# For $1\le p < +\infty$ $L^p$ is a Banach space: Real and abstract analysis, Hewitt - Stromberg

I have some doubts about the proof of this theorem. From time to time I will put my justification.

For $$1\le p < +\infty$$, $$L^p$$ is a Banach space

Let $$(f_n)_n$$ be a Cauchy sequence in $$L^p$$, i.e., $$(f_n)$$ has the property that $$\lim_{m,n\to \infty} \lVert f_n-f_m\rVert_p=0.$$ We can find a subsequence of $$(f_n)$$ that does converge $$\mu-$$almost averywhere. In fact, chosse $$(f_{n_k})_k$$ as any subsequence of $$(f_n)$$ such that $$n_1 and $$\sum_{k=1}^\infty \lVert f_{n_{k+1}}-f_{n_k}\rVert_p=\alpha<\infty$$ This is possiblr: e.g., we can select increasing $$n_k's$$ such that $$\lVert f_m-f_{n_k}\rVert_p<2^{-k}$$ for all $$m\ge n_k$$.

My justification for this fact: For the definition of Cauchy sequence we have that $$(\forall k\in\mathbb{N})\quad (\exists n(k)\in\mathbb{N})\quad (\forall m,n \ge n(k))\quad \lVert f_m-f_n \rVert_p<2^{-k}$$ I choose $$n_1\ge n(1)$$, then from the above definition results $$\lVert f_m - f_{n_k}\rVert_p<2^{-1}$$ for all $$m\ge n_1$$.

Now, I choose $$n_2:=\max\{n_1+1, n(2)\}$$, then $$n_2>n_1$$ and in particular $$n_2\ge n(2)$$ then from the definition we have that $$\lVert f_m-f_{n_2}\rVert_p<2^{-2}$$ for all $$m\ge n_2$$ and since $$n_2>n_1\ge n(1)$$ from the previous step results that $$\lVert f_{n_2}-f_{n_1}\lVert _p<2^{-1}$$. In this way we make an increasing sequence $$n_k$$ such that $$\lVert f_{n_{k+1}}-f_{n_k}\rVert_p<2^{-k}\quad \forall k\in\mathbb{N}.$$ Question 1 Is this enough to justify what the author said?

Now, we define $$g_k= \lvert f_{n_1}-f_{n_0}\rvert+ \lvert f_{n_2}-f_{n_1}\rvert+\cdots+\lvert f_{n_{k+1}}-f_{n_k} \rvert,\quad \text{for}\ k=1,2,3,\dots\quad f_{n_0}=0$$. It is clear that from the Minkowski's inequality we have that $$\begin{eqnarray*} \lVert g_k^p \rVert_1=\lVert g_k\rVert_p^p&=&\left(\left\rVert \sum_{j=0}^k \lvert f_{n_{j+1}}-f_{n_j}\rvert\right\rVert_p\right)^p\\ &\le&\left ( \lVert f_{n_1}\rVert_p+\sum_{j=1}^k\lVert f_{n_{j+1}}-f_{n_j}\rVert_p\right)\le (\lVert f_{n_1}\rVert_p)+\alpha)^p<\infty \end{eqnarray*}$$

Let $$g=\lim_{k\to\infty} g_k$$. By B. Levi's theorem and the above estimate we have $$\int g^p\;d\mu=\int\lim_{k\to\infty}g_k^p\;d\mu=\lim_{k\to\infty}\int g_k^p\;d\mu<\infty.$$ Hence $$g$$ is in $$L^p$$; i.e $$\int\left [\sum_{j=0}^\infty \lvert f_{n_{j+1}}-f_{n_j}\rvert\right]^p\;d\mu<\infty$$ The nonnegative integrand above must be finte a.e, and so the series $$\sum_{j=1}^\infty \lvert f_{n_{j+1}}(x)-f_{n_j}(x)\rvert$$ converges a-e and therefore also the series $$\sum_{j=0}^\infty(f_{n_{j+1}}(x)-f_{n_j}(x))$$ converges a.e. The $$k$$ th partial sum of this serie is $$f_{n_{k+1}}(x)$$ and so the sequence $$(f_{n_k}(x))$$ converges to a complex number $$f(x)$$ for all $$x\in A$$, where $$A\in\mathcal{A}$$ and $$\mu(X\setminus A)=0$$. Define $$f(x)$$ as $$0$$ for all $$x\in X\setminus A$$.

