Whether the function is integrable or not?

Question

Let $H$ be a Hilbert space and $f:[0,\tau]\times H\to H$ be a function such that $s\mapsto f(s,\phi)$ is in $L^1([0,\tau];H)$ for each $\phi\in H$. Is it true that $s\mapsto f(s,g(s))$ is in $L^1([0,\tau];H)$, where $g:[0,\tau]\to H$ is a continuous function?

Answer 1

Define the function $f: [0, \tau] \times \mathbb{R} \to \mathbb{R}$ By

$$f(s,x) = |s-x|^{-1/2}, s\neq x \quad \text{and} \quad f(x,x) =0.$$

Then $f(.,x) \in L^1([0, \tau]; \mathbb{R})$ for any $x \in \mathbb{R}.$

Choose $g(s)= s+s^2$ Then, $s\mapsto f(s, g(s))$ is not integrable on $[0, \tau].$

Based on your next question, the example in this answer provides a counterexample... You can compute:

$$\int_0^\tau f(s,x) ds = \left\{\begin{align} 2\left(\sqrt{x}-\sqrt{x-\tau}\right), \text{if}\quad x\geq \tau\\ 2\left(\sqrt{\tau -x}-\sqrt{-x}\right), \text{if}\quad x\leq 0\\ 2\left(\sqrt{\tau -x}+\sqrt{x}\right), \text{if}\quad 0<x<\tau \end{align} \right.$$ Thus, we have $$\lim_{x\to \pm \infty} \int_0^\tau f(s,x) ds = 0,$$ and $\sup_{x\in \mathbb{R}} \int_0^\tau f(s,x) ds <\infty.$