For $ {\lambda} > 0$, let
\begin{equation}{h}_{{\lambda}} \left(x\right) = \sqrt{\frac{2}{{\pi}}} \frac{{\lambda}}{{{\lambda}}^{2}+{x}^{2}} , \qquad {H}_{{\lambda}} \left(t\right) = {e}^{{-{\lambda}} \left|t\right|}\end{equation}
as in Rudin's paragraph 9.7. It is easy to check that
\begin{equation}{\hat{h}}_{{\lambda}} = {H}_{{\lambda}} , \qquad {\hat{H}}_{{\lambda}} = {h}_{{\lambda}}\end{equation}
Now we have the following
Lemma A: if $ g \in {L}^{1} \cap {L}^{2}$ and $ {g}_{{\lambda}} = g \ast {h}_{{\lambda}}$ then $ {g}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$ and
$ {\hat{g}}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$.
Indeed, by theorem 9.10 in Rudin we have $ {g}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$ (see also exercise 8.4)
and by theorem 9.2, we have $ {\hat{g}}_{{\lambda}} = \hat{g} {H}_{{\lambda}}$, but
$ \hat{g} \in {C}_{0} \subset {L}^{\infty }$ (theorem 9.6) and
$ {H}_{{\lambda}} \in {L}^{1} \cap {L}^{2}$.
Lemma B:
The set
$ X = \left\{g \ast {h}_{{\lambda}} \ \mid \ g \in {L}^{1} \cap {L}^{2} , {\lambda} > 0\right\}$ is dense in $ {L}^{2}$.
Indeed, let $ f \in {L}^{2}$ and $ {\varepsilon} > 0$.
There exists $ A > 0$ such that $ {\left\|{f}_{A}-f\right\|}_{{L}^{2}} < {\varepsilon}$
where $ {f}_{A} = f\, \mathbf{1}_{\left[{-A} , A\right]}$. We have
$ {f}_{A} \in {L}^{1} \cap {L}^{2}$ and by theorem 9.10, there exists some
$ {\lambda} > 0$ such that $ {\left\|{f}_{A} \ast {h}_{{\lambda}}-{f}_{A}\right\|}_{{L}^{2}} < {\varepsilon}$. Hence $ {\left\|{f}_{A} \ast {h}_{{\lambda}}-f\right\|}_{{L}^{2}} < 2 {\varepsilon}$ and $ {f}_{A} \ast {h}_{{\lambda}} \in X$.
Let $ {\Phi}$ be the Fourier transform defined on $ {L}^{1}$ and also on $ {L}^{2}$ by the density argument given by Rudin in the
proof of parts a) b) c) of theorem 9.13, and let $ \mathcal{R}$ be
the mapping $ \mathcal{R} \left(f\right) \left(x\right) = f \left({-x}\right)$.
It is clear that $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$ for every function $ f \in X$ because by lemma A, these functions satisfy
the conditions of theorem 9.11. As $ \mathcal{R} {\Phi} {\Phi}$ is an isometry
of $ {L}^{2}$, the relation $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$ extends to
all $ f \in {L}^{2}$ by continuity.
Now in the proof of part d) of theorem 9.13 we have that
$ {\left\|{{\psi}}_{A}- \mathcal{R} {\Phi} {\Phi} \left(f\right)\right\|}_{{L}^{2}} \rightarrow 0$, which proves the result
since $ \mathcal{R} {\Phi} {\Phi} \left(f\right) = f$.
Best Answer
If $a,b:\mathbb R\to\mathbb C$ are functions in $L^2$ and $a\,\bar b=0$, then this occurs pointwise a.e. If now $a,b\in\mathbb C$ and $a\bar b=0$, then either $a=0$ or $b=0$, so $ab=0$.