$L^2$-convergence of a sequence of step functions of differences Suppose $f\in L^2=L^2([0,1])$ (that is, $f$ is square-integrable). Let $f_n$ be an approximation of $f$ by steps of differences of $\int f$. Formally,
$$f_n = \sum_{i=1}^n\mathbf{1}_{((i-1)/n,i/n]}\left(n\int_{(i-1)/n}^{i/n}f(t)dt\right)$$
Explanation: I have divided $[0,1]$ into $n$ equal parts, and let $f_n$ be constant on each part, with height equal to $\int f$'s difference on the edges of that part, divided by its length. In fact, it looks like there's nothing special about dividing into $n$ equal parts - as long as the maximal part's length tends to 0 with $n$, it should probably be all right.
My question is this: does it follow that $f_n$ converges to $f$ in $L^2$?
 A: I assume that you are working only on $[0,1]$, i.e. $L^2 = L^2([0,1])$.
Let us consider the linear operators $T_n : L^2([0,1]) \rightarrow L^2([0,1]), f \mapsto f_n$.
First calculate
\begin{eqnarray*}
\Vert T_n f \Vert _2^2 = \left\Vert f_{n}\right\Vert _{2}^{2} & = & \sum_{i=1}^{n}\int_{\left(i-1\right)/n}^{i/n}\left|n\cdot\int_{\left(i-1\right)/n}^{i/n}f\left(t\right)\, dt\right|^{2}\, dx\\
 & \overset{\text{Hölder}}{\leq} & \sum_{i=1}^{n}\int_{\left(i-1\right)/n}^{i/n}\left[n\cdot\sqrt{\int_{\left(i-1\right)/n}^{i/n}1^{1}\, dt}\cdot\sqrt{\int_{\left(i-1\right)/n}^{i/n}\left|f\left(t\right)\right|^{2}\, dt}\right]^{2}\, dx\\
 & = & \sum_{i=1}^{n}\int_{\left(i-1\right)/n}^{i/n}n\cdot\int_{\left(i-1\right)/n}^{i/n}\left|f\left(t\right)\right|^{2}\, dt\, dx\\
 & = & \int_{0}^{1}\left|f\left(t\right)\right|^{2}\, dt=\left\Vert f\right\Vert _{2}^{2}.
\end{eqnarray*}
Then note that for $f \in C([0,1])$ we have $f_n \rightarrow f$ uniformly and hence in particular in $L^2$.
Because $C([0,1])$ is dense in $L^2([0,1])$, you can verify your claim using the above estimate on the operator norms of $T_n$ (how?).
