I am listening to Functional Analysis and I don't really understand the Lp norm or how to work with it. In the script is the following equality:
$$\frac{d}{dt} \int_{\Omega} |u(x,t)|^p dx = p \int_{\Omega} u_t u |u|^{p-2} dx$$
Here $\Omega$ is a bounded interval in $\mathbb{R}$ and u is continuously differentiable. Why is the derivative to t defined in this way? So if I ignore the absolute value and derive partially according to the normal knowledge, i come of course to the result but here the function is in the absolute value.
Note that $\dfrac{d}{dx} |x| = \dfrac{x}{|x|}$ if $x \not= 0$.
Next apply the chain rule: if $p$ is a real number then $\dfrac{d}{dx} |x|^p = p|x|^{p-1} \dfrac{x}{|x|} = px|x|^{p-2}$ provided that $x \not= 0$.
If $p > 1$ you have $\left.\dfrac{d}{dx} |x|^p\right|_{x=0} = \displaystyle \lim_{x \to 0} \frac{|x|^p - 0}{x-0} = \lim_{x \to 0} \frac{|x|^p}{x} = 0$.
Thus it is safe to assert $\dfrac{d}{dx}|x|^p = px|x|^{p-2}$ for all $x$ whenever $p > 1$, because despite the possibility that $2-p < 0$ you have $$|px|x|^{p-2}| = p|x|^{p-1}$$ which is $0$ when $x=0$.
Now do the chain rule again: $$\frac{\partial}{\partial t} |u(x,t)|^p = p u(x,t) |u(x,t)|^{p-2} \frac{\partial}{\partial t} u(x,t) = pu_t u |u|^{p-2}$$ where the last equality is just for notational brevity.
Finally convince yourself that $$\frac {d}{dt} \int_\Omega |u(x,t)|^p \, dx = \int_\Omega \frac{\partial}{\partial t} |u(x,t)|^p \, dx.$$ You'll need some type of bounded convergence result which will utilize the fact that $u_t$ is continuous.