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

Question

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?

Answer 1

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.

Answer 2

Suppose $f_i$ is a Cauchy sequence in $L^{p_0}+L^{p_1}$.

By definition, for every $\epsilon > 0$, there is an $N$, such that $\|f_n - f_m \|_{L^{p_0}+L^{p_1}} < \epsilon$ for any $n,m \geq N$. For concreteness, we can take $n = N$.

We can decompose $f_m - f_N = \alpha_m + \beta_m$ where $\alpha_m \in L^{p_0}$ and $\beta_m \in L^{p_1}$, and $\|\alpha_m\|_{p_0} \leq 1.1 \epsilon$ and $\|\beta_m\|_{p_1} \leq 1.1 \epsilon$. (I am using 1.1 but you can replace it by any number greater than 1.)

Now if we decompose $f_N = A_N + B_N$ where $A_N \in L^{p_0}$ and $B_N \in L^{p_1}$, we have $f_m = (A_N + \alpha_m) + (B_N + \beta_m)$ for any $m > N$.

For any $m,n > N$, we have $$ \| (A_N + \alpha_m) - (A_N + \alpha_n) \|_{L^{p_0}} = \| \alpha_m - \alpha_n \|_{L^{p_0}} \leq \| \alpha_m \|_{L^{p_0}} + \| \alpha_n \|_{L^{p_0}} \leq 2.2 \epsilon, $$

and similarly for $ \| (B_N + \beta_m) - (B_N + \beta_n) \|_{L^{p_1}} $.

Now we can do the following.

• Take $\epsilon_1 = 1$ and find a corresponding $N_1$ and decompose $f_n$ accordingly for any $n > N_1$.
• Take $\epsilon_2 = 1/2$ and find a corresponding $N_2 > N_1$ and re-decompose $f_n$ accordingly for any $n > N_2$. (We keep the decomposition of $f_{N_2}$ which we got from the previous step.)
• Take $\epsilon_3 = 1/2^2$ and find a corresponding $N_3 > N_2$ and re-decompose $f_n$ accordingly for any $n > N_3$. (We keep the decomposition of $f_{N_3}$ which we got from the previous step.)
• Et cetera, et cetera.

The end result is that we have an unambiguous decomposition of $f_i = f'_i + f''_i$ where $f'_i$ is a Cauchy sequence in $L^{p_0}$ and $f''_i$ is a Cauchy sequence in $L^{p_1}$.

Finally, we have the desired result because $L^{p_0}$ and $L^{p_1}$ are complete.