Let $\mu$ be a positive measure, $f,g \in L^p(\mu)$,$ \parallel f\parallel _p \leq R$, $ \parallel g \parallel _ p \leq R$
I have to prove :
$$\int ||f|^p-|g|^p|d\mu \leq 2pR^{p-1} \parallel f-g\parallel _p$$
for $ p>1$.
I was given a hint , that says that, for $x,y \geq 0$ ,
$$|x^p-y^p| \leq p |x-y|(x^{p-1}+y^{p-1})$$
I have been trying to do it, it seems easy with the hint, but I can't find a way to do it.
Applying the last inequality :
$||f|^p-|g|^p| \leq p|f-g|(|f|^{p-1} +|g|^{p-1})$
Integrating both sides:
$\int ||f|^p-|g|^p|d\mu \leq p\int|f-g|(|f|^{p-1} +|g|^{p-1}) d \mu$
but I can't see how I can get from there to the result. I think I have to use some kind of inequality like Minkowski's.
Any help would be helpful.
If $p>1$:
By Holder's inequality, \begin{align*} \int\bigg||f|^{p}-|g|^{p}\bigg|\leq p\|f-g\|_{L^{p}}\left(\int(|f|^{p-1}+|g|^{p-1})^{q}\right)^{1/q}, \end{align*} where $1/p+1/q=1$. But we also have $(a+b)^{q}\leq 2^{q-1}(a^{q}+b^{q})$ for $a,b\geq 0$, so \begin{align*} \left(\int(|f|^{p-1}+|g|^{p-1})^{q}\right)^{1/q}&\leq 2^{(q-1)/q}\left(\int|f|^{(p-1)q}+|g|^{(p-1)q}\right)^{1/q}\\ &=2^{1/p}\left(\int|f|^{p}+|g|^{p}\right)^{1/q}\\ &\leq 2^{1/p}(\|f\|_{L^{p}}^{p}+\|g\|_{L^{p}}^{p})^{1/q}\\ &=2^{1/p}2^{1/q} R^{p/q}\\ &=2 R^{p-1}. \end{align*}