Spring 2019 - Problem 3

Solution.

1. Let $E_K = \set{hf \mid h \in L^1\p{\R},\ \norm{h}_{L^1} \leq K}$, and suppose $\set{h_nf}_n \subseteq E_K$ is a sequence of functions which converges to some $g$ in $L^1\p{\R}$. Since $f$ is non-zero, we see that $h_n \to \frac{g}{f} \eqqcolon h$ pointwise. Thus, by Fatou's lemma,

   $$\norm{h}_{L^1} \leq \liminf_{n\to\infty}\,\norm{h_n}_{L^1} \leq K \implies hf \in E_K.$$

   Finally, by Hölder's inequality, because $f$ is vanishes at infinity, it is bounded, which gives

   $$\norm{hf}_{L^1} \leq \norm{h}_{L^1} \norm{f}_{L^\infty} \leq K \norm{f}_{L^\infty} < \infty,$$

   so $E_K$ is a closed subset of $L^1\p{\R}$. For the second claim, let $hf \in E_K$ and $\epsilon > 0$. We need to find $g \in L^1\p{\R}$ which approximates $hf$ well in $L^1\p{\R}$ but has $\norm{\frac{g}{f}}_{L^1} > K$. For constants $0 < R < M$ which we will choose, consider

   $$g = hf\chi_{B\p{0,R}} + \p{K + 1}f\chi_{\br{M, M+1}}.$$

   We choose this form since we can approximate $hf$ very well by $g$, but still have issues in the second term after dividing by $f$. Since $hf \in L^1\p{\R}$, we may pick $R > 0$ so large so that $\norm{hf\chi_{B\p{0,R}^\comp}}_{L^1} < \epsilon$. Pick $M > R$ so large so that $\abs{f\p{t}} \leq \frac{\epsilon}{K + 1}$ for $t \geq M$. Then

   $$\begin{aligned} \norm{hf - g}_{L^1} &\leq \norm{hf_{\chi_{B\p{0,R}^\comp}}}_{L^1} + \p{K + 1}\norm{f\chi_{\br{M,M+1}}}_{L^\infty} \norm{\chi_{\br{M,M+1}}}_{L^1} \\ &\leq \epsilon + \p{K + 1} \frac{\epsilon}{K + 1} \\ &= 2\epsilon, \end{aligned}$$

   but

   $$\norm{\frac{g}{f}}_{L^1} \geq \p{K + 1}\norm{\chi_{\br{M,M+1}}} = K + 1 > K,$$

   so $g \notin E_K$. Thus, $E_K$ is a closed, nowhere dense subset of $L^1\p{\R}$.

2. Since $L^1\p{\R}$ is complete, the Baire category theorem tells us that

   $$L^1\p{\R} \neq \bigcup_{n, K \in \N} \set{hf_n \mid h \in L^1\p{\R},\ \norm{h}_{L^1} \leq K}.$$

   Thus, there exists $g \in L^1\p{\R}$, which is not in the right-hand side, i.e., for all $n, K \in \N$, we have $\frac{g}{f_n} \notin L^1\p{\R}$ or $\norm{g}_{L^1} > K$. Thus, if we fix $K$ large enough, we know that $\norm{g}_{L^1} \leq K$, which means that $\frac{g}{f_n} \notin L^1\p{\R}$ for all $n$.