Sum of Banach Spaces is complete Let $A_1, A_2,...$ be a sequence of Banach spaces with $\|\cdot\|_n$ denoting the norm on $A_n$. Let $p\in[1,\infty)$ and
$$\sum\limits_pA_n:=\{(a_n)_{n=1}^{\infty}| a_n\in A_n \text{ and } \sum\limits_{n=1}^\infty\|a_n\|_n^p<\infty\}.$$
Show that $\sum\limits_p A_n$ is a Banach space when equipped with the norm
$$\|(a_n)\|:=\left(\sum\limits_{n=1}^\infty\|a_n\|_n^p\right)^{\frac{1}{p}}$$
I'm having trouble with what to do as well as the notation.
Since I want to prove this is a Banach space, I want to show that the space is complete.
So let $((a_{j,n}))_{n=1}^\infty$ (This is a sequence of sequences?) be a Cauchy sequence in $\sum\limits_p A_j$. Then for all $\epsilon>0$ there exists an $N\in\mathbb{N}$ such that for all $m,n\geq N$, $\|(a_{j,n})-(a_{j,m})\|=\left(\sum\limits_{j=1}^\infty\|a_{j,n}-a_{j,m}\|_j^p\right)^{\frac{1}{p}}<\epsilon$.
This implies that $\|a_{j,n}-a_{j,m}\|_j<\epsilon$ and hence $(a_{j,n})_{j=1}^\infty$ is Cauchy in $A_j$ and because $A_j$ is complete the sequence also converges to a point call it $a_n$. 
I'm not sure where to go from here. Or if I am even on the right track.
I would appreciate suggestions. 
 A: Let $(a_n)_j$ be a Cauchy sequence of sequences. 
Now, the "obvious" limit choice would be the sequence $(a_n)$ where $a_n = \lim_{j \to \infty} (a_n)_j$. To show this, you have to prove that all those $(a_n)$ exist for any $j$, and then that that is the actual limit. 
For the first part, it's enough to observe that
$$ \|(a_n)_j - (a_n)_k\|_n^p \leq \sum_{n=1}^\infty \|(a_n)_j-(a_n)_k\|^p_n < \varepsilon$$
The second part is a bit tricky. To be a bit more light on notations, we put $a^j_n = (a_n)_j$.
$$\|(a_n)_j -(a_n)\|^p = \sum_{n=1}^\infty \|a_n^j - a_n\|^p_n \leq \sum_{n=1}^\infty \|a_n^j - a_n^{k(n)}\|^p_n + \|a_n^{k(n)} - a_n\|^p_n$$
where we choose $k(n)$ such that both norms are smaller than $\displaystyle \frac{\varepsilon}{2^n}$. We can do that for any $n$ because $a_n$ is the limit of $a_n^j$.
Now,
$$\sum_{n=1}^\infty \|a_n^j - a_n^{k(n)}\|^p_n + \|a_n^{k(n)} - a_n\|^p_n \leq \varepsilon \sum_{n=1}^\infty \frac{1}{2^n} = \varepsilon$$
and we are done.
A: You are definitely on the right track and nearing a complete proof! I'm going to write $\mathcal{A}_p$ for what you're denoting by $\sum_p A_n$ (since I find the placement of the $p$ rather unsettling in your notation ;)).
If you're comfortable with the idea of a direct product of vector spaces, you can think of $\mathcal{A}_p$ (before assigning it a topology) as a vector subspace of $\prod_{n=1}^\infty A_n$, specifically, the subspace of all sequences (think "infinite tuples", if you prefer) $x = (a_k)$ such that $a_k \in A_k$ for all $k \in \mathbb{N}$ and $\|x\|_p := \sum_{n=1}^\infty \|a_n\|_n^p < \infty$. Here $\|\cdot\|_n$ denotes a norm defining the Banach topology on $A_n$ for each $n$. One checks (as you should, if you haven't already) that $\|\cdot\|_p$ defines a norm on $\mathcal{A}_p$, and then, as you've observed, it is enough to show that $\mathcal{A}_p$ is sequentially complete.
Your argument for sequential completeness is sound, although it might be cleaner to name the sequences $(a_{j,n})$. Indeed, you need to consider Cauchy sequences $(x_n)$ whose entries are themselves in $\mathcal{A}_p$. Hence you're correct that each $x_n$ should take the form $x_n = (a_{n,k})_{k=1}^\infty$ for some choice of $a_{n,k} \in A_k$ ($k \in \mathbb{N}$). That $(x_n)$ is Cauchy in $\mathcal{A}_p$ implies for all $\epsilon > 0$ there exists $N_\epsilon \in \mathbb{N}$ such that if $m,n \in \mathbb{N}$ with $m,n \geq N_\epsilon$ then $\|x_m - x_n\|_p < \epsilon$. Expanding the definition as you have shows that
$$
    \sum_{k=1}^\infty \|x_{m,k} - x_{n,k}\|_k^p = \|x_m - x_n\|_p^p < \epsilon^p,
$$
whence, as the leftmost series contains only non-negative summands, it must be that $\|x_{m,k}-x_{n,k}\|_k^p < \epsilon^p$ for all $k$. It follows that for all $k \in \mathbb{N}$, for all $\epsilon > 0$ there exists $N_\epsilon$ such that $\|x_{m,k} - x_{n,k}\|_k < \epsilon$ whenever $m,n \geq N_\epsilon$, whence, by definition, $(x_{n,k})_{n=1}^\infty$ is a Cauchy sequence in $A_k$.

