Derivative of integrals of $L^p([a,b])$ functions

Question

Question: Let $f,g \in L^p([a,b]),1 < p < \infty$. Show that the function $I \colon \mathbb{R} \to \mathbb{R}$ defined by $$I(t) := \int_a^b |f(x) + t g(x)|^p\, dx$$ is differentiable at $t = 0$ and compute its derivative.

My attempt: Using Minkowski's inequality the function $I$ is defined for all $t \in \mathbb{R}$ and $I(0) = \int_a^b |f(x)|^p \, dx$. I have consider using Lebesgue-dominated convergence theorem by letting $\{t_n\}\subset \mathbb{R} - \{0\}$ so that $t_n \to 0$ and defining $$I_n(t) := \int_a^b \frac{|f(x) + t_n g(x)|^p - |f(x)|^p}{t_n}\, dx$$ however I am not too sure how to carry on from here. Alternatively, I suspect the derivative at $t = 0$ should be the following expression $$A := \int_a^b p |f(x)|^{p-1} |g(x)| \, dx$$ so i tried to consider proving the following $$\lim_{t \to 0} \int_a^b \frac{|f(x) + t g(x)|^p - pt |f(x)|^{p-1}|g(x)| - |f(x)|}{t}\, dx = 0.$$ However, I am not sure of any inequalities to estimate the integrand. Any hints will be appreciated.

Answer 1

The result follows by dominated convergence for differentiation of integrals. One version, perhaps the weakest, is discussed here, works for the problem in the OP.

Define $F(t,x)=|f(x)+t g(x)|^p$ for $(t, x)\in\mathbb{R}\times [a,b]$.

1. It is clear that for each $t$ fixed, $F(t,\cdot)\in L_1([a,b])$.
2. As the map $\phi:s\mapsto |s|^p$, $1<p<\infty$, is differentiable on $\mathbb{R}$ and $\phi'(s)=p\,\operatorname{sign}(s)|s|^{p-1}$, we have that for each $x\in[a,b]$, $t\mapsto F(t, x)$ is differentiable for all $t\in\mathbb{R}$, and $$\partial_t F(t, x)=p\,\operatorname{sign}(f(x)+tg(x))|f(x)+t g(x)|^{p-1}g(x)$$
3. For all $|t|<1$, $$|\partial_tF(t,x)|\leq \phi(x):=p2^{p-1}\big(|f(x)|^{p-1}|g(x)|+|g(x)|^p \big)\in L_1([a,b])$$ for by assumption, $f,g\in L_p([a,b])$. Integrability of $|f(x)|^{p-1}|g(x)|$ follows from Hölder's inequality since $|f|^{p-1}\in L_{p/(p-1)}$, $g\in L_p$ and $\frac{p-1}{p}+\frac1p=1$.

Then, for each $|t|<1$, the function $x\mapsto \partial_tF(t, x)$ is integrable over $[a,b]$, $$\Phi(t)=\int_{[a,b]}F(t, x)\,dx,\qquad |t|<1$$ is differentiable, and \begin{align} \Phi'(t)&=\int_{[a,b]}\partial_t F(t, x)\,dx\\ &=\int_{[a,b]}p\,\operatorname{sign}(f(x)+t g(x))|f(x)+ tg(x)|^{p-1} g(x)\,dx\qquad |t|<1 \end{align} In particular, $$\Phi'(0)=p\int_{[a,b]}\operatorname{sign}(f(x))|f(x)|^{p-1} g(x)\,dx$$

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To emphasize how dominated convergence is used, notice that by the mean value theorem, for fixed $(t_0,x)\in(-1,1)\times[a,b]$, and all $|h|<1-|t_0|$, $$\Big|\frac{F(t_0+h, x)-F(t_0, x)}{h}\Big|=\Big|\partial_t{F}(t_0+\theta_{t_0,x}h,x)\Big|\leq |\phi(x)|$$ where $|\theta_{t_0,x}|<1$. Thus, moving the limit $h\rightarrow0$ inside the integral in $$\int_{[a,b]}\frac{F(t_0+h,x)-F(t_0,x)}{h}$$ is justified by dominated convergence.