# Sufficiency in the proof that $L^p(\mu)$ is complete

In the proof that $$L^p(\mu)$$ is complete for $$p\in[1,\infty]$$ (as done in Saxe, Theorem 3.21 or in Folland, Theorem 6.6, the latter of which is outlined here) we make use of the following completeness criterion:

Lemma: Let $$(X,\|\cdot\|_X)$$ be a normed vector space and consider a sequence $$(x_n)_{n\in\mathbb N}$$ in $$X$$. If $$\sum_{i=1}^\infty x_i$$ converges whenever $$\sum_{i=1}^\infty \|x_i\|_X$$ converges, then $$(X,\|\cdot\|_X)$$ is complete.

In order to show completeness, we start with a Cauchy sequence $$(f_k)_{k\in\mathbb N}$$ in $$L^p(\mu)$$ and we aim to show, with the help of the above result, that $$(f_k)_{k\in\mathbb N}$$ converges in $$L^p(\mu)$$. In particular, we suppose that Cauchy $$(f_k)_{k\in\mathbb N}$$ is such that $$\sum_{i=1}^\infty \|f_k\|_p$$ converges. The authors then go to show that the corresponding series converges, which yields the result.

What I don't understand is why is this sufficient to show that any Cauchy sequence of elements converges in $$L^p(\mu)$$? In considering those Cauchy sequences $$(f_k)_{k\in\mathbb N}$$ which satisfy that $$\sum_{i=1}^\infty \|f_k\|_p$$ converges, are we not restricting our attention to a select few Cauchy sequences? Why is focusing on this subset of Cauchy sequences enough?

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I had thought, perhaps, that there was a relation between Cauchy sequences and their corresponding series' converging absolutely, but that has been indicated to not be true here.

• I think you completely misunderstood the proof, but I am not sure where exactly. The lemma translates "every Cauchy sequence converges" to "every absolutey convergent series converges", and that's what you prove to establish completeness of $L^p$. – MaoWao Mar 14 '19 at 11:30
• May be duplicated for math.stackexchange.com/questions/2388647/… – S. Maths Mar 14 '19 at 11:39
• @S.Cho, similar but not a duplicate; my question is specifically on the relation to considering Cauchy sequences and sufficiency. – Jeremy Jeffrey James Mar 14 '19 at 11:54

Suppose that in some normed space $$X$$ the following condition holds:

Whenever $$\sum_{n=1}^{\infty}\|x_n\|$$ converges, the series $$\sum_{n=1}^{\infty}x_n$$ converges in $$X$$.

Then, actually, $$X$$ is complete with respect to the given norm.

Now let's take any Cauchy sequence $$\{x_n\}_{n=1}^{\infty}$$ in $$X$$. Our aim is to use the condition above, in order to prove that $$\{x_n\}_{n=1}^{\infty}$$ is convergent. Note that we are not claiming that $$\sum_{n=1}^{\infty}\|x_n\|<\infty$$ for the specific Cauchy sequence in hand. Indeed, this may not be the case. But we are claiming that we can use the above condition to prove that $$\{x_n\}_{n=1}^{\infty}$$ converges. We do this as follows. Pick $$k_1$$ such that for every $$n,m\geq k_1$$, we have $$\|x_n-x_m\|<\frac{1}{2}$$. Next pick $$k_2>k_1$$ such that for every $$n,m\geq k_2$$, we have $$\|x_n-x_m\|<\frac{1}{4}$$. Continue inductively in this way. Put $$y_i=x_{k_{i+1}}-x_{k_i}$$. The series $$\sum_{i=1}^{\infty}y_i$$ converges absolutely, because $$\|y_i\|<2^{-i}$$. Also, $$\sum_{i=1}^my_i=x_{k_2}-x_{k_1}+\cdots + x_{k_{m+1}}-x_{k_m}=x_{k_{m+1}}-x_{k_1}$$ The condition above implies that $$\sum_{i=1}^{\infty}y_i$$ converges in $$X$$, hence the sequence $$x_{k_{m+1}}$$ converges. Being a subsequence of a Cauchy sequence, however, this implies that the original sequence $$\{x_n\}_{n=1}^{\infty}$$ converges too, because it is well known that if a Cauchy sequence has a convergent subsequence, then it converges as well.

What is usually done, and what at least Folland does (I don't have the other book), is consider an arbitrary sequence such that $$\sum_i \|x_i\| < \infty$$ and shows that $$\sum_{i=1}^n x_i$$ converges in $$L^p(\mu)$$ as $$n \to \infty$$. By the Lemma you state, this shows $$L^p(\mu)$$ is complete. One never considers Cauchy sequences explicitly here.

The consideration of Cauchy sequences is hidden in the Lemma, since given a Cauchy sequence $$(y_i)$$, one can construct an associated absolutely convergent series. See e.g. the proof given here.

• "One never considers Cauchy sequences explicitly here" - I see I should have gone with Folland from the outset; the other source I indicated specifies that the sequence we take be Cauchy, hence my confusion. – Jeremy Jeffrey James Mar 14 '19 at 11:48