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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 $L^2$?

Bach
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1 Answers1

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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?).

PhoemueX
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  • I have a feeling that this is an overkill... – Bach May 25 '14 at 11:10
  • That may be the case, but the technique of proof (show that $T_n f \rightarrow T f$ for $f$ from some dense subset and use that $\Vert T_n \Vert$ is uniformely bounded) used here is relatively common. It is for example used to show that $\Vert \tau^x f - f \Vert_{L^p} \rightarrow 0$ for $x \rightarrow 0$ and $f \in L^p(\Bbb{R}^d)$. Here, $(\tau^x f)(y) := f(y-x)$. – PhoemueX May 27 '14 at 08:48