For a function \(f\in L^1[0,2\pi]\), its Fourier coefficient at frequency \(n\in\mathbb{Z}\) is

$$ \widehat{f}(n) = \frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx. $$

Now define the Fourier operator by

$$ \mathcal{F}f=\bigl(\widehat{f}(n)\bigr)_{n\in\mathbb{Z}}. $$

It takes a function and returns its sequence of Fourier coefficients. This operator is linear by linearity of the integral. In the next four examples, we studying different spaces and norms for this operator.

Example 1: Continuous Functions and Square-Summable Sequences

Give \(C[0,2\pi]\) the supremum norm and define the Fourier coefficient operator by

$$ \begin{aligned} \mathcal{F}\colon C[0,2\pi]\to\ell^2(\mathbb{Z}) \end{aligned} $$

For every \(f\in C[0,2\pi]\), if we use the definition of the squared \(\ell^2\) norm, then

$$ \begin{aligned} \|\mathcal{F}f\|_{\ell^2}^2 &= \sum_{n\in\mathbb{Z}}|\widehat{f}(n)|^2 \end{aligned} $$

Recall Parseval’s identity:

$$ \begin{aligned} \sum_{n\in\mathbb{Z}}|\widehat{f}(n)|^2 &= \frac{1}{2\pi}\int_0^{2\pi}|f(x)|^2\,dx \end{aligned} $$

Therefore,

$$ \begin{aligned} \|\mathcal{F}f\|_{\ell^2}^2 &= \frac{1}{2\pi}\int_0^{2\pi}|f(x)|^2\,dx \end{aligned} $$

Finally, Since \(|f|\) is continuous on the compact interval \([0,2\pi]\), it attains a finite maximum there. Thus \(\|f\|_\infty\) is finite, and \(|f(x)|\le\|f\|_\infty\) for every \(x\in[0,2\pi]\).

$$ \begin{aligned} \|\mathcal{F}f\|_{\ell^2}^2 &\le \frac{1}{2\pi}\int_0^{2\pi}\|f\|_\infty^2\,dx = \|f\|_\infty^2. \end{aligned} $$

Taking square roots gives \(\|\mathcal{F}f\|_{\ell^2}\le\|f\|_\infty\) for every \(f\in C[0,2\pi]\). Hence \(\mathcal{F}\) is bounded with constant \(1\).

2. Square-Integrable Functions and Square-Summable Sequences

Now consider the same Fourier coefficient operator with domain \(L^2[0,2\pi]\), equipped with the normalized \(L^2\) norm:

$$ \begin{align*} \mathcal{F}\colon L^2[0,2\pi] &\to \ell^2(\mathbb{Z}), \\ \mathcal{F}f &= \bigl(\widehat{f}(n)\bigr)_{n\in\mathbb{Z}}. \end{align*} $$

For every \(f\in L^2[0,2\pi]\), the definition of the squared \(\ell^2\) norm gives

$$ \|\mathcal{F}f\|_{\ell^2}^2 = \sum_{n\in\mathbb{Z}}|\widehat{f}(n)|^2. $$

By Parseval’s identity,

$$ \sum_{n\in\mathbb{Z}}|\widehat{f}(n)|^2 = \frac{1}{2\pi}\int_0^{2\pi}|f(x)|^2\,dx. $$

The integral on the right, including the factor \(1/(2\pi)\), is precisely the definition of the squared normalized \(L^2\) norm of \(f\). Therefore,

$$ \|\mathcal{F}f\|_{\ell^2}^2 = \|f\|_{L^2}^2. $$

Taking square roots gives \(\|\mathcal{F}f\|_{\ell^2}=\|f\|_{L^2}\) for every \(f\in L^2[0,2\pi]\). Thus \(\mathcal{F}\) preserves the norm and is bounded with constant \(1\).

3. Integrable Functions and Bounded Sequences

Now consider the same Fourier coefficient operator with domain \(L^1[0,2\pi]\), equipped with the normalized \(L^1\) norm:

$$ \begin{align*} \mathcal{F}\colon L^1[0,2\pi] &\to \ell^\infty(\mathbb{Z}), \\ \mathcal{F}f &= \bigl(\widehat{f}(n)\bigr)_{n\in\mathbb{Z}}. \end{align*} $$

Let \(f\in L^1[0,2\pi]\) and fix \(n\in\mathbb{Z}\). By the definition of the Fourier coefficient and the triangle inequality for integrals,

$$ \begin{aligned} |\widehat{f}(n)| &= \frac{1}{2\pi}\left|\int_0^{2\pi}f(x)e^{-inx}\,dx\right| \\ &\le \frac{1}{2\pi}\int_0^{2\pi}|f(x)|\,|e^{-inx}|\,dx. \end{aligned} $$

Since \(|e^{-inx}|=1\), the right-hand side is exactly the normalized \(L^1\) norm of \(f\). Thus

$$ |\widehat{f}(n)| \le \frac{1}{2\pi}\int_0^{2\pi}|f(x)|\,dx = \|f\|_{L^1}. $$

The same bound holds for every \(n\in\mathbb{Z}\). Taking the supremum over \(n\) gives the bound on the entire sequence:

$$ \|\mathcal{F}f\|_{\ell^\infty} = \sup_{n\in\mathbb{Z}}|\widehat{f}(n)| \le \|f\|_{L^1}. $$

This inequality holds for every \(f\in L^1[0,2\pi]\). Hence \(\mathcal{F}\colon L^1[0,2\pi]\to\ell^\infty(\mathbb{Z})\) is bounded with constant \(1\).

4. Fourier Coefficients Tend to Zero

Riemann–Lebesgue Lemma
If \(f\in L^1[0,2\pi]\), then \(\widehat{f}(n)\to 0\) as \(|n|\to\infty\).

The lemma places the Fourier coefficient sequence in the smaller space

$$ c_0(\mathbb{Z}) = \left\{ (a_n)_{n\in\mathbb{Z}}\in\ell^\infty(\mathbb{Z}) \,\middle|\, \lim_{|n|\to\infty}a_n=0 \right\}. $$

Since \(c_0(\mathbb{Z})\) carries the same supremum norm as \(\ell^\infty(\mathbb{Z})\), the preceding estimate still applies. Thus \(\mathcal{F}\colon L^1[0,2\pi]\to c_0(\mathbb{Z})\) is bounded.

References

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