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I have studied the Fourier transform and the inverse Fourier transform for functions in $L^1(\mathbb{R})$, so in 1D. In my book, I see the following definition:

If $f \in L^1(\mathbb{R})$ and its Fourier transform $\hat{f} \in L^1(\mathbb{R})$, then the inverse Fourier transform of $\hat{f}(\omega)$ is defined by: $g(t) = \mathcal{F}^{-1}\{\hat{f}(\omega)\} = \dots $.

Additionally, $g=f$ for almost every $t \in \mathbb{R}$. If the original $f$ is continuous, then $g=f$ for every $t \in \mathbb{R}$.

I don't understand how that is possible. I imagine that a function in $f \in L^1(\mathbb{R})$ could be discontinuous almost everywhere. But by applying Fourier transform and inverse Fourier transform, I will receive a function $g$ that is continuous on $\mathbb{R}$.

By the proposition above, this would be a fatal issue.

What am I thinking wrong?

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  • $\begingroup$ Part of the assertion of the Riemann-Lebesgue lemma is that the Fourier transform of an $L^1$ function is continuous... $\endgroup$ – paul garrett Mar 5 '18 at 1:39
  • $\begingroup$ @paulgarrett exactly. But the original f that has been transformed could be discontinuous? So how does that fit together? $\endgroup$ – andreas Mar 5 '18 at 1:45
  • $\begingroup$ Yes, but... as in @PhoemueX' answer... $\endgroup$ – paul garrett Mar 5 '18 at 13:18
  • $\begingroup$ ... as in @AOrtiz' answer... $\endgroup$ – paul garrett Mar 5 '18 at 13:18
  • $\begingroup$ @paulgarrett yeah, my issue was that I couldn't believe that $\hat f$ makes the original $f$ continuous almost everywhere. Now it is all clear. $\endgroup$ – andreas Mar 5 '18 at 21:08
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You are absolutely right that there are functions $f\in L^1$ sucht that there is no continuous function $g$ with $f=g$ almost everywhere.

Nevertheless, the statement that you cite is true: If in addition to $f\in L^1$ you assume $\hat{f}\in L^1$, then $f=\mathcal{F}^{-1}\hat{f}$ almost everywhere, where the right hand side is continuous.

In particular, this shows that if $f$ is such that there is no continuous $g$ with $f=g$ almost everywhere, then you must have $\hat{f}\notin L^1$.

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  • $\begingroup$ Okay. So is it correct to say that only because of $\hat{f} \in L^1$ the original function $f$ MUST be almost everywhere continuous? Or in other words: In the definition there is a hidden "theorem" about continuity of $f$? $\endgroup$ – andreas Mar 5 '18 at 14:09
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    $\begingroup$ @andreas: Yes, precisely. $\endgroup$ – PhoemueX Mar 5 '18 at 18:45
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The proper statement of this is that if $f\in L^1$ and $\hat f \in L^1$, then $\mathcal F^{-1}(\hat f)(x) = f(x)$ for almost every $x$. Since the inverse Fourier transform of an $L^1$ function is continuous, $f$ can be modified on a set of measure zero so that it is continuous, but it need not be continuous, as you already noted.

If the original function $f$ is also continuous, then we in fact gain that $\mathcal F^{-1}(\hat f)(x) = f(x)$ for every $x$ because two continuous functions that agree almost everywhere must agree everywhere.

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The Fourier inversion theorem holds for all Schwartz functions (roughly speaking, smooth functions that decay quickly and whose derivatives all decay quickly). I think this is one of the regularity conditions f can obey.

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