at most \(N\), it belongs to

$$ \begin{align*} W_N=\operatorname{span}\{e^{inx}:-N\le n\le N\}. \end{align*} $$

Recall that \(S_Nf\) is the orthogonal projection of \(f\) onto \(W_N\), so it is the closest element of \(W_N\) to \(f\) in the \(L^2\) norm. Therefore,

$$ \begin{align*} \boxed{\|f-S_Nf\|_2\le \|f-\sigma_Nf\|_2.} \end{align*} $$

Uniform convergence implies \(L^2\) convergence here:

$$ \begin{align*} \|f-\sigma_Nf\|_2\le \|f-\sigma_Nf\|_\infty\longrightarrow0. \end{align*} $$

Hence

$$ \begin{align*} 0\le \|f-S_Nf\|_2\le \|f-\sigma_Nf\|_2\longrightarrow0. \end{align*} $$

So the Fourier partial sums \(S_Nf\) converge to \(f\) in \(L^2\). Consequently,

$$ \begin{align*} \int_0^{2\pi}|f(x)-S_Nf(x)|^2\,dx =2\pi\|f-S_Nf\|_2^2\longrightarrow0, \end{align*} $$

which proves the first conclusion. Next, we use Lemma 0.3:

$$ \begin{align*} \|f\|_2^2 = \|f-S_Nf\|_2^2 +\sum_{n=-N}^{N}|\widehat f(n)|^2. \end{align*} $$

Since we just showed \(\|f-S_Nf\|_2\to0\), taking \(N\to\infty\) gives

$$ \begin{align*} \boxed{\|f\|_2^2=\sum_{n\in\mathbb Z}|\widehat f(n)|^2.} \end{align*} $$

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi