Analysis Problem of the Day 13

Today’s problem appears as Problem 2 on the UCLA Fall 2024 Analysis Qual:

Problem 2: Let $f \in L^1(\mathbb{R})$ and let $\{a_n\}$ be a bounded sequence in $\mathbb{R}.$ Assume that

$g(x) = \lim_{n \to \infty} f(x-a_n)$

exists for almost all $x \in \mathbb{R}$ and that $g$ does not vanish almost everywhere. Show that the sequence $\{a_n\}$ converges.

Solution: For sake of contradiction, suppose $\{a_n\}$ does not converge, and let $b := \liminf_{n \to \infty} a_n < \limsup_{n \to \infty} a_n := c.$ Let $b_n \to b, c_n \to c$ be two subsequences of $\{a_n\}.$ Then, if $T_a: L^1 \to L^1,$ $T_a f(x) = f(x-a)$ is the family of translation operators, $a \in \mathbb{R},$ we are given that $T_{a_n} f \to g$ a.e., i.e. $T_{b_n} f \to g, T_{c_n} f \to g$ a.e. But by the continuity of translation operators on $L^p,$ $T_{b_n} f \to f(x-b), T_{c_n} f \to f(x-c)$ in $L^1.$

Now, recall that if $f_n \to f$ in $L^1,$ there exists a subsequence which converges to $f$ a.e. In particular, there exist subsequences of $b'_n, c'_n$ of $b_n,c_n$ such that $T_{b'_n} f \to f(x-b), T_{c'_n} f \to f(x-c)$ a.e., respectively. But since $T_{a_n} f \to g$ a.e, this implies that $g(x)=f(x-b)=f(x-c)$ a.e., i.e $f$ is a.e. periodic (with period $c-b$ ). The only periodic $L^1$ function is the zero function, so $f$ is zero a.e., which contradicts the assumption that $g$ does not vanish a.e. Thus, the sequence $\{a_n\}$ converges.

Leave a Reply Cancel reply