Let \(f:[0,1]\times[0,1]\to\mathbb R\) be continuous. Assume that for each \(x\), the map \(t\mapsto f(x,t)\) is differentiable and that \[ \left|\frac{\partial f}{\partial t}(x,t)\right|\le g(x) \] for some measurable \(g\in L^1([0,1])\). Define \[ F(t)=\int_0^1 f(x,t)\,dx. \] Prove that \[ F'(t)=\int_0^1\frac{\partial f}{\partial t}(x,t)\,dx. \]
Solution
Fix \(t\in(0,1)\). For \(h\ne0\) small enough that \(t+h\in[0,1]\), define \[ q_h(x)=\frac{f(x,t+h)-f(x,t)}{h}. \] For each fixed \(x\), differentiability in \(t\) gives \[ q_h(x)\longrightarrow \frac{\partial f}{\partial t}(x,t) \qquad(h\to0). \] By the one-variable mean value theorem, for each \(x\) and such \(h\), there exists \(\theta\) between \(t\) and \(t+h\) such that \[ q_h(x)=\frac{\partial f}{\partial t}(x,\theta). \] Hence \[ |q_h(x)|\le g(x). \] Since \(g\in L^1\), dominated convergence yields \[ \begin{aligned} F'(t) &=\lim_{h\to0}\frac{F(t+h)-F(t)}h\\ &=\lim_{h\to0}\int_0^1q_h(x)\,dx\\ &=\int_0^1\frac{\partial f}{\partial t}(x,t)\,dx. \end{aligned} \] The same argument with one-sided difference quotients proves the corresponding one-sided derivative statements at \(t=0\) and \(t=1\).