Differentiability at $0$

Question

Let $\phi:[0,1]\to\mathbb R$ be a Lebesgue measurable and integrable function. Consider: $$F:t\in [0,\infty ) \to \int_0^1 \sqrt{t+\phi^2(x)}dx.$$ What is a necessary and sufficient condition for the function $F$ to be differentiable at $t=0$?

Answer 1

This answer just fills in a few details of Did's succinct answer.

A necessary & sufficient condition is that $x \mapsto \frac{1}{\phi(x)}$ is integrable.

Let $f(x,t) = \sqrt{t+\phi(x)^2}$, $E = \phi^{-1}\{0\}$. If $x \in E^c$, we have $\frac{\partial f(x,t)}{\partial t} = \frac{1}{2 \sqrt{t+\phi(x)^2}}$ and $|\frac{\partial f(x,t)}{\partial t}| \le \frac{1}{2 |\phi(x)|}$.

Suppose the integrability condition holds. Then $mE = 0$, and $F(t) = \int_{E^c} f(x,t) dx$. Since $\frac{f(x,t)-f(x,0)}{t} = \frac{1}{|\phi(x)| + \sqrt{t+\phi(x)^2}} \le \frac{1}{2 |\phi(x)|}$, we can use the dominated convergence theorem with

$\frac{F(t)-F(0)}{t} = \int_{E^c} \frac{f(x,t)-f(x,0)}{t} dx$ to get $\frac{dF(0)}{dt} = \int_{E^c} \frac{\partial f(x,0)}{\partial t} dx = \int_0^1 \frac{dx}{2 |\phi(x)|}$.

Now suppose $F$ is differentiable at $t=0$. If $t>0$, we have $\frac{F(t)-F(0)}{t} = \int_0^1 \frac{1}{|\phi(x)| + \sqrt{t+\phi(x)^2}} dx$. We note that the integrand is non-negative and a non-increasing function of $t$. Hence the monotone convergence theorem give $\frac{dF(0)}{dt} = \int_0^1 \frac{dx}{2 |\phi(x)|}$, from which the integrability condition follows.

Answer 2

$$\frac{F(t)-F(0)}t=\int_0^1\frac{\mathrm dx}{|\phi(x)|+\sqrt{t+\phi^2(x)}}\ \underset{t\to0+}{\longrightarrow}\ \int_0^1\frac{\mathrm dx}{2|\phi(x)|}$$