I am looking for conditions or at least one example such that the following holds true.
Let $K_t^{\epsilon}:[0,T]\to\mathbb R$ be an smooth function such that $K_t^{\epsilon}\to \mathbf{1}_{[0,t]}$ in $L^2([0,T])$ (as $\epsilon\to 0$) where $\mathbf{1}_{[0,t]}$ denotes the indicator function of the interval $[0,t], t\in [0,T]$.
From the triangular inequality we know that for any $t\in[0,T]$, $\|K_t^{\epsilon}\|^2\to t$, but I need a stronger condition, namely that $\frac{d}{dt}\|K_t^{\epsilon}\|^2\to 1$.
I do know that in general the uniform convergence of the derivatives implies under certain conditions the uniform convergence of the sequence of functions, but there's no general condition for the converse to hold true. Nevertheless I was wondering if in this particular example we are able to find an example or conditions such that this holds true.
In principle I need this in order to prove a result, and I could assume this holds true but I don't know if it's too strong as an assumption.
Thanks in advance.
Hint Take $K_t^\epsilon(x) = k_1^\epsilon(x/t)\mid_{[0,T]}$ where $k_1^\epsilon = \rho_\epsilon * \mathbf{1}_{[0,1]}$ and $\rho_\epsilon$ is a suitable approximation of $\delta_0$.
We have \begin{equation} \|K_t^\epsilon\|^2 = t\int_0^{T/t}(k_1^\epsilon(u))^2 d u \end{equation}