Analysis Problem of the Day 89

Today’s problem appeared as Problem 3 on UCLA’s Spring 2019 Analysis Qual:

Problem 3. a) Let $f$ be a continuous positive function on $\mathbb{R}$ with $\lim_{|t| \to \infty} f(t)=0.$ Show that the set $\{hf: h \in L^1(\mathbb{R}), \|h\|_1 \leq K\}$ is a nowhere dense closed subset of $L^1(\mathbb{R})$ for any positive $K.$

b) Let $\{f_n\}$ be a sequence positive continuous functions as in (a). Show that there exists $g \in L^1(\mathbb{R})$ such that $\frac{g}{f_n} \not \in L^1(\mathbb{R})$ for all $n.$

Solution: a) Denote the desired set $A^f_K.$ We first show $A^f_K$ is closed. Let $h_n f \to g$ in $L^1, h_n f \in A_K.$ Since $f$ is positive everywhere, note that division by $f$ is allowed. Moreover, notice that

$\int_F \left|\frac{g}{f}\right| dx \leq \int_F \left|\frac{g}{f}-h_n\right|+|h_n|dx$

on any compact set $F.$ Now, on any compact set $F,$ $|f| \geq \epsilon$ for some $\epsilon>0,$ so that $\int_F |h_n f-g| = \int_F |f| |h_n - \frac{g}{f}| \geq \epsilon \int_F |h_n - \frac{g}{f}|.$ By sending $n \to \infty,$ we conclude that $h_n \to \frac{g}{f}$ on $F,$ i.e. $h_n \to \frac{g}{f}$ in $L^1_{loc}.$ Returning to our first use of triangle inequality, we note that on any compact set $F,$ $\int |h_n| \leq K$ and $\int_F |h_n - \frac{g}{f}| \to 0$ as $n \to \infty.$ Thus, returning to the original inequality, by taking $n$ large enough, we obtain that $\int_F |\frac{g}{f}| \leq K+\epsilon$ for any $\epsilon>0$ and any compact set $F.$ This immediately implies that $\frac{g}{f} \in L^1$ and $\|\frac{g}{f}\|_1 \leq K,$ so $g = \frac{g}{f} f \in A^f_K.$ Thus, $A^f_K$ is closed.

We now show that $A^f_K$ is nowhere dense. Since $A^f_K$ is closed, this is equivalent to demonstrating that $A^f_K$ has empty interior. Fix $h \in L^1, \|h\|_1 \leq K,$ and fix $\epsilon>0.$ Define $I_n^\epsilon := \{x: \frac{\epsilon}{2^{n+1}} \leq f(x) < \frac{\epsilon}{2^n}\},$ and note that $|I_n|<\infty$ for all $n$ since $f$ converges to $0$ at infinity. Now, define $g(x) = \sum_{n=1}^\infty \frac{\epsilon}{2^n}\frac{\chi_{I_n}(x)}{|I_n|},$ and note that $\int |g(x)| dx \leq \epsilon.$ Now, pick a large compact set $F$ so that $\|hf\|_{L^1(F^c)} \leq \epsilon,$ and define $G$ to be $hf$ on $F$ and $g$ on $F^c.$ Then, clearly $G \in L^1$ and $\|G-hf\|_1 < 2\epsilon.$ However, $\frac{G}{f} \geq \frac{1}{|I_n|}\frac{\epsilon}{2^{n}} \frac{2^{n}}{\epsilon} = \frac{1}{|I_n|}$ on each $I_n$ for sufficiently large $n,$ and so $\int_{F^c} \frac{G}{f} \geq \sum_{I_n \cap F = \varnothing} 1 = \infty,$ as $f$ is continuous and attains a minimum on any compact set. Thus, $\frac{G}{f} \not \in L^1,$ so $G \not \in A^f_K,$ as otherwise $G=hf, h \in L^1 \implies h = \frac{G}{f} \in L^1.$ We thus have shown that no open ball around any $hf \in A^f_K$ is contained in $A^f_K,$ i.e. $A^f_K$ is closed and has empty interior. Thus, $A^f_K$ is a closed nowhere dense set.

b) By the Baire category theorem, since $L^1$ is complete, a countable union of closed nowhere dense sets in $L^1$ cannot be all of $L^1.$ Denoting $A^f:=\bigcup_{K=1}^\infty A^f_K$ and $A:=\bigcup_{n=1}^\infty A^{f_n},$ we note that $A \not = L^1,$ i.e. there exists $g \in L^1$ such that $g \not \in A^{f_n}$ for all $n,$ i.e. $g$ does not satisfy $\|\frac{g}{f_n}\|_1 \leq K$ for any $K>0$ and any $n,$ i.e. $\frac{g}{f_n} \not \in L^1$ for all $n,$ as desired.

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