Derivative of $t \mapsto \int_\mathbb{R} (u+tv)^p$ Let $u,v: \mathbb{R} \rightarrow \mathbb{R}_+$ be two Lebesgue-mesurable functions, p > 1. If we have $u^p, v^p \in \mathscr{L}^1(\lambda)$ then


*

*$(u+tv)^p$ is a $\mathscr{L}^1$-function and

*$t \mapsto \int_\mathbb{R} (u+tv)^p$ is differentiable in $(0, \infty)$.


I was able to show the first point but I am stuck at the second since I don't know when I am allowed to move the $\frac{d}{dt}$ under the integral sign.
 A: We just apply Leibniz integral rule. Let $f(t):=\int_{\mathbb{R}}g(s,t)ds$ where $g:\mathbb{R}\times (0,\infty)\to \mathbb{R}$ is given by
$$g(s,t):=\left(u(s)+tv(s)\right)^p $$
First off, $s\mapsto g(s,t)\in L^1(\mathbb{R})$ for all $t\in (0,\infty)$. This is because it is a linear combination of $u^p$ and $v^p$ which are in $L^1$. Moreover, $g(s,t)$ is differentiable in $t$ for all $t,s$, with
$$g_t(s,t)=p\cdot v(s)\left(u(s)+tv(s)\right)^{p-1} $$
Now let $t_0\in (0,\infty)$ and let $\varepsilon\in (0,t_0)$. Since $g_t$ is increasing in $t$, we have
$$g_t(s,t)\leq h(s):=g_t(s,t_0+\varepsilon),\qquad \forall (s,t)\in \mathbb{R}\times [t_0-\varepsilon,t_0+\varepsilon] $$
Last thing we need to prove $h(s)\in L^1(\mathbb{R})$. To see this, notice that $v(s)\in L^p(\mathbb{R})$ and $(u(s)+(t_0+\varepsilon)v(s))^{p-1}\in L^{\frac{p}{p-1}}(\mathbb{R})=L^{p'}(\mathbb{R})$. Hence by Holder's inequality, $h(s)=p\cdot v(s)\left(u(s)+(t_0+\varepsilon)v(s)\right)^{p-1}\in L^1(\mathbb{R})$. This allows us to conclude
$$f'(t)=\int_{\mathbb{R}}g_t(s,t)ds,\qquad \forall t\in (t_0-\varepsilon,t_0+\varepsilon) $$
by Leibniz integral rule.
Since $t_0$ is arbitrary in $(0,\infty)$, we are done.
