Analysis Problem of the Day 33

Today’s problem is Problem 3 on the Texas A&M August 2017 Analysis Qual:

Problem 3: Construct a sequence $f_n \geq 0$ of $L^1$ functions on $[0,1]$ such that $f_n \to 0$ a.e. and $f_n \rightharpoonup 1$ in $L^1,$ i.e. for any interval $[a,b],$ $\lim_{n \to \infty} \int_a^b f(x)dx = b-a.$

Solution: Our motivation comes from a famous example of a weakly converging sequence – namely, $\sin(nx) \rightharpoonup 0.$ We will thus try to create a function rapidly oscillating on every interval, yet which tends pointwise a.e. to 0. Define $g_j(x) = 2^j \chi_{[1-2^{-j},1]}(x)$ and $h_n(x) := \sum_{i=0}^{2^n-1} g_n(2^n x-i).$ That is, split $[0,1]$ into intervals of length $2^n$ and put a copy of $g_n(x)$ in each interval. Then, the measure of the support of $h_n$ is $2^{n} (2^{-n} 2^{-n}) = 2^{-n} \to 0,$ so $h_n \to 0$ a.e. However, for any interval $[a,b]$ and large $n$ there are $2^n (b-a)+O(1)$ subintervals of length $2^{-n}$ contained inside it, over each of which the integral is $2^{-n},$ so

$\int_a^b h_n(x) dx \to \sum_{I \cap [a,b] \not = \varnothing} \int_I h_n(x)dx= 2^n (b-a) 2^{-n} +O(2^{-n}) \to b-a,$

showing the required properties.

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