# Prove normed vector space is a Banach space

I'm having an issue with a problem about proving a normed vector space is a Banach space. First, I'll just state the introduction to the problem:

Let $(V,\|\cdot\|_V)$ and $(W,\|\cdot\|_W)$ be complex normed vector spaces. Define the product $V\oplus W$ of the vector spaces by giving the set

$$V\oplus W=\{(v,w)|v\in V,w\in W\}$$

the obvious entrywise defined vector space structure:

$$(x,y)+(v,w):=(x+v,y+w)$$ $$\alpha(v,w):=(\alpha v,\alpha w)$$

for $x,v\in V,y,w\in W$, and $\alpha \in \mathbb{C}$. The norm used with this vector space is

$$\|(v,w)\|_1=\|v\|_V+\|w\|_W,\text{ for }v\in V,w\in W$$

The problem is the following:

Now assume that $(V,\|\cdot\|_V)$ and $(W,\|\cdot\|_W)$ are Banach spaces. Prove that $V\oplus W$ equipped with the norm $\|\cdot\|_1$ is a Banach space.

I know that for $V\oplus W$ to be a Banach space that means I need to show that every Cauchy sequence in $V\oplus W$ is convergent. The problem hints that I consider the Cauchy sequence $\{(v_k,w_k)\}_{k=1}^{\infty}$ in $V\oplus W$.

I'm not that sure how I would show the Cauchy sequence is convergent. I know that since $\{(v_k,w_k)\}_{k=1}^{\infty}$ is a Cauchy sequence that means for each $\epsilon>0$ there exists some $N\in \mathbb{N}$ such that

$$\|(v_k-v_l,w_k-w_l)\|\leq\epsilon\text{ whenever }k,l\geq N$$

I'm pretty sure it's a combination of that and the fact that both $(V,\|\cdot\|_V)$ and $(W,\|\cdot\|_W)$ are Banach spaces (since that means every Cauchy sequence is convergent in $V$ and $W$) but I can't quite figure it out. Any help/hints are be highly appreciated!

• I replaced $\otimes$ with $\oplus$, as the notation $\otimes$ is universally reserved for the tensor product of vector spaces, which is quite different than the structure you are using. – Aweygan May 24 '18 at 18:55

You can almost mimic the proof that $(\mathbb R^2, \Vert \cdot \Vert_1)$ is a Cauchy space for the norm $\Vert (x,y)\Vert_1 = \vert x \vert + \vert y \vert$.
The idea is that if $\{(v_k,w_k)\}_{k=1}^{\infty}$ is a Cauchy sequence, then $\{(v_k)\}_{k=1}^{\infty}$ and $\{(w_k)\}_{k=1}^{\infty}$ are Cauchy sequences of their respective Banach space. Therefore $\{(v_k,w_k)\}_{k=1}^{\infty}$ converges to $(v,w)$ where $v$ is the limit of $\{(v_k)\}_{k=1}^{\infty}$ and $w$ the limit of $\{(w_k)\}_{k=1}^{\infty}$.
Hint: In order to show a sequence is convergent, it's helpful to first guess what it should converge to. What would be your guess in this case? (This is the step where you'll use the condition that $V$ and $W$ are Banach spaces.)
Hint: Since we have that $$\|(v_k-v_l,w_k-w_l)\|\leq\epsilon\text{ whenever }k,l\geq N$$ we know $$\|v_k-v_l\|+\|w_k-w_l\|\leq\epsilon\text{ whenever }k,l\geq N$$ and so $$\|v_k-v_l\|\leq \epsilon,\|w_k-w_l\|\leq\epsilon\text{ whenever }k,l\geq N.$$ Thus, $(v_n)$ and $(w_n)$ are Cauchy sequences in $V$ and $W$ respectively, so they both converge.