A diagonal differentiation theorem Suppose that $f \in L^2([0,1])$. For each $n\in \mathbb{N}$ and each $k\in\{1,\dots,n\}$, define $I_{n,k}:=(\frac{k-1}{n}, \frac{k}{n}]$. Is it true that
\begin{equation*}
   n\cdot\sum_{k=1}^n\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|\mathrm{d}s\mathrm{d}t \to 0, \qquad n\to\infty \quad?
\end{equation*}
I stumbled into this expression while doing some calculations involving $L^2$-approximation of $f$ via a sequence of piecewise constant functions, and I'm not sure if it is true or not. Intuitively, it seems a "diagonal" differentiation theorem.
I tried to use first Cauchy-Schwarz inequality (with $|f(s)-f(t)|$ vs $1_{I_{n,k}}(s)1_{I_{n,k}}(t)$), then upper bound $\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|^2\mathrm{d}s\mathrm{d}t$ with $\frac{2}{n} \int_{I_{n,k}}|f(s)|^2\mathrm{d}s$, and finally apply Jensen's inequality for concave functions (on $x\mapsto\sqrt{x}$), but this leads just to the previous quantity bounded above by $\sqrt{2} \|f\|_2$.
Since I've upper bounded the original quantity three times, probably I must have lost a lot of room along the way.
Any smarter idea on how to prove or disprove the claim?
 A: For $f \in L^2([0,1])$, let
$$T_n(f):= n\cdot\sum_{k=1}^n\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|\, ds\, dt \,.$$
As noted in the problem, applying Cauchy-Schwarz twice gives
$$\frac{T_n(f)^2}{n^2} \le n\sum_{k=1}^n \Bigl(\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|\, ds\, dt\Bigr)^2 $$ $$\le \frac1n\sum_{k=1}^n \Bigl(\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|^2\, ds\, dt\Bigr) $$
$$ \le 
\frac4{n^2}\sum_{k=1}^n \Bigl(\int_{I_{n,k}}  f(t)^2\,   dt\Bigr) =\frac4{n^2}\int_0^1  f(t)^2\,   dt \,,$$
so
$$T_n(f) \le 2\|f\|_2 \tag{1} \,.$$
Given  $f \in L^2([0,1])$ and $\epsilon>0$, there exists a continuous function $g \in C[0,1]$ such that $h=f-g$ satisfies $\|h\|_2 <\epsilon\,.$
Choose $n_\epsilon$ large enough so that $$|g(t)-g(s)|<\epsilon \quad \text{if} \quad   |t-s| \le  1/n_\epsilon \,.$$
Clearly, for all  $n \ge n_\epsilon$, we have $|T_n(g)| \le \epsilon\, . $
Moreover, $$\forall t,s  \quad |f(t)-f(s)| \le |g(t)-g(s)|+|h(t)-h(s)| \,, $$  so for all   $n \ge n_\epsilon$, we obtain
$$T_n(f) \le  T_n(g)+ T_n(h)  \le \epsilon+2\|h\|_2 <3\epsilon\,.
$$
This proves that $T_n(f) \to 0$ as $n \to \infty$.
