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?

  • $\begingroup$ 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

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 $\hat{f}\notin L^1$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @andreas: Yes, precisely. $\endgroup$ – PhoemueX Mar 5 '18 at 18:45

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.


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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