If $f : [0,1] \to H$ has $t$-derivative with respect to the norm of $H$, and $H=L^2[0,1]$ itself, does the $t$-derivative exist in ordinary sense?

Question

The question is as in the title.

Let $H$ be a separable Hilbert space and $f : [0,1] \to H$ be a continuous mapping such that \begin{equation} f'(t):=\lim\limits_{\alpha \to 0} \frac{f(t+\alpha)-f(t)}{\alpha} \in H \end{equation} exists for almost every $t \in [0,1]$ and $f' : [0,1] \to H$ is Borel-measurable. In the above formula, convergence of the limit is with respect to the norm of $H$.

However, let us further suppose that $H=L^2\bigl([0,1],\mathbb{R}\bigr)$ itself, so that $f(t) : [0,1] \to H$ can be in fact written as $f(t,x) : [0,1] \times [0,1] \to \mathbb{R}$

Then, I am curious about the following:

1. Is it possible for the above derivative $f'(t) : [0,1] \to H$ to be defined almost everywhere on $[0,1] \times [0,1]$ as a real-valued Borel-measurable function?

2. If so, does $f'(t)$ coincide almost everywhere on $[0,1]^2$ with the usual definition of partial derivative of $f(t,x)$ with respect to $t$?

This seems more subtle than I expected and I am quite confused about dealing weith almost everywhere equalities... I posted this question on ME but haven't received any answer yet. all my attempts to figure out myself have failed.. Could anyone please help me?

Answer 1

The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.

For example, consider the case where $f(t,x) = \mathbb{1}_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.

Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.

Other possible argument: call $g$ the limit in $L^2[0,1]$ of $[f(t,\cdot)-f(0,\cdot)]/t$ as $t \to 0$. If you choose a sequence $(t_n)$ tending to $0$ sufficiently fast so that the series $\sum \Vert [f(t_n,\cdot)-f(0,\cdot)]/t_n - g \Vert_2$ converges, then the series $\sum [f(t_n,\cdot)-f(0,\cdot)]/t_n$ converges almost surely (and in $L^2$), so $[f(t_n,\cdot)-f(0,\cdot)]/t_n - g\to 0$ almost surely.