Ok, so far I've just written out what you've already shown, but with a few points of clarification and some cleaner language. Going forward, I'll leave you with hints/a roadmap of how to reach the desired conclusion. So, using what we have done so far, how can we conclude from the above that $(x_n)$ has a limit $L$ in $\mathcal{A}_p$? Well, $L$ will have to be of the form $(L_k)_{k=1}^\infty$ with $L_k \in A_k$ for all $k$, and we have a good candidate for what each $L_k$ should be based on the above arguments (what is it?). Once you know that, you need to answer two more questions before you know $\mathcal{A}_p$ is complete:


*

*Is $L \in \mathcal{A}_p$? Since $L_k \in A_k$ for each $k \in \mathbb{N}$, you just need to verify $\|L\|_p < \infty$ to confirm that $L \in \mathcal{A}_p$.

*Is $L$, in fact, the limit of $(x_n)$ in $\mathcal{A}_p$? To prove this, you need to show $\|x_n - L\|_p \to 0$ as $n \to \infty$.


What you need to show to answer both of these questions is a direct computation using the norm $\|\cdot\|_p$ and familiar arguments from analysis. I can be of further assistance if needed. Good luck!

Edit in Response to Your Comment: You've now shown that $L = \lim_{n\to\infty}x_n$ should be of the form $L = (L_k)$ where for each $k \in \mathbb N$ the $k$-th entry $L_k$ is the limit of the Cauchy sequence $(x_{n,k})_{n=1}^\infty$ in $A_k$. One way of doing this, which essentially knocks out both of the steps (1. and 2. above) that I provided in my original response is the following.


*

*First, we may exploit the relationship between $L$ and the $x_n$: While we don't know the value of $\|L\|_p$ is finite right off the bat, we should expect that (1) $\|L\|_p$ should be intimately related to $\|x_n\|_p$ for large $n$ and (2) $\|x_n\|_p$ is finite for all $n$. So we can try to force $x_n$ to appear in the expression $\|L\|_p$ (or, for convenience, $\|L\|_p^p$) in the usual way: adding zero. Since $\|\cdot\|_p$ is a norm, it satisfies the triangle inequality, so $\|L\|_p \leq \|L-x_n\|_p + \|x_n\|_p$. Thus, since $\|x_n\|_p$ is finite for every value of $n \in \mathbb{N}$, the norm $\|L\|_p < \infty$ if we can show that $\|L-x_n\|_p$ for a sufficiently large choice of $n$.

