Let the function $f(x)=x$ if $x\in[0,1]$ and $f(x)=0$ otherwise. The function is in $L_2$.
Unless mistaken, the Fourier transform of f is $\hat f(\xi)=\int_0^1{xe^{ix\xi}dx}=e^{i\xi}(1/\xi^2-i/\xi) - 1/\xi^2$.
$\hat f$ explodes around zero as $\xi$ appears in the denominator. I thought that Fourier transform was an isometry on $L_2$ (using Parseval's theorem), but $\hat f$ does not seem to be square-integrable.
Can you please explain to me what I am missing?
The Fourier transform of $f(x)=x\cdot\mathbb{1}_{(0,1)}(x)$ is given by:
$$\widehat{f}(\xi)=\int_{0}^{1} x\, e^{-i x \xi}\,dx=\frac{(1+i\xi)e^{-i\xi}-1}{\xi^2} $$ hence: $$\widehat{f}(\xi)\cdot\overline{\widehat{f}(\xi)}=\frac{2+\xi^2-2\cos\xi-2\xi\sin\xi}{\xi^4}=\frac{1}{4}-\frac{\xi^2}{72}+O(\xi^4)\quad\left(\xi\to 0\right)$$ and that function is clearly integrable.