# Completeness of the sum of two $L^p$ spaces

Q:Suppose $L^{p_0}+L^{p_1}$ is defined as the vector space of measurable functions $f$ on a measure space $X$,that can be written as a sum $f=f_0+f_1$ with $f_0\in L^{p_0}$ and $f_1\in L^{p_1}$. Consider $$\|f\|_{L^{p_0}+L^{p_1}}=inf\{\|f_0\|_{L^{p_0}}+\|f_1\|_{L^{p_1}}\},$$ where the infimum is taken over all decomposition $f=f_0+f_1$ with $f_0\in L^{p_0}$ and $f_1\in L^{p_1}$. Show that $\|\cdot\|_{L^{p_0}+L^{p_1}}$ is a norm, and that $L^{p_0}+L^{p_1}$ with this norm is a Banach space.

It is just an exercise on the functional analysis of Elias Stein. It is not difficult to prove that it is a norm. But I wonder how to deal with the completeness. If I start from the decomposition of each $f_n$ of a Cauchy sequence from the infimum. I can't get back to the norm of the difference $f_n-f_{n+p}$,and similarly if I start from the difference. I'm so confused now. May someone help me?

• maybe try to find an almost everywhere convergent subsequence for all the $f_0^n$s and then another subsequence to have that for the $f_1^n$s, too and go on from there? (just an idea...) – Max Oct 19 '15 at 15:38
• @Max Maybe that can do.I'll try it. – T.Jassen Oct 19 '15 at 16:00

Often (also in this case), the easiest way to show that a norm ed vector space is complete is not to use the definition, but to use the following characterization of completeness:

A normed vector space $X$ is complete iff for each sequence $(x_n )$ with $\sum_n \|x_n\|<\infty$, the series $\sum_n x_n$ converges (in $X$!).

If you need further details, I will happily provide them, but I think you should first try it yourself!

EDIT: Here are some ideas for proving the above equivalence. You still have to fill in some gaps. For the "only if" direction, show that the sequence of partial sums is Cauchy. For the "if" direction, proceed as follows:

1. Show that a Cauchy sequence converges of and only if it has a convergent subsequence.

2. Let $(y_n)_n$ be a Cauchy sequence. Show that there is a subsequence (which I call $(z_n)_n$) with $\| z_{n+1}-z_n\|<2^{-n}$ for all $n$. By step 1, it suffices to show that $(z_n )_n$ converges.

3. Let $x_n := z_{n+1}-z_n$ and note $\sum_n \|x_n\|<\infty$. By assumption, this shows that the limit $$\sum_n x_n =\lim_N \sum_{n=1}^N x_n = \lim_N \sum_{n=1}^N (z_{n+1}-z_n)=\lim_N z_{N+1}-z_1$$ exists in $X$.

4. Conclude the proof.

• It becomes a so easy question if this equivalence is satisfied.But can you tell me how to prove the 'if' part in this equivalence? – T.Jassen Oct 19 '15 at 17:47
• @T.Jassen: I added a proof sketch. – PhoemueX Oct 19 '15 at 20:57
• Great! I understand it.But why such a strong theorem hardly appears in a book?(I've only seen some proof using it implicitly.Ok maybe just I've read so little.) – T.Jassen Oct 19 '15 at 23:26