The Fourier coefficient map sends a function to its sequence of Fourier coefficients. Which sequences arise this way? We first answer this question for \(L^2(\mathbb{T})\) and then for \(L^1(\mathbb{T})\). Throughout, we use normalized measure \(dx/(2\pi)\) on the circle, so

$$ \mathcal{F}f=\bigl(\widehat f(n)\bigr)_{n\in\mathbb{Z}}, \qquad \widehat f(n)=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}\,dx. $$
Definition — Isometric Isomorphism
Two normed spaces \((X,\|\cdot\|_X)\) and \((Y,\|\cdot\|_Y)\) are isometrically isomorphic if there is a map \(T\colon X\to Y\) satisfying all three properties:
  1. \(T\) is linear.
  2. \(T\) is bijective.
  3. \(\|Tx\|_Y=\|x\|_X\) for every \(x\in X\).
Such a map \(T\) is called an isometric isomorphism.

The Fourier Coefficient Map on \(L^2(\mathbb{T})\)

Theorem
The Fourier coefficient map \(\mathcal{F}\colon L^2(\mathbb{T})\to\ell^2(\mathbb{Z})\) is an isometric isomorphism.

Proof

Linearity follows from linearity of the integral defining each Fourier coefficient. To see that the map preserves the norm, apply Parseval’s identity:

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

Taking square roots gives \(\|\mathcal{F}f\|_{\ell^2}=\|f\|_{L^2}\). This also proves injectivity. If \(\mathcal{F}f=0\), then \(\|f\|_{L^2}=0\), so \(f=0\) almost everywhere. Thus the kernel of \(\mathcal{F}\) consists only of the zero element of \(L^2(\mathbb{T})\).

To prove surjectivity, start with an arbitrary sequence \((c_n)_{n\in\mathbb{Z}}\in\ell^2(\mathbb{Z})\). We want a function \(f\in L^2(\mathbb{T})\) whose Fourier coefficients are exactly the numbers \(c_n\). Form the finite sums

$$ p_N(x)=\sum_{n=-N}^{N}c_ne^{inx}. $$

For \(M>N\), orthonormality of the exponentials gives

$$ \|p_M-p_N\|_{L^2}^2 =\left\|\sum_{N\lt |n|\le M}c_ne^{inx}\right\|_{L^2}^2 =\sum_{N\lt |n|\le M}|c_n|^2. $$

Since \(\sum_{n\in\mathbb{Z}}|c_n|^2\) converges, its tails tend to zero. Thus \((p_N)\) is Cauchy in \(L^2(\mathbb{T})\). By completeness, there exists \(f\in L^2(\mathbb{T})\) such that \(p_N\to f\) in the \(L^2\) norm.

Fix \(m\in\mathbb{Z}\). For every \(N\ge |m|\), we have \(\widehat{p_N}(m)=c_m\). By Cauchy–Schwarz and the fact that \(\|e^{imx}\|_{L^2}=1\),

$$ |\widehat f(m)-c_m| =|\langle f-p_N,e^{imx}\rangle| \le\|f-p_N\|_{L^2}\to 0. $$

Hence \(\widehat f(m)=c_m\) for every \(m\in\mathbb{Z}\). This proves surjectivity and completes the proof.

The Fourier Coefficient Map on \(L^1(\mathbb{T})\)

Since the circle has normalized measure \(1\), Cauchy–Schwarz gives \(\|f\|_{L^1}\le\|f\|_{L^2}\). Thus \(L^2(\mathbb{T})\subseteq L^1(\mathbb{T})\). For an arbitrary \(f\in L^1(\mathbb{T})\), the triangle inequality gives

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

The Riemann–Lebesgue lemma also gives \(\widehat f(n)\to 0\) as \(|n|\to\infty\). Consequently, the coefficient sequence belongs to \(c_0(\mathbb{Z})\), the space of two-sided sequences tending to zero. Taking the supremum over \(n\) gives

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

Thus \(\mathcal{F}\colon L^1(\mathbb{T})\to c_0(\mathbb{Z})\) is bounded. Is it surjective? In other words, does every sequence tending to zero arise as the Fourier coefficient sequence of some integrable function?

Application — Which Sequences Are Fourier Coefficients?

Theorem
The Fourier coefficient map \(\mathcal{F}\colon L^1(\mathbb{T})\to c_0(\mathbb{Z})\) is not surjective.

Proof

Suppose \(\mathcal{F}\) were surjective. By the uniqueness theorem for Fourier coefficients, if \(f\in L^1(\mathbb{T})\) and all its Fourier coefficients vanish, then \(f=0\) almost everywhere. Thus \(\mathcal{F}\) is injective as well. Under our assumption, it would therefore be a bounded linear bijection between the Banach spaces \(L^1(\mathbb{T})\) and \(c_0(\mathbb{Z})\).

The Bounded Inverse Theorem would then give a constant \(A>0\) such that

$$ A\|f\|_{L^1} \le\|\mathcal{F}f\|_\infty =\sup_{n\in\mathbb{Z}}|\widehat f(n)| \qquad\text{for every }f\in L^1(\mathbb{T}). $$

Test this inequality on the Dirichlet kernel \(D_N(x)=\sum_{k=-N}^{N}e^{ikx}\). Reading off its Fourier coefficients gives

$$ \widehat{D_N}(n)= \begin{cases} 1, & |n|\le N,\\ 0, & |n|>N. \end{cases} $$

Thus \(\|\mathcal{F}D_N\|_\infty=1\) for every \(N\). The proposed lower bound would imply

$$ A\|D_N\|_{L^1}\le 1 \qquad\text{for every }N. $$

But we already know that \(\|D_N\|_{L^1}\to\infty\). Since \(A\) is a fixed positive constant, the inequality fails for sufficiently large \(N\). This contradiction proves that some sequences in \(c_0(\mathbb{Z})\) are not Fourier coefficient sequences of integrable functions.

Example 2: Two Complete Norms on the Same Vector Space

Suppose both \((X,\|\cdot\|_1)\) and \((X,\|\cdot\|_2)\) are Banach spaces, and suppose there exists \(B>0\) such that \(\|x\|_2\le B\|x\|_1\) for every \(x\in X\). To obtain the inequality in the other direction, consider the identity map

$$ I\colon (X,\|\cdot\|_1)\to(X,\|\cdot\|_2), \qquad I(x)=x. $$

This map is a linear bijection. The assumed inequality says exactly that \(I\) is bounded. By the Bounded Inverse Theorem, \(I^{-1}\) is also bounded, so there exists \(C>0\) such that \(\|x\|_1\le C\|x\|_2\) for every \(x\in X\). Taking \(A=1/C\) gives

$$ A\|x\|_1\le\|x\|_2\le B\|x\|_1 \qquad\text{for every }x\in X. $$

Hence the two norms are equivalent.

References

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