function $f\in L^1(\mathbb R) \cup L^2(\mathbb R)$ with $\hat{f}=\sin$

Question

Is there a function $f\in L^1(\mathbb R) \cup L^2(\mathbb R)$ with $\hat{f}(x)=\sin(x)$? I think yes because the Fourier transform maps a $L^1$ function into $L^\infty (\mathbb R) \cap C^0(\mathbb R)$ and the function $\sin(x)$ is in $L^\infty (\mathbb R) \cap C^0(\mathbb R)$ Is this true?

Answer 1

In fact there is no such $f$ in $L^1+L^2$, a stronger result.

Prop. If $f\in L^1+L^2$ and $\delta>0$ then $\lim_{A\to\infty}\int_A^{A+\delta}\hat f=0$.

Proof: If $f\in L^1$ this follows from Riemann-Lebesgue, while if $f\in L^2$ then $\hat f\in L^2$. Hence for $f\in L^1+L^2$.

Answer 2

No, since $\mathcal{F}(\delta_a) = e^{-2i \pi a}$ in the sense of distributions, it follows that $$ \mathcal{F}^{-1}(\sin) = \tfrac{1}{2i} \mathcal{F}^{-1}(e^{ix}-e^{-ix}) = \frac{\delta_{-1/(2\pi)}-\delta_{1/(2\pi)}}{2i} $$ which is a measure but is not a locally measurable function.

Answer 3

If such a function $f$ had been in $L^1(\Bbb R)$, then by Riemann-Lebesgue lemma one would have $\lim_{x \to \pm \infty} \sin(x)=0$, a contradiction! If such a function $f$ had been in $L^2(\Bbb R)$, then by Plancherel theorem one would have $\int_{-\infty}^\infty |\sin(x)|^2 dx < \infty$, a contradiction!