Speed of convergence of $\frac1n\int_0^1 f(x/n)dx\to0$? Let $f\in L^2(0,1)$, this implies that
$$\frac1n\left|\int_0^1 f(x/n) dx\,\right| =\left|\int_0^{1/n} f(x) dx\,\right|=\left|\langle \chi_{[0,1/n]}, f\rangle_{L^2}\right|≤\|\chi_{[0,1/n]}\|_{L^2}\cdot \|f\|_{L_2}=\frac{\|f\|_{L^2}}{\sqrt{n}}$$
My question is whether there is a better bound on the speed of convergence than a $\frac1{\sqrt n}$ factor.
I ask because playing around with different $f$ it looks like
$$\sum_n \frac1{n^2}\left|\int_0^1 f(x/n) dx\,\right|^2$$
is always summable, and I would like to find a bound like $\sum_n\frac1{n^2}\left|\int_0^1 f(x/n)\, dx\right|^2≤M\|f\|_{L^2}^2$.
 A: This argument owes a large debt to the comments of Daniel Fischer and D. Thomine!
I think the following argument works. I'll assume $f$ is real-valued to avoid a lot of conjugation signs.
\begin{align*}
\sum_{n=1}^\infty \frac1{n^2}\bigg(\int_0^1 f(x/n) \,dx\,\bigg)^2 &= \sum_{n=1}^\infty \bigg(\int_0^{1/n} f(x) \,dx\bigg)^2 \\
&= \sum_{n=1}^\infty \int_0^{1/n} \int_0^{1/n} f(x) f(y) \,dx\,dy \\
&= \int_0^1 \int_0^1 f(x) f(y) \bigg( \sum_{\substack{n\ge 1 \\ 1/n \ge x \\ 1/n \ge y}} 1 \bigg) \,dx\,dy \\
&= \int_0^1 \int_0^1 f(x) f(y) \min\{\lfloor 1/x \rfloor,\lfloor 1/y \rfloor\} \,dx\,dy \\
&= 2 \int_0^1 \int_0^y f(x) f(y) \lfloor 1/y \rfloor \,dx\,dy \\
&\le 2 \int_0^1 f(y) \frac1y \int_0^y f(x) \,dx\,dy \\
&=2 \int_0^1 f(y) g(y) \,dy,
\end{align*}
where we have defined $g(y) = \frac1y \int_0^y f(x) \,dx$. Hardy's inequality tells us that $g$ is square-integrable and $\|g\|_2 \le 2\|f\|_2$. Therefore by Cauchy–Schwarz,
\begin{align*}
\sum_{n=1}^\infty \frac1{n^2}\bigg(\int_0^1 f(x/n) \,dx\,\bigg)^2 &\le 2 \|f\|_2 \|g\|_2 \le 4\|f\|_2^2.
\end{align*}
