I've been trying to attempt the following question, but can't quite seem to make any progress on it:
Suppose that $f_n$ is a sequence of Lebesgue measurable function such that $$ \sup_{n \geq 1} \int_{1}^{\infty} f_n^2(x) dx \leq 1 $$ and suppose $f_n$ converges to some function $f$ pointwise. Show that $$ \lim_{n \to \infty} \int_1^{\infty} \frac{f_n(x)}{x} dx = \int_{1}^{\infty} \frac{f(x)}{x} dx. $$
I've used some limiting arguments and the Holder inequality to show that $f_{n}(x)/x$ is integrable, however, this hasn't really gotten me too far. Does anyone have any suggestion for how to proceed?
By Fatou's Lemma, $f$ is $L^{2}$ with $||f||_{2}\leq$1. For $\int_{1}^{\infty}f^{2}\leq\liminf_{n}\int_{1}^{\infty}f_{n}^{2}\leq1$. Note that $f_{n}(x)$ and $\frac{1}{x}$ are $L^{2}[1,\infty)$, so $\frac{f_{n}(x)}{x}$ is $L^{1}$ by Holder inequality (or Cauchy inequality). Define $g_{n}=|f_{n}-f|,$ then $||g_{n}||_{2}\leq2$ and $g_{n}\rightarrow0$ pointwisely. We assert that: $\int_{1}^{\infty}\frac{g_{n}(x)}{x}dx\rightarrow0$.
Sketch of proof: Let $\epsilon>0$ be given. Choose $M>0$ such that $\int_{M}^{\infty}\frac{1}{x^{2}}dx<\epsilon^{2}$. By Cauchy inequality, $\int_{M}^{\infty}\frac{g_{n}(x)}{x}dx\leq||g_{n}||_{2}\cdot\epsilon\leq2\epsilon$.
$\int_{1}^{M}\frac{g_{n}(x)}{x}dx\leq\int_{1}^{M}g_{n}(x)dx\rightarrow0$ as $n\rightarrow\infty$.
For the last step, we recall two facts:
Let $(X,\mathcal{F},\mu$) be a measure space with $\mu(X)<\infty$. Let $p\in(1,\infty)$. Let $\{f_{n}\}\subseteq L^{p}$. If $\sup_{n}||f_n||_{p}<\infty$, then the family $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly integrable.
If a sequence $\{f_{n}\}$ is uniformly integrable and $f_{n}\rightarrow0$ pointwisely a.e., then $\int f_{n}\rightarrow0$ (a theorem due to Vitali).
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For the sake of self-containedness, I add the definition of uniform integrability and proofs for the two claims in above.
We recall the definition of uniform integrability: Let $(X,\mathcal{F},\mu)$ be a measure space with $\mu(X)<\infty$. Let $\{f_{i}\mid i\in I\}$ be a family of real-valued, measurable functions (possibly uncountably many) defined on $X$. We say that the family is uniformly integrable if for each $\epsilon>0$, there exists $c>0$ such that for each $i\in I$, $\int_{[|f_{i}|\geq c]}|f_{i}|\,d\mu<\epsilon$.
Proposition 1: Let $(X,\mathcal{F},\mu)$ be a measure space with $\mu(X)<\infty$. Let $p\in(1,\infty)$. Let $\{f_{i}\mid i\in I\}$ be an arbitrary family of real-valued, measurable functions. If $\sup_{i}||f_{i}||_{p}=M<\infty$, then the family $\{f_{i}\mid i\in I\}$ is uniformly integrable.
Proof of Proposition 1: By replacing $f_{i}$ with $|f_{i}|$, we may assume that $f_{i}$ is non-negative. Note that if $f\geq0$ with $||f||_{p}\leq M$, we have: $$ M^{p}\geq\int f^{p}\geq\int_{[f\geq c]}f^{p}\geq c^{p}\mu([f\geq c]). $$ That is, $\mu([f\geq c])\leq\frac{M^{p}}{c^{p}}$. Let $q\in(1,\infty)$ be such that $\frac{1}{p}+\frac{1}{q}=1.$ By Holder inequality, $$ \int_{[f\geq c]}f=\int1_{[f\geq c]}f\leq||f||_{p}\left(\mu([f\geq c])\right)^{\frac{1}{q}}\leq M\cdot\frac{M^{\frac{p}{q}}}{c^{\frac{p}{q}}}. $$ Let $\epsilon>0$ be given. Choose $c>0$ such that $M\cdot\frac{M^{\frac{p}{q}}}{c^{\frac{p}{q}}}<\epsilon$. Then $\int_{[f_{i}\geq c]}f_{i}\,d\mu<\epsilon$ for any $i\in I$.
Proposition 2: Let $(X,\mathcal{F},\mu)$ be a measure space with $\mu(X)<\infty$. Let $\{f_{n}\mid n\in\mathbb{N}\}$ be uniformly integrable. Suppose that $f_{n}\rightarrow0$ pointwisely a.e, then $\int f_{n}\rightarrow0$.
Proof of Proposition 2: By replacing $f_{n}$ with $|f_{n}|$, we may assume that $f_{n}\geq0$. Let $\epsilon>0$ be arbitrary. Choose $c>0$ such that $\int_{[f_{n}\geq c]}f_{n}<\epsilon$ for each $n$. Since $\mu(X)<\infty$, constant functions are integrable. Note that $f_{n}1_{[f_{n}<c]}\leq c$. By Lebesgue Dominated Convergence Theorem, we have $\int f_{n}1_{[f_{n}<c]}\rightarrow0$. Therefore, there exists $N$ such that $\int f_{n}1_{[f_{n}<c]}<\epsilon$ whenever $n\geq N$. For any $n\geq N$, we have $\int f_{n}=\int_{[f_{n}\geq c]}f_{n}+\int_{[f_{n}<c]}f_{n}<2\epsilon$. That is, $\lim_{n}\int f_{n}=0$.