*Second, we may reduce to proving $\|L-x_n\|_p \to 0$ as $n \to \infty$: Note that $\|L - x_n\|_p$ being finite for a sufficiently large choice of $n$ follows immediately if $\|L-x_n\|_p \to 0$ as $n \to \infty$. Hence the latter assertion actually implies both 1. and 2. (really, it confirms that $L \in \mathcal{A}_p$ and then, knowing that, also verifies it is the limit of $(x_n)$).

*Third, and this is the trick, to show that this limit is $0$, we need to bear in mind what might be best summarized as the Cauchy-ness of $(x_{n,k})_{n=1}^\infty$ in $A_k$ is independent of the Cauchy-ness of $(x_{n,l})_{n=1}^\infty$ in $A_l$ for $k \neq l$: In other hand-wavy words, the $\epsilon$ and $N_\epsilon$ occuring the definition of a Cauchy sequence (and the limit of a sequence) can be defined separately for each $k$. Thus, for instance, for all $k \in \mathbb{N}$, for all $\epsilon_k > 0$, there exists an $N_{k,\epsilon_k} \in \mathbb{N}$ such that for all $m,n \geq N_{k,\epsilon_k}$ we have $\|x_{m,k}-x_{n,k}\|_k < \epsilon_k$. In fact, we may assume for all $k$, for all $\epsilon_k > 0$, and for all $m,n \geq N_{k,\epsilon_k}$ that both $\|x_{m,k}-x_{n,k}\|_k < \epsilon_k$ and (since $x_{n,k} \to L_k$) $\|L_k-x_{n,k}\|_k < \epsilon_k$.

*Finally, employ the triangle inequality for $\|\cdot\|_k$ for each $k$, separately: Why all the added notation in the third bullet? Well, note that for any choice of $n$ and for any choice of $m_k \in \mathbb{N}$ (for each $k$) that
$$
    \|L-x_n\|_p = \sum_{k=1}^\infty \|L_k-x_{n,k}\|_k
\leq \sum_{k=1}^\infty \left(\|L_k - x_{m_k,k}\|_k+\|x_{m_k,k}-x_{n,k}\|_k\right).
$$
Again this follows by the triangle inequality (but now for each $\|\cdot\|_k$). So the idea is that if $n$ and $m_k$ can be chosen such that $n,m_k \geq N_{k,\epsilon_k}$ for all $k \in \mathbb{N}$, then this shows that
$$
    \|L-x_n\|_p \leq \sum_{k=1}^\infty 2\epsilon_k.
$$
Hence, for a given $\epsilon > 0$, if we select $\epsilon_k$ for each $k \in \mathbb{N}$ in such a way that $\sum_{k=1}^\infty 2\epsilon_k = \epsilon/2$, then we have shown that $\|L-x_n\|_p \leq \epsilon/2 < \epsilon$ for all $n \geq \sup_{k \in \mathbb{N}}N_{k,\epsilon_k}$, completing the proof.

So I wound up writing more of the proof than I intended, but I think it will help show you how you can reason through an argument like this. There are still a couple small but important points that I've left out that need to be addressed. Specifically,


*

*For each $\epsilon > 0$, what is a good choice of $(\epsilon_k)_{k=1}^\infty$ satisfying $\sum_{k=1}^\infty 2\epsilon_k = \epsilon/2$? There is a natural way to do this using geometric series. (Hint: Start with $\epsilon = 1$)

*Given these $\epsilon_k$, why is it true that I choose $N_{k,\epsilon_k}$ for each $k$ in such a way that $\sup_{k \in \mathbb{N}}N_{k,\epsilon_k}$ is finite? (Hint: Look carefully at and elaborate on the argument showing that $(x_n)$ Cauchy in $\mathcal{A}_p$ implies $(x_{n,k})_{n=1}^\infty$ Cauchy in $A_k$ for each $k$.)
Hope this clarifies things!
