If $f,g\in L^1(\mathbb{R})$, it is not hard to show by definition that $$(\hat{f\ast g)}(t)=\hat{f}(t)\hat{g}(t).$$ But what about if $f,g\in L^2(\mathbb{R})$? The Fourier transform on $L^2(\mathbb{R})$ is defined in as an extension of the Fourier transform in the Schwartz class. Then it's hard to work with the definition…
Fourier Analysis – Fourier Transform of Convolution for L² Functions
convolutionfourier analysislp-spaces
Related Solutions
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
For $f,g\in L^2$, the convolution $f*g$ belongs to $C_0(\mathbb R)$. Taking the Fourier transform of a $C_0$ function is problematic: such transforms are generally not functions, but merely distributions (as Jose27 points out).
It is better to work from right to left. The product $\hat f \hat g$ is in $L^1$. Apply the inverse Fourier transform to it (essentially same as Fourier transform, but with opposite sign in the exponential, and maybe different normalization). The result is a $C_0$ function. You want to show that this function is $f*g$.
Argue by density. For $f,g\in \mathcal S$ the result is known. The map $(f,g)\mapsto f*g$ is a continuous map from $L^2\times L^2$ into $C_0$. Also, $(f,g)\mapsto \hat f \hat g$ is continuous from $L^2\times L^2$ to $L^1$. The inverse Fourier transform is continuous from $L^1$ to $C_0$. The conclusion follows.
In the preceding paragraph, $C_0$ can be replaced by $L^\infty$ throughout (the norm is the same), saving you the trouble of proving that $f*g$ vanishes at infinity.