I've come across two contradicting statements which I'd be glad if you could help me resolve:

Theorem: if $f\left( x \right)$ is continuous and absolutely integrable ($\int\limits_{ - \infty }^\infty {\left| {f(x)} \right|dx} < \infty $) and suppose $\widehat f\left( \omega \right) = \int\limits_{ - \infty }^\infty {f\left( x \right){e^{ - i\omega x}}dx} $ is absolutely integrable ($\int\limits_{ - \infty }^{ - \infty } {{{\left| {\widehat f\left( \omega \right)} \right|}}d\omega } < \infty $) then $$F\left\{ {F\left\{ {f\left( x \right)} \right\}} \right\} = 2\pi f\left( { - x} \right)$$ Just to clarify notation: $F\left\{ {f\left( x \right)} \right\} = \widehat f\left( \omega \right)$.

So this theorem assumes $f(x)$ is continuous on the real line.

But I really think we can demand less, that $f(x)$ be only piecewise continuous and get the continuity of $f(x)$ as a result of this theorem.

My reasoning:

if $f(x)$ is piecewise continuous and absolutely integrable then we know its fourier transform is continuous.

By the same thinking, since $F\left\{ {f\left( x \right)} \right\}$ is continuous and absolutely integrable then its fourier transform, $2\pi f\left( { - x} \right)$, is also continuous.

Am I correct?

  • $\begingroup$ Yes. But the Fourier inversion theorem for $f,\hat{f} \in L^1$ is what you need to prove the rest. $\endgroup$ – reuns Dec 7 '18 at 22:00
  • $\begingroup$ You mean that I first need to prove that if $f,\widehat f \in {L^1}$ then $f\left( x \right) = \int\limits_{ - \infty }^\infty {\widehat f\left( \omega \right){e^{i\omega x}}d\omega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense? $\endgroup$ – zokomoko Dec 7 '18 at 22:16
  • $\begingroup$ $\lim_{n \to \infty}\int\limits_{ - \infty }^\infty {\widehat f\left( \omega \right)e^{-\omega^2/n^2}e^{i\omega x}d\omega }$ converges to $2\pi f$ in quite all the normed space where $f$ belongs to. When $\widehat{f} \in L^1$ we can easily link it to $\lim_{n \to \infty}\int\limits_{ -n}^n {\widehat f\left( \omega \right)e^{i\omega x}d\omega }$ and obtain uniform convergence, $L^2$ convergence ... $\endgroup$ – reuns Dec 8 '18 at 20:30
  • $\begingroup$ What are the domains $\mathbb{R},\mathbb{C} ?$ and range $\mathbb{R},\mathbb{C} ?$ of your $f$ ? $\endgroup$ – Jean Marie Dec 14 '18 at 18:34
  • $\begingroup$ A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ? $\endgroup$ – Jean Marie Dec 14 '18 at 23:17

Suppose $f(x)= 0$ for $x\ne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $\mathbb R.$ Since $f=0$ a.e., we have $F[f]\equiv 0,$ and hence $F[F[f]](x)\equiv 0.$ Thus $F[F[f]](x)$ does not equal $2\pi f(-x)$ everywhere.

If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.

But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:\mathbb R\to \mathbb R$ is continuous on $(-\infty,0)\cup(0,\infty)$ and $f\in L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.

Claim: If $F[f]\in L^1,$ then $f$ has a removable singularity at $0.$

Proof: The well known inversion theorem for $L^1$ shows

$$f(x)= -\frac{F[F[f])(-x)}{2\pi}$$

for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-\infty,0).$ But two continuous functions that agree a.e. on $(-\infty,0)$ actually agree everywhere on $(-\infty,0).$ (Nice exercise) The same holds on $(0,\infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.

The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.


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