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?
Best Answer
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.