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Suppose that $\{f_n\}_{n=1}^{\infty}\in L^2(R)$ is a sequence that converges to 0 in $L^2$ norm; in other words, $$ \left(\int_{-\infty}^{\infty}|f_n|^2dx\right)^{1/2} \to 0. $$ Prove that there exists a subsequence ${f_{n_k}}\to 0$ almost everywhere.

I feel I am stuck for a long time, can someone tell me how to prove it? I am thinking about how to construct the subsequence, but still have no idea how to find the subsequence.

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  • $\begingroup$ Can you prove convergence in $L^p$ implies convergence in probability? Then, use this. $\endgroup$ – Clement C. Sep 2 '16 at 14:37
  • $\begingroup$ But this is not a consequence of the completeness theorem for Lebesgue spaces? $\endgroup$ – user288972 Sep 2 '16 at 14:42
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Since $(f_n)$ is a Cauchy sequence, consider a subsequence $(f_{n_k} )$ denoted by $(f_k)$ such that

$$\|f_{k+1}-f_k\|_2 \leq \frac{1}{2^k}$$

Let $$g_n(x)=\sum_{k=1}^n|f_{k+1}(x)-f_k(x)|$$ so that $\|g_n\|_2 \leq 1$, by the monotone convergence theorem, $g_n(x)$ tends to a finite limit, say $g(x)$, a.e. $R$, with $g \in L^2$. On the other hand, for $m \geq n \geq 2$ we have that

$$|f_m(x)−f_n(x)| \leq |f_m(x)−f_{m−1}(x)|+· · ·+|f_{n+1}(x)−f_n(x)| \leq g(x)−g_{n−1}(x).$$

It follows that a.e. on $R$, $f_n(x)$ is Cauchy and converges to a finite limit, say $f(x).$ We have a.e. on $R$,

$$ |f (x) − f_k(x)| ≤ g(x), \quad k \geq 2$$

and in particular $f \in L^2$. Note that by dominated convergence we know that $f_k \rightarrow f$ in $L^2$ and thus $f = 0$ a.e.

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$L^2$ convergence implies convergence in measure. Thus $\mu(|f_n|>\epsilon)\to0$ as $n\to\infty$.

So taking a sequence $\epsilon_n$ to be $\dfrac{1}{n}$, you have there exists $N_n$ (where $N_n$ can be taken such that $N_1<N_2<N_3<...)$ for which $\mu(|f_{N_n}|>\dfrac{1}{n})<\dfrac{1}{2^n}$. Now use Borel Cantelli to show that this subsequence $\{f_{N_n}\}$ is the one you desire.

EDIT: Please do fill in the details. It's important.

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This is a consequence of the completeness theorem for the Lebsgue spaces, i.e. $L^p(\Omega)$ is a Banach space.

$Theorem$. If $\lbrace f_n \rbrace \subset L^p(\Omega)$ is a Cauchy sequence, then there is $f \in L^p(\Omega)$ such that $||f_n - f||_p \rightarrow 0$. More precisely there is $\lbrace f_{n_k} \rbrace$ and a function $F \in L^p(\Omega)$ $F(x) \geq 0$ such that

$(1)$ $|f_{n_k}(x)| \leq F(x)$ a.e in $\Omega$

$(2)$ $f_{n_k}(x) \rightarrow f(x)$ a.e. in $\Omega$.

In the proof of this theorem you built this subsequence.

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  • $\begingroup$ But isn't that basically invoking a stronger theorem than necessary to solve a standard question? $\endgroup$ – Clement C. Sep 2 '16 at 14:48
  • $\begingroup$ I do not understand your comment. This theorem in functional analysis should be well known to all. $\endgroup$ – user288972 Sep 2 '16 at 14:50
  • $\begingroup$ It should, after taking the course. But while taking the course, it's not the first result shown -- at least when I took it, we proved that "convergence in $L^p$ implies convergence in probability implies convergence a.e. of a subsequence" quite before proving completeness of $L^p$ spaces. And given the question, which asks to prove this very standard result, I would wager the OP is not done with the course yet... $\endgroup$ – Clement C. Sep 2 '16 at 14:52
  • $\begingroup$ This depends on how they are structured the various functional analysis courses. For me it was one of the first things that we have shown. $\endgroup$ – user288972 Sep 2 '16 at 14:54
  • $\begingroup$ (In addition, it's a consequence of the proof of the completeness rather than of the completeness itself -- not really the same point I'm making above, but just knowing that "$L^p$ spaces are complete" does not trivially lead to the answer to the OP's question) $\endgroup$ – Clement C. Sep 2 '16 at 14:54
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Another solution (Suitable for all $p\geq 1$):

$f_n \xrightarrow{L^p} 0$ hence we can extract a subsequence such that $ \| f_{n_k} \| < 2^{-k}$

We then define $g=\sum_{k=1}^\infty |f_{n_k}|$ and notice that $\|g\| \leq 1$. This implies that $g$ is finite almost everywhere, or in other words: the sum $\sum_{k=1}^\infty f_{n_k}$ converge absolutely a.e., which in turn implies $f_{n_k}(x)\xrightarrow{\text{a.e.}} 0$.

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