If a $2\pi$-periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue integrable in $[-\pi,\pi]$, and the series $\frac{a_0}{2}+\sum_{n=1}^\infty [a_n \cos{nx}+b_n \sin{nx}] $, where $(a_n), (b_n)$ are some real sequences, is convergent to $f$ uniformly or in $L_p$ norm or pointwise, then it is known that $a_n$, $b_n$ are Fourier coefficients of $f$. (The last part of this statement it is du Bois-Reymond theorem).

My question concerns analogue of such type theorems for Fourier integrals. Namely:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue integrable and let

$f(x)=\lim_{T\rightarrow \infty} \int_0^T [a(\omega) \cos (\omega x)+b(\omega )\sin(\omega x)]d\omega$ for $x \in \mathbb{R}$.

Under what conditions:

$a(\omega)=\frac{1}{\pi} \int_{-\infty}^\infty f(x) \cos(\omega x)dx$,

$b(\omega)=\frac{1}{\pi} \int_{-\infty}^\infty f(x) \sin(\omega x)dx$ ?



Excuse my different conventions, but working with $\sin$ and $\cos$ is tedious. Here $$ \hat{f}(k) := \int_{\mathbb{R}} dx \, e^{-ikx} \, f(x). $$

We have the Riemann-Lebesgue Lemma (Reed & Simon: Methods of Modern Mathematical Physics Vol II, Theorem IX.7):

The Fourier transform [on the Schwartz space $\mathcal{S}$ of functions of rapid decay] extends uniquely to a continuous map from $L^{1}$ into (NOT necessarily onto) $C_{\infty}$ the continuous functions vanishing at infinity.

Sketch of the proof:

For $f \in \mathcal{S}$ $$ \|\hat f\|_{\infty} \leq \|f\|_{1} $$ is obvious and $\mathscr{S}$ is dense in $L^{1}$.

Thus a necessary condition is $a(k) \in C_{\infty}$.

From now on let $a(k) \in C_{\infty}$ and $$ f (x) = \mathcal{S}'-\lim_{T \rightarrow \infty} \int_{-T}^{T} \frac{dk}{2 \pi} \, e^{ikx} \, a(k), $$ i.e. I assume weak convergence as a tempered distribution, if you want to: $L^{1}$ limit implies that (you should specify the convergence in your question).

Then for all test functions $\varphi \in \mathcal{S}$ we have $$ \int_{\mathbb{R}} dx f(x) \hat{\varphi}(x) = \lim_{T} \int_{\mathbb{R}} dx \left\{ \int_{-T}^{+T} \frac{dk}{2 \pi} \, e^{ikx} \, a(k) \right\} \hat{\varphi}(x). $$ The regularity of $a$ and $\varphi$ allows us to apply Fubini's theorem and take the limit: $$ \int f \hat{\varphi} = \int a \varphi. $$ Thus $a$ coincides with the Fourier transform of $f$ as a tempered distribution and also on $L^1$. By the Riemann Lebesgue lemma we have as desired $$ a(k) = \int f(x) e^{-ikx}. $$


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