Show that $\lim\limits_{t\rightarrow 0}\frac{\psi(u+th)-\psi(u)}{t}=p\int_\Omega|u|^{p-2}uh$

Question

Let $u,h\in L^p(\Omega)$ and $\psi(u)=\int_\Omega|u|^p$.

Show that $\lim\limits_{t\rightarrow 0}\frac{\psi(u+th)-\psi(u)}{t}=p\int_\Omega|u|^{p-2}uh$

If I could ,I hope getting a detail answer or hint ,so thanks.

I have try to get that: $$ \lim\limits_{t\rightarrow 0}\frac{\psi(u+th)-\psi(u)}{t}= \int_\Omega \lim\limits_{t\rightarrow 0}\frac{|u+th|^p-|u|^p}{t} $$

But I don't know how to computing $\lim\limits_{t\rightarrow 0}\frac{|u+th|^p-|u|^p}{t}$

Answer 1

By the chain rule,

$$\frac{d}{dt}|u + th|^p = \frac{d}{dt}(|u + th|^2)^{p/2} = \frac{p}{2}(|u + th|^2)^{\frac{p}{2}-1}\frac{d}{dt}(|u + th|^2), $$

and

$$\frac{d}{dt}(|u + th|^2) = 2(u + th)h.$$

So

$$\frac{d}{dt}\bigg|_{t = 0} |u + th|^p = \frac{p}{2}|u|^{p-2}(2uh) = p|u|^{p-2}uh,$$

that is,

$$\lim_{t\to 0} \frac{|u + th|^p - |u|^p}{t} = p|u|^{p-2}uh.$$