a L^p space problem

liyi

Suppose $f_k\in L^p([a,b]) (1\leq p \leq \infty)$ and
$\sum\limits_{k=1}^\infty \|f_k\|_p < \infty$ .
Show that there exists $f\in L^p([a,b])$ which satisfies the following two conditions:
(i) $\sum_{k=1}^\infty f_k(x)=f(x),\quad \mbox{ a.e. } x\in [a,b]$ ;
(ii) $\sum_{k=1}^N f_k(x)$ converges to $f$ in $L^p([a,b])$ sense.

**Posted:** Thu Jan 27, 2005 12:30 am

grobber

The condition states that $g_n=\sum_{i=1}^n f_i$ is a Cauchy sequence in $L^p$ , so it must have a limit in $L^p$ , as $L^p$ is complete. The limit $f$ satisfies (ii) by construction, and the following is a well known theorem, and the proof is not tough at all (standard boring argument Smile

): a Cauchy sequence in $L^p$ has a subsequence converging to the limit of the sequence a.e.

**Posted:** Thu Jan 27, 2005 9:41 am

liyi

Yes, you are right. (L^p convergence implies convergence in measure, hence (Riesz theorem) there is a subsequence which is convergent a.e.)
but it is just a subsequence... we want the convergence of the whole sequence...

If $|f_k|\geq 0$ for all $k$ , the conclusion follows easily..
but if not....

**Posted:** Thu Jan 27, 2005 5:21 pm

Kent Merryfield

You can't have a.e. convergence of the whole sequence. I don't have time right now to write out the "rotating tooth" example of a sequence of functions that converges in norm (and in measure) but not pointwise - but that example applies here.

**Posted:** Thu Jan 27, 2005 5:26 pm

grobber

But maybe the given condition is strictly stronger than the mere convergence in norm of $g_n=\sum_{i=1}^n f_i$ , so maybe pointwise convergence holds in this case.

**Posted:** Thu Jan 27, 2005 5:28 pm

liyi

**Posted:** Thu Jan 27, 2005 5:33 pm

Kent Merryfield

And any sequence can be turned into a series simply by taking differences of adjacent elements of the sequence as the terms of the series.

**Posted:** Thu Jan 27, 2005 7:01 pm

liyi

You are right.
What I said is: If $f_k\geq 0$ , then $\sum f_k$ is convergent almost everywhere, because $g_n = \sum_{k=1}^n f_k$ is increasing for a fixed $x$ .

It is easy to prove that
$\{f_n(x)\}$ converges to $f$ in measure, and $\{f_n(x)\}$ is increasing for $x$ . Then it converges to $f$ almost everywhere.

**Posted:** Thu Jan 27, 2005 9:20 pm