I'm trying to prove that for p $\in$ (1,2] and for some $C_p$ > 0,
$$\sum_{n \in \mathbb Z} |\hat {f}(n)|^p |n|^{p-2} \leq C_p \|f\|^p_{L^p} \text{ for all } f \in L^p(\Bbb T)$$
Now we first note that if we take p $\in$ (1,2]
Then $|n|^{p-2} = \frac{1}{|n|^{2-p}} \leq 1$
Thus, $\sum_{n \in \mathbb Z} |\hat {f}(n)|^p |n|^{p-2} \leq \sum_{n \in \mathbb Z} |\hat {f}(n)|^p$ for all $f \in L^p(\Bbb T)$
For $p = 2$,
we know that $\|f\|^2_{L^2(\Bbb T)} = \sum_{n\in \Bbb Z} |\hat f(n)|^2$ for all $f \in L^2(\Bbb T)$
So $\sum_{n \in \mathbb Z} |\hat {f}(n)|^p |n|^{p-2} \leq C_p \|f\|^p_{L^{p}}$ for $p = 2$.
For $p = 1$
$\sum_{n \in \mathbb Z} |\hat {f}(n)|^p |n|^{p-2}$ with p = 1 is $\sum_{n \in \mathbb Z} \frac{|\hat {f}(n)^p|}{|n|}$ and according to this example here , there exists such examples where $\sum_{n \in \mathbb Z} |\hat{f}(n)|^p |n|^{p-2}= \infty$ for $p = 1$.
So how do I prove the cases for $p\in (1,2)$?