Denote by $$w_{\lambda}(t):=\int_0^t f_{\lambda}(t,\sigma) d\sigma, \quad t \in \mathbb{R},$$ where $f_{\lambda}(t,\sigma)$ is a continuously differentiable function on $\mathbb{R}$ in respect to $t$ and $f_{\lambda}(t,t)=0$, $\lambda \in \mathbb{R}$.
The aim is to prove that $v(t):=\displaystyle\lim_{\lambda \to \infty} w_{\lambda}(t)$ defines a continuously differentiable function on $\mathbb{R}$.
Using Leibniz integral rule, we can prove that $$w_\lambda'(t)=\int_0^t \frac{\partial}{\partial t}f_{\lambda}(t,\sigma) d\sigma.$$ I also proved using some results on the function $f_{\lambda}(t,\sigma)$ that there exists $\omega \geq 0$ such that for all $\lambda>\omega$ we have $$\|w_\lambda'(t)\| \leq \frac{\lambda}{\lambda-\omega} K(T), \quad \forall t \in [-T,T],$$ where $K(T)$ is a positive constant independent of $t$, this means that the limit of $w_\lambda'(t)$ converge uniformly on compacts of $\mathbb{R}$. Can we deduce then that $v(t)$ is a continuously differentiable function on $\mathbb{R}$?
