I was reading a proof about the differentiability of the function:
$N(\lambda) = \int_{x} | f + \lambda g|^p dm$ with $ \lambda \in [0,1]$. Where $(X, M, m)$ is a measure space and $f, g \in L^p(m)$ with $p>1$
In the proof it is said that the function defined by $F(\lambda, y) = | f(y) + \lambda g(y)|^p$ is $C^1$ with respect to $\lambda$ for almost every $y \in X$ I don't understand how one can see it. I tried to get a counterexample and I thought about:
$X=[0,1]$ with lebesgue measure. $f,g:[0,1] \rightarrow \mathbb{R}$ defined by $f(x) = -1/2$ if $x \in \mathbb{R}-\mathbb{Q}$ and $f(x) =3$ if $x \in \mathbb{Q}$. And $g(x) = 1$ if $x \in \mathbb{R}-\mathbb{Q}$ and $g(x) = 0$ if $x \in \mathbb{Q}$. Thus I get:
$F(\lambda, x)=|-1/2 + \lambda|^p$ if $x \in \mathbb{R}-\mathbb{Q}$ and $F(\lambda, x)=|3|^p$ if $x \in \mathbb{Q}$. Thus i get that $F$ is not differentiable in $\lambda = 1/2$ for every $x \in R-Q$. Is it a counterexample?
No that's no counterexample. Note that because of $p > 1$, the function $f \colon \xi \mapsto |\xi|^p$ is differentiable on the whole of $\mathbf R$, with derivative $$ f'(\xi) = p\cdot \mathrm{sgn}\,\xi \cdot |\xi|^{p-1} $$ which is continuous. Hence $f \in C^1(\mathbf R)$. Therefore also $\lambda \mapsto \left|-1/2 + \lambda\right|^p$ is $C^1$.