We will show that $$f$$ is the limit in $$L^p$$ of the sequence $$(f_n)$$, and this will of course prove that $$L^p$$ is complete in the metric induce by the $$L^p-$$norm. Given $$\varepsilon > 0$$, then from definition of Cauchy sequence exists $$n(\varepsilon)\in \mathbb{N}$$ such that $$\lVert f_s -f_t\rVert_p <\varepsilon$$ for all $$s,t\ge n(\varepsilon)$$, so let $$l$$ so large such that $$\lVert f_s - f_t\rVert_p<\varepsilon\quad\text{for}\quad s, t\ge n_l$$ (practically $$n_l> n(\varepsilon)$$, this is possible, we remember that $$\{n_k\}$$ is increasing). Then for $$k\ge l$$ and $$m>n_l$$, we have $$\lVert f_m-f_{n_k}\rVert_p<\varepsilon.\tag1$$ By Fatou's lemma we have $$\begin{eqnarray*} \int \lvert f-f_m \rvert^p\;d\mu&=&\int\liminf_{k\to\infty}\lvert f_{n_k}-f_m\rvert^p\;d\mu\\ &\le&\liminf_{k\to\infty}\lVert f_{n_k}-f_m\rVert_p^p\color{red}{\le} \varepsilon^p \end{eqnarray*}$$

Question 2. The inequality in red comes from $$(1)$$, but I don't know how to see it formally, could you explain it to me?

Thus for each $$m>n_l$$, the function $$f-f_m$$ is in $$L^p$$ and then chosen any $$m>n_l$$ we have that $$f=(f-f_m)+f_m$$ is in $$L^p$$. This implies that $$f$$ is in $$L^p$$ and we have done. Thanks for the attention.

• For your second question: Do you know that $\liminf$ is monotone, i.e. $a_n \leq b$ for all $n$ implies $\liminf_n a_n \leq b$? Commented Jun 5 at 7:44
• @HyperbolicPDEfriendYes, so you say that since for a fixed $m>n_l$ we have that $$\lVert f_m -f_{n_k}\rVert_p<\varepsilon,\quad\forall k\ge l$$ then $\liminf_k \lVert f_m -f_{n_k}\rVert_p^p\le \varepsilon^p$? Commented Jun 5 at 8:02
• @HiberpolicPDEfriendIs the first justification correct? Commented Jun 5 at 9:52
• It seems correct to me. Commented Jun 5 at 10:02

For your first question, what you did to find the first two terms was fine. Since you started the argument, you may as well finish it and describe how to find all of the terms. You can find a strictly increasing sequence $$(n_{k})_{k\in\mathbb{N}}$$ of natural numbers such that for any $$k\in\mathbb{N}$$, if $$m\in\mathbb{N}$$ with $$m\geq n_{k}$$, then $$\|f_{m} - f_{n_{k}}\|_{p} < 2^{-k}$$. This can be done recursively. You already described how to choose $$n_{1}$$. Given $$k\in \mathbb{N}$$ and supposing that $$n_{1}, \ldots , n_{k}$$ have been chosen with the desired properties, set $$n_{k+1} := \max\{n_{k}+1, n(k+1)\}$$. Then $$n_{j} < n_{k+1}$$ for all $$j\in \{1, \ldots , k\}$$. If $$m\in\mathbb{N}$$ with $$m\geq n_{k+1}$$, then $$m\geq n(k+1)$$ and $$n_{k+1} \geq n(k+1)$$, so $$\|f_{m} - f_{n_{k+1}}\|_{p} < 2^{-(k+1)}$$. By recursive definition, a sequence $$(n_{k})_{k\in\mathbb{N}}$$ is obtained with the desired properties. That is what you should add if you want to complete the recursive definition argument.
For your second question, recall that if $$(a_{n})_{n\in\mathbb{N}}$$ is a bounded sequence of real numbers, then $$\liminf_{n\to\infty} a_{n}$$ is equal to $$\sup \{ \inf \{a_{k} : k\in\mathbb{N} , n\leq k \} : n\in\mathbb{N} \}$$. So if you find some $$b\in\mathbb{R}$$ such that $$\inf\{a_{k} : k\in\mathbb{N} , n\leq k \} \leq b$$ for all $$n\in\mathbb{N}$$, it follows that $$b$$ is an upper bound for $$\inf\{\{a_{k} : k\in\mathbb{N} , n\leq k \} : n\in\mathbb{N}\}$$ and so $$\liminf_{n\to\infty} a_{n} \leq b$$. However, in your case, you have something much stronger. You have some $$l\in\mathbb{N}$$ such that if $$k\in\mathbb{N}$$ and $$k\geq l$$, then $$\|f_{n_{k}} - f_{m}\|_{p}^{p} < \varepsilon^{p}$$. As a result, because $$\max\{\ell, l\} \geq l$$ for any $$\ell\in\mathbb{N}$$, $$\inf \{\|f_{n_{k}} - f_{m}\|_{p}^{p} : k\in\mathbb{N}, \ell\leq k \} < \varepsilon^{p}$$ for all $$\ell \in \mathbb{N}$$. It then follows from what was stated above that $$\liminf_{k\to\infty} \|f_{n_{k}} - f_{m}\|_{p} \leq \varepsilon^{p}$$.