I'm studying the theory of Bessel potentials $J_\alpha$ and I'm reading the proof of their strong(p,p) regularity:
(Theorem) Let $\alpha>0$. Then, $J_\alpha : L^p\rightarrow L^p$ for all $p\in[1,+\infty]$ and $||J_\alpha||_{op}=1$.
where $J_\alpha(f)=G_\alpha\ast f$ (initially for all $f$ Schwartz function) and $G_\alpha(x)$ as the inverse Fourier transform of the tempered distribution induced by $(1+4\pi^2|\cdot|^2)^{-\alpha/2}$.
The only problem I got is that they justify the fact that $||J_\alpha||_{op}=1$ with the only fact that "$G_\alpha>0$ for $\alpha>0$" (and I know this is true). But I can't see how these two thing are related.
We know that for $\alpha>0$, $G_\alpha\in C^\infty(R^n\setminus\{0\})$ and \begin{equation} G_\alpha(x)\leq C_\alpha e^{-|x|/2} \end{equation} for $|x|\geq1$. Also, for $\alpha>n$, $G_\alpha$ is continuous and, hence, limited and $\forall 0<\alpha\leq n$ $G_\alpha$ has the same order of growth of \begin{equation} \begin{cases} -log|x| & \text{$\alpha=n$}\\ 1/|x|^{n-\alpha} & \text{$0<\alpha<n$} \end{cases} \end{equation} for $|x|\rightarrow0$.