Analysis Problem of the Day 96

Today’s problem appeared as Problem 3 on the Texas A&M August 2021 Analysis Qual:

Problem 3: Show that given any $f \in L^\infty([0,1]),$ there exists a sequence of polynomials converging to $f$ weak $-*$ in $L^\infty([0,1]).$

Solution: Unraveling definitions, we are asked to show that for some sequence of polynomials $p_n,$ $\int_0^1 p_n g \to \int_0^1 fg$ for all $g \in L^1([0,1]).$ The difficulty with the argument is that $g \in L^1$ is a very weak condition that prevents us from using any kind of Hölder’s inequality. Thus, we attempt an approximation argument. We first claim that there exists a uniformly bounded sequence of continuous functions converging to $f$ in $L^1.$ Indeed, by density, there exists a sequence of continuous functions $f_n \to f$ in $L^1$ (with $f_n$ not necessarily uniformly bounded), and we note that replacing $f_n$ with $g_n = \min(\max(f_n,-2\|f\|_\infty),2\|f\|_\infty))$ is continuous and satisfies $\|g_n -f\|_\infty \leq \|f_n -f\|_\infty,$ so that $g_n \to f$ in $L^1$ and is uniformly bounded by $2\|f\|_\infty.$ Next, by the Weierstrass approximation theorem, there exists a polynomial $p_n$ such that $\|p_n - g_n\|_\infty < \epsilon.$ Thus, there exists a sequence $p_n$ of uniformly bounded polynomials converging to $f$ in $L^1.$ It follows that for $[0,1] = A_M \cup A_M^c,$ where $A_M:=g^{-1}([-M,M])$ and large enough $M$ such that $\mu(A_M^c)<\epsilon,$ we have

$\left|\int (p_n -f)g\right| \leq \int_{A_M} |(p_n-f)g|+ \int_{A_M^c} |(p_n-f)g| < \epsilon + \epsilon = 2\epsilon,$

where the first integral is small by Hölder’s inequality since $p_n \to f$ in $L^1$ and $|g| \leq M$ on $A_M,$ and the second integral is small since $\int_{A_M^c} |g|<\epsilon$ and $\|p_n-f\|_\infty$ is uniformly bounded in $n.$ Sending $\epsilon \to 0$ and $M \to \infty$ thus yields $\int p_n g \to \int fg$ (and in fact $p_n g \to fg$ in $L^1$ ), which shows that $p_n$ converges weak $-*$ in $L^\infty$ to $f.$

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