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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $\hat{f} \not \in L^1$.

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