Let $1\leq p\leq q \leq \infty$ and let $f\in L^p([0,2\pi])$ such that $\hat{f}(k)=0$ for all $\vert k\vert > N$. I would like to show that
$$\Vert f \Vert_p \leq (2N-1)^{1/p-1/q} \Vert f \Vert_q.$$
I managed to show the case $p=\infty$, $q=2$ using Young's Inequality for $f= f * g$, where $g$ is a function such that $\hat{g}(k)=1$ for all $k= -N,\ldots, N$ and zero otherwise, and Parseval Identity. How can I obtain the general case?
Thank you very much for any help.
Your proof generalises with little effort to a large number of pairs $(p,q)$ but not all of them. Young's inequality gives $$ \| f * g \|_{L^p} \le \|f\|_{L^q} \|g\|_{L^r} $$ where $1+\frac1p = \frac1q +\frac1r$. The natural replacement for Parseval is (the dual inequality of) Hausdorff-Young, which works for $r\ge 2$ i.e. $\frac12 + \frac1p \le \frac1q$, giving $$ \|g\|_{L^r} = \|\mathcal F^{-1} \hat g\|_{L^r} \le \| \hat g\|_{\ell^{r'}}, \quad \frac1r + \frac1{r'}=1 \iff \frac1{r'} = \frac1q - \frac1p$$ which gives $$\|g\|_{L^r} \le (2N+1)^{1/r'} = (2N+1)^{1/q - 1/p} $$ since $\hat g$ is supported on the $2N+1$ integers $|n|\le N$ where it takes the value $1$. (PS I think your inequality is not quite right)
This proof doesn't work when $\frac1q-\frac1p < \frac12$, and I suspect the result isn't true with constant 1 in this case since all sources I look at do not phrase it (Bernstein's inequality on the torus, see this answer and the referenced "EBP lecture notes on singular stochastic PDEs" by Gubinelli and Perkowski, pages 37 and onwards) with a constant 1. If one is prepared to lose a constant, then the computation here also finishes the proof, since your $g$ is nothing but the Dirichlet kernel $g=\sum_{|k|\le N} e^{ikx}$.