The Fourier transform of an $L^1$ function is defined by $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$
Is it true that for functions $f,g\in L^1$, if $\hat{f}=\hat{g}$, then $f=g$?
Real Analysis – Uniqueness of Fourier Transform in L^1
fourier analysisreal-analysis
Related Solutions
Hint: If $\hat{f}\in L^1$, then by Fourier Inversion, $f$ is continuous.
Let us, to distinguish between the two Fourier transforms, denote the one by $\mathscr{F}_1$,
$$\mathscr{F}_1(f)(y) = \int_\mathbb{R} f(x)e^{-ixy}\,dx\tag{1}$$
for $f\in L^1(\mathbb{R})$, and the other by $\mathscr{F}_2$,
$$\mathscr{F}_2(g) = \lim_{n\to\infty} \mathscr{F}_1(g_n)\tag{2}$$
for $g\in L^2(\mathbb{R})$ where $g_n$ is a sequence of functions in $S$ that converges to $g$ in $L^2$, and the limit in $(2)$ is the $L^2$-limit.
For $f \in L^1 \cap L^2$, find a sequence $(g_n)$ in $S$ such that not only $\lVert g_n - f\rVert_2 \to 0$ but also $\lVert g_n - f\rVert_1 \to 0$.(1)
Then you can conclude
$$\left\lVert \mathscr{F}_1(g_n) - \mathscr{F}_1(f)\right\rVert_\infty \leqslant \lVert g_n - f\rVert_1 \to 0,$$
i.e. the Fourier transforms of the $g_n$ converge uniformly to the $L^1$ Fourier transform of $f$, and by definition you have
$$\mathscr{F}_1(g_n) \xrightarrow{L^2} \mathscr{F}_2(f),$$
and there is a subsequence of the $\mathscr{F}_1(g_n)$ that converges pointwise almost everywhere to $\mathscr{F}_2(f)$ (well, we know it converges uniformly, so the entire sequence converges pointwise), so you have $\mathscr{F}_1(f) = \mathscr{F}_2(f)$ almost everywhere (and you can choose $\mathscr{F}_1(f)$ as a representative, so then you have equality everywhere). In that sense, the two definitions coincide, for $f \in L^1\cap L^2$, the $L^1$-Fourier transform of $f$ is a representative of the $L^2$-Fourier transform of $f$.
If you choose a sequence $(g_n)$ in $S$ such that $\lVert g_n - f\rVert_2 \to 0$, but not $\lVert g_n -f \rVert_1 \to 0$, then by the above you have convergence $\hat{g}_n \xrightarrow{L^2} \hat{f}$, and that implies that for a subsequence $g_{n_k}$, you have pointwise convergence $\hat{g}_{n_k}(y) \to \hat{f}(y)$ almost everywhere, but there is no guarantee (known to me) that you have pointwise a.e. convergence for the full sequence, nor that you have pointwise convergence everywhere for any subsequence. But it would be difficult at least, I think, to come up with a concrete example of a function $f\in L^1\cap L^2$ and a sequence $g_n$ of Schwartz functions converging to $f$ in $L^2$ such that you don't have pointwise convergence almost everywhere or even everywhere for the full sequence.
(1) That is always possible, let $f_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x)$. Then $f_m \to f$ both in $L^1$ and $L^2$. $f_m$ is a bounded function with compact support, so convolving it with a compactly supported approximation of the identity produces compactly supported smooth functions converging to $f_m$ in both $L^1$ and $L^2$.
Best Answer
$$ \int_{\mathbb R} e^{-a|x|^2/2 + i x x_0} \hat f(x) \, dx = \int_{\mathbb R} f(y) \int_{\mathbb R} \frac1{2\pi} e^{-a|x|^2/2} e^{-ix(y-x_0)} \, dx \, dy = \int_{\mathbb R} \frac1{\sqrt{2 \pi a}} e^{-|y-x_0|^2/2a} f(y) \, dy .$$ The first equality is by substitution and Fubini. The second uses standard tables of Fourier transforms. The last quantity converges to $f(x_0)$ in $L_1$ as $a\to 0^+$ (see first that it converges if $f$ is a continuous and compactly supported, then use the usual tricks of approximating $f$ by a such a function). I might be off by a factor of $2\pi$.