Let's say I have a sequence of functions $f_n\in L^1(\Bbb{R})$ such that

  1. $f_n\rightarrow f$ in $L^1(\Bbb{R})$, and
  2. $\hat{f}_n\in L^1(\Bbb{R})$ for all $n$ (where hat is the Fourier transform), and
  3. $\hat{f}_n\rightarrow g$ for some $g\in L^1(\Bbb{R})$.

Then it should be the case that $g = \hat{f}$, correct? Certainly it suffices to show that $\hat{f}_n\rightarrow \hat{f}$ in $L^1$. By the boundedness of $\mathcal{F}:L^1\rightarrow L^\infty$, we have trivially that

$$ \|\hat{f}_n-\hat{f}\|_\infty\leq\|f_n-f\|_1 $$ and so $\hat{f}_n\rightarrow \hat{f}$ pointwise almost everywhere (actually, everywhere since $f_n$ are continuous by virtue of $\hat{f}_n\in L^1$). From here I can think of a few complicated directions, but can you think of an elegant proof?

  • $\begingroup$ That's assuming $\hat{f}\in L^1(\mathbb{R})$ in the first place. If that is the case, then note that convergence of $\hat{f_n}$ to $g$ in $L^1$ implies the existence of a subsequence $\hat{f_{n_k}}$ converging a.e. to $g$, but that subsequence also converges a.e. to $\hat{f}$. $\endgroup$ – Joey Zou Jul 8 '16 at 6:04
  • $\begingroup$ OK, that's effectively what I was thinking but I was hoping for something more enlightening. $\endgroup$ – icurays1 Jul 8 '16 at 6:06
  • $\begingroup$ On second thought you don't need to assume $\hat{f}\in L^1$ either, since that would actually follow from $\hat{f} = g$ a.e. $\endgroup$ – Joey Zou Jul 8 '16 at 6:07

Let $g_n = \hat f_n$. We have that $g_n \to g$ in $L^1$, so by the proof that $L^1$ is a Banach space, there is a subsequence $(g_{{n_k}})$ converging to $g$ a.e. Now, $(g_{n_k})$ is a subsequence of $(g_n)$, so it must also converge to $\hat f$ a.e. as so does $(g_n)$. Thus, $g = \hat f$ a.e. By continuity, we get $g = \hat f$ on $\Bbb R$.


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