# Fourier Transform - Duality formula. What are the necessary conditions?

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?

• Yes. But the Fourier inversion theorem for $f,\hat{f} \in L^1$ is what you need to prove the rest. – reuns Dec 7 '18 at 22:00
• 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? – zokomoko Dec 7 '18 at 22:16
• $\lim_{n \to \infty}\int\limits_{ - \infty }^\infty {\widehat f\left( \omega \right)e^{-\omega^2/n^2}e^{i\omega x}d\omega }$ 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 ... – reuns Dec 8 '18 at 20:30
• What are the domains $\mathbb{R},\mathbb{C} ?$ and range $\mathbb{R},\mathbb{C} ?$ of your $f$ ? – Jean Marie Dec 14 '18 at 18:34
• 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 ? – 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.