Fourier transform of $f=e^{inx}$ on $L^2[-\pi,\pi]$ for $1\leq p\leq2$

Question

If $f\in L^2[-\pi,\pi]$, let $\hat f$ be the Fourier transform of $f$: $$\hat f=\frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-i\xi x}dx \, , \quad \xi\in\mathbb R\, . $$ If $f=e^{inx}$, then $$\hat f=\frac{1}{2\pi} \int_{-\pi}^\pi e^{i(n-\xi)x}dx=\frac{e^{i(n-\xi)\pi}-e^{-i(n-\xi)\pi}}{2\pi i (n-\xi)}=\frac{\sin [(n-\xi)\pi] }{\pi (n-\xi)}\, . \quad \quad (1)$$ Moreover, $f\in L^p[-\pi,\pi]$ for $1\leq p\leq 2$. Does result (1) also apply to the $L^p$ case ($1\leq p\leq 2$)?

Answer 1

Since in the compact setting the $L^p$ spaces are nested $$ L^\infty \subset L^q \subset L^2 \subset L^p \subset L^1 \;\;\;\;(\text{for}\; 1 \le p \le 2 \le q \le \infty)$$ the Fourier-Transform of any $L^p$-function corresponds to its Fourier-Transform of it interpreted as an $L^1$-function. So the answer to your question is yes.

More generally, "the Fourier-Transforms" of a function that lives in at least two $L^p$-spaces coincide, since they are both usually defined as the limit of Fourier-Transforms of functions also present in $L^1$.