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Spring 2015 - Problem 1

density argument, Lp spaces

Let f∈L1(R)f \in L^1\p{\R}. Show that

lim⁡n→∞∑k=−n2n2∣∫k/n(k+1)/nf(x) dx∣=∫∣f(x)∣ dx.\lim_{n\to\infty} \sum_{k=-n^2}^{n^2} \abs{\int_{k/n}^{\p{k+1}/n} f\p{x} \,\diff{x}} = \int \abs{f\p{x}} \,\diff{x}.
Solution.

First, let ff be a step function, i.e., f=∑j=1NajχIjf = \sum_{j=1}^N a_j \chi_{I_j}, where the IjI_j are disjoint closed intervals and have finite measure. Let δ=min⁡ {d(Ij,Ik)∣j≠k}>0\delta = \min\,\set{d\p{I_j, I_k} \mid j \neq k} > 0, since there are only finitely many of these. Hence, if 1n<δ\frac{1}{n} < \delta, then each [kn,k+1n]\br{\frac{k}{n}, \frac{k+1}{n}} intersects at most one IkI_k. Hence,

∣∫k/n(k+1)/nf(x) dx∣=∫k/n(k+1)/n∣f(x)∣ dx,\abs{\int_{k/n}^{\p{k+1}/n} f\p{x} \,\diff{x}} = \int_{k/n}^{\p{k+1}/n} \abs{f\p{x}} \,\diff{x},

since on each of these intervals, ff is either non-negative or non-positive. Consequently, we get

∑k=−n2n2∣∫k/n(k+1)/nf(x) dx∣=∑k=−n2n2∫k/n(k+1)/n∣f(x)∣ dx=∫−nn+1n∣f(x)∣ dx→n→∞∫∣f(x)∣ dx\begin{aligned} \sum_{k=-n^2}^{n^2} \abs{\int_{k/n}^{\p{k+1}/n} f\p{x} \,\diff{x}} &= \sum_{k=-n^2}^{n^2} \int_{k/n}^{\p{k+1}/n} \abs{f\p{x}} \,\diff{x} \\ &= \int_{-n}^{n + \frac{1}{n}} \abs{f\p{x}} \,\diff{x} \xrightarrow{n\to\infty} \int \abs{f\p{x}} \,\diff{x} \end{aligned}

by monotone convergence. Thus, the result holds for step functions. For a general f∈L1(R)f \in L^1\p{\R}, let ε>0\epsilon > 0 and recall that step functions are dense in L1(R)L^1\p{\R}, so there exists a step function gg with ∥f−g∥L1<ε\norm{f - g}_{L^1} < \epsilon. Write

An(f)=∑k=−n2n2∣∫k/n(k+1)/nf(x) dx∣A_n\p{f} = \sum_{k=-n^2}^{n^2} \abs{\int_{k/n}^{\p{k+1}/n} f\p{x} \,\diff{x}}

which gives

An(f)≤∑k=−n2n2∫k/n(k+1)/n∣f(x)∣ dx=∫−nn+1n∣f(x)∣ dx≤∥f∥L1.A_n\p{f} \leq \sum_{k=-n^2}^{n^2} \int_{k/n}^{\p{k+1}/n} \abs{f\p{x}} \,\diff{x} = \int_{-n}^{n + \frac{1}{n}} \abs{f\p{x}} \,\diff{x} \leq \norm{f}_{L^1}.

Hence,

∣An(f)−∥f∥L1∣≤∣An(f)−An(g)∣+∣An(g)−∥g∥L1∣+∣∥f∥L1−∥g∥L1∣≤2∥f−g∥L1+∣An(g)−∥g∥L1∣.\begin{aligned} \abs{A_n\p{f} - \norm{f}_{L^1}} &\leq \abs{A_n\p{f} - A_n\p{g}} + \abs{A_n\p{g} - \norm{g}_{L^1}} + \abs{\norm{f}_{L^1} - \norm{g}_{L^1}} \\ &\leq 2\norm{f - g}_{L^1} + \abs{A_n\p{g} - \norm{g}_{L^1}}. \end{aligned}

Since the claim is true for gg, taking lim sup⁡n→∞\displaystyle\limsup_{n\to\infty}, we end up with

lim sup⁡n→∞ ∣An(f)−∥f∥L1∣≤2∥f−g∥L1≤2ε,\limsup_{n\to\infty}\,\abs{A_n\p{f} - \norm{f}_{L^1}} \leq 2\norm{f - g}_{L^1} \leq 2\epsilon,

which completes the proof.