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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.

gt6989b
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python3
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4 Answers4

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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.

Landon Carter
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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.

user288972
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  • But isn't that basically invoking a stronger theorem than necessary to solve a standard question? – Clement C. Sep 02 '16 at 14:48
  • I do not understand your comment. This theorem in functional analysis should be well known to all. – user288972 Sep 02 '16 at 14:50
  • 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... – Clement C. Sep 02 '16 at 14:52
  • 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. – user288972 Sep 02 '16 at 14:54
  • (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) – Clement C. Sep 02 '16 at 14:54
  • In fact, I specified the statement, and that in the proof of this theorem we build this subsequence. – user288972 Sep 02 '16 at 14:56
  • I know, I have seen. I believe there is some communication barrier here: all I'm saying is that the statement that the OP asks to prove is strictly weaker than the one you use to prove it, and that arguably the OP may not have seen it yet. Now, that last part is just a guess based on my own experience, and that is for the OP herself to clarify. – Clement C. Sep 02 '16 at 14:58
  • This we can not know, in any case the questioner can go to study the proof of this theorem. My answer may also be considered as a suggestion. – user288972 Sep 02 '16 at 15:01
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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.

ferangares
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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 $f_{n_k}(x)\xrightarrow{\text{a.e.}} 0$.

Ranc
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