I am trying to prove that the Fourier Transform of $f=\chi_{[-1,1]}$ is not in $L^1[\mathbb{R}]$.

I computed $\hat{f}$ and got $\hat{f}(w)=\sqrt{\frac{2}{\pi}}\frac{\sin w}{w}$.

My problem with this is that $\hat{f}$ is not even defined at 0. We certainly define it to be 1 at 0 in order to obtain continuity, but this is convenient, not mandatory. And we should not integrate over a set that supersedes the domain of $\hat{f}$, I believe.

And a related question is: where does the Fourier transform map to? In this, case, since $f\in L^1[\mathbb{R}]$ and $f\in L^2[\mathbb{R}]$, then $f\in L^2[\mathbb{R}]$. Can we say anything else?

If I was not clear, please let me know.

  • $\begingroup$ What definition are you using for $\hat f$? (what's the normalization?) $\endgroup$ – user384138 May 7 '17 at 12:35
  • $\begingroup$ @OpenBall Probably you're wondering if there isn't a $\sqrt{\frac{2}{\pi}}$ missing. There is! I will edit. $\endgroup$ – Soap May 7 '17 at 14:20
  • 1
    $\begingroup$ Then now I'm okay. $\hat f$ is indeed defined at $0$. We have: $$\hat f(w) = \frac1{\sqrt{2\pi}} \int_{\mathbb R} f(x)e^{-iwx}dx$$ When you integrated to obtain your $\hat f$, you implicitly assumed $w \neq 0$. Now if $w = 0$, then replacing $w$ in the original expression of $\hat f$, we get: $$\frac1{\sqrt{2\pi}} \int_{\mathbb R} f(x) dx$$ Which gives $\hat f(0) = \sqrt{\frac2{\pi}}$. $\endgroup$ – user384138 May 7 '17 at 14:23

Of course $\hat{f}(0) = \int_{-\infty}^\infty f(x)dx = 2$ is well-defined, and $\hat{f}(\omega) = 2 \frac{\sin \omega}{\omega}$.

The Fourier transform is a bounded (continuous) operator $L^1 \to C_0^0$ (where $C_0^0$ are the uniformly continuous functions vanishing at $\infty$ with the $\sup$ norm) and $L^2 \to L^2$. Since $f \in L^1 \cap L^2$ then $\hat{f} \in C_0^0 \cap L^2$.

Conversely, if $\hat{f} \in L^1$ then $\check {\hat{f}}=f \in C_0^0$. But here $f = \chi_{[-1,1]} \not \in C_0^0$ thus $\hat{f} \not \in L^1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.