Show that $\lim_{n\to\infty}\|f_*^{(n)}\|_1^{\frac{1}{n}}=\|\hat{f}\|_\infty$ Let $1<p\leq 2$ and $f\in L^p(\mathbb{T})$, i.e. $f$ is $p-$th power integrable and is $1-$periodic. Define
$$f_*^{(n)}=f*f*\dots*f\quad n\text{ times}$$
Show that
$$\lim_{n\to\infty}\|f_*^{(n)}\|_1^{\frac{1}{n}}=\|\hat{f}\|_\infty$$
where
$$\|\hat{f}\|_\infty=\max_{j\in\mathbb{Z}}|\hat{f}(j)|$$
My attempt:
By Hausdorff-Young inequality, we have
$$\|f_*^{(n)}\|_1\geq \|\widehat{f_*^{(n)}}\|_\infty=\|\hat{f}\|_\infty^n$$
Therefore
$$\liminf_{n\to\infty}\|f_*^{(n)}\|_1^{\frac{1}{n}}\geq\|\hat{f}\|_\infty$$
The difficult part is to prove that
$$\limsup_{n\to\infty}\|f_*^{(n)}\|_1^{\frac{1}{n}}\leq\|\hat{f}\|_\infty$$
which I have no idea how to deal with. Any suggestions for this? Thanks in advance.

To get your second inequality, do the following.
First, show that $$\lim_{n \to \infty} ||f^{(n)}_*||_2^{1/n} = ||\hat{f}||_\infty.$$
To do this, note that $$||f^{(n)}_*||_2^{1/n} = ||(\hat{f})^n||_2^{1/n} = ||\hat{f}^2||_n^{1/2} \to ||\hat{f}^2||_\infty^{1/2}$$
as $n \to \infty$. We've used norm equivalence in $L^2$, properties of transforms of convolutions, and the fact that $||g||_n \to ||g||_\infty$ as $n \to \infty$.
Now can you show the $L^2$ norm controls the $L^1$ norm (i.e., find the useful inequality here to finish)?
