Subsequence Properties and Lp Spaces

Question

Exercise:

Let $E$ be a measurable set in $\mathbb{R}$ under Lebesgue measure, and let $1<p<\infty$. Suppose $\{f_n\}$ is a bounded sequence in $L^p(E)$ and $f \text{ belongs to } L^p(E) $. Consider the following 4 properties:

(i) $\{f_n\} \rightarrow f$ pointwise a.e. on $E$.

(ii) $\{f_n\} \rightharpoonup f$ in $L^p(E)$.

(iii) $\{\|f_n\|_p\}$ converges to $\{\|f\|_p\}$.

(iv) $\{f_n\} \rightarrow f$ in $L^p(E)$.

If $\{f_n\}$ possesses two of these properties, does a subsequence posses all four properties?

I'm not quite sure where to start. My intuition tells me that that this is false. But I'm not sure as to which properties to start with. Any thoughts would be helpful.

Note that you can use \{ and \} for braces in LaTeX code. — Caleb Stanford, May 10 '14 at 22:34
No. It should say "measurable set." This is not in terms of general measure theory. — user0430, May 10 '14 at 22:47
A measurable set is a measure space, and vice versa. But do you mean that you are considering only Lebesgue measure? — Caleb Stanford, May 10 '14 at 22:47
You know that (iv) implies (iii) and (ii), as well as (i) for a subsequence. So if one of the two properties is (iv), a subsequence has all properties. That leaves $\binom{3}{2} = 3$ configurations to check. Some of them imply (iv). — Daniel Fischer, May 10 '14 at 22:54

score 1 · Answer 1 · edited Apr 13 '17 at 12:19

(Not a full answer.)

First of all, as Daniel Fischer notes in the comments,

$L^p$ convergence implies convergence in measure. Convergence in measure implies there is some subsequence converging pointwise.
Strong ($L^p$) convergence implies weak convergence.
$\newcommand{\norm}[1]{\left\| #1 \right\|_p} 0 \le \left| \norm{f_n} - \norm{f} \right| \le \norm{f_n - f}$ by Triangle inequality.

Therefore, if you want to find a counterexample (which you do), you should steer clear of $L^p$ convergence. But there is more:

Convergence a.e. and of norms implies strong convergence. (Uses $p < \infty$) (Weaker form: Pointwise convergence and bounded norms imply weak convergence)

Therefore there are two cases left: Convergence of norms + weak convergence, and Pointwise a.e. convergence + weak convergence.

Convergence of norms plus weak convergence implies norm-convergence for $1 < p < \infty$ by the uniform convexity of $L^p$, by the way. — Daniel Fischer, May 11 '14 at 00:13

score 0 · Answer 2 · answered Aug 27 '14 at 20:08

if (iv) is one of the two assumptions, then all four are true for some subsequence.

To check the other three case,

(i)+(ii), for $1<p<\infty$, $f_n$ bounded in $L^p$, pointwise a.e convergence implies weak convergence, thus the only asumption we have is basically $f_n \rightarrow f$ pointwise a.e. This would make the statement false, take $f_n(x) = \chi_{[n,n+1]}$ as a counter example.

Both (i)+(iii) and (ii) + (iii) implies strong convergence; the statement is true.

It would be a good exercise to check for the case when $p=1$.

Subsequence Properties and Lp Spaces

2 Answers2