We have the operator $T: L^p(\mathbb{R}^+) \to L^p(\mathbb{R}^+) $ with $p \in (1,+\infty)$, defined by $T(f):=\frac{1}{x}\int_{0}^{x}{f(t)dt}$. We define $\tilde{f}(x)=e^{x/p}f(e^x)$ for all $f \in L^p(\mathbb{R}^+)$.
We want calculate $\|T\|$.
I showed that:
$i)$ $f \in L^p(\mathbb{R}^+) \iff \tilde{f} \in L^p(\mathbb{R})$ and in this case $\|f\|_p=\|\tilde{f}\|_p$ (with change of variable $x=e^t)$
$ii)$ let $g(x)=e^{-x/q}\chi_{[0,+\infty)}$ with $q=1-1/p$ , we have $g \in L^1(\mathbb{R})$ and $\tilde{T(f)}=\tilde{f}\star g$ (where $\star$ is the convolution) (we can see that, with other change of variable $t=e^{x-y}$ starting from $\tilde{f}\star g$).
So
$\|T\|=\sup_{\|f\|_p=1} \|T(f)\|_p=\sup_{\|f\|_p=1}\|\tilde{T(f)}\|_p=\sup_{\|f\|_p=1}\|\tilde{f}\star g\|_p \leq \sup_{\|f\|_p=1}\|\tilde{f}\|_p\|g\|_1=\sup_{\|f\|_p=1}\|f\|_p\|g\|_1=\|g\|_p=q$
Now: $\|T\|=q$ ? I tried to find a function $f$ such that $\|f\|_p=1$ and $\|\tilde{f}\star g\|_p =q$ but I didn't find it