Completeness of a sequence space c(V)

Question

I have an exercise given by:

With $V \equiv \mathbb{R}^{n}$ being furnished with Euclidean rorm $\|\cdot\|_{2}$, define a normed space $$ c(V) \equiv\left\{\left(v_{j} \in V\right)_{j \in \mathbb{N}}: \exists \lim _{j \rightarrow \infty} v_{j}\right\} $$ with a norm $$ \|v\| \equiv \sup _{j}\left\|v_{j}\right\|_{2} $$ Is the space complete?

Now I know how to show convergence:

Suppose $v^{(n)}$ is a Cauchy sequence in c(V). Then we have for all $\epsilon>0$ there's an N s.t. n,m> N gives:

$||v_j^{(n)} - v_j^{(m)}||_2 \leq ||v^{(n)} - v^{(m)}|| < \epsilon$

Hence for all j, $v_j^{(n)}$ is a Cauchy sequence and has a limit $v_j \in \mathbb{R}^n$, set $v = (v_i)_{i \in \mathbb{N}}$. Then again for all j:

$||v_j - v_j^{(m)}||_2 = lim_n ||v_j^{(n)} - v_j^{(m)}||_2 < \epsilon$

Taking sup over j, we get:

$||v - v^{(m)}|| < \epsilon$

Hence we get convergence. Now we have to show that $v \in c(V)$. I tried to do this like this:

$||v_j - v_i|| \leq ||v_j - v_j^{(m)} + v_j^{(m)} - v_i^{(m)} + v_i^{(m)} - v_i|| \leq ||v_j - v_j^{(m)} || + ||v_j^{(m)} - v_i^{(m)} || + ||v_i^{(m)} - v_i||$

Now I can find an N, which will let me bound the first and the last expressions by $\epsilon/3$, when n, m> N. For that N $v_i^{(N)}$ is Cauchy over i. Then we can find J s.t. i, j> J gives $||v_i^{(N)} - v_j^{(N)}|| < \epsilon/3$ let k>N, we get:

$||v_i^{(k)} - v_j^{(k)}|| \leq ||v_j - v_j^{(N)} || + ||v_j^{(N)} - v_i^{(N)} || + ||v_i^{(N)} - v_i|| < \epsilon$

Hence we take that N and J and we are done. Is all of that correct? That was a terribly long and complicated proof, is there a simpler way to do this?

Answer 1

There are a few little mistakes. It seems to me that you are not using the uniform convergence in a correct way. Try like this:

• Let $v^n = \{ v^n_j\}_{j \in \mathbb{N}}$ be your Cauchy sequence of bounded sequences with limit points $x^n$. Since $\ell^\infty(\mathbb{N})$ is complete, $v^n \to v$ uniformly for some bounded sequence $v$.

• By $ \Vert x^n - x^m \Vert = \lim_j \Vert v_j^n - v_j^m \Vert \leq \Vert v^n - v^m \Vert_\infty$ you have that $x^n$ is a Cauchy sequence as well, thus $x^n \to x$ for some $x$.

• Given $\epsilon >0$, pick $n>0$ such that $\Vert v^i - v^j \Vert_\infty \leq \epsilon$ for $i, j \geq n$. By the previous point, after a limit you have $ \Vert x^n - x \Vert \leq \Vert v^n - v \Vert_\infty \leq \epsilon$. Now consider $k=k_n$ such that $ \Vert v^n_j - x^n \Vert \leq \epsilon$ for $j \geq k$. Finally, you have that $ \Vert v_j - x \Vert \leq \Vert v_j - v_j^n \Vert + \Vert v_j^n - x^n \Vert + \Vert x^n - x \Vert \leq 3 \epsilon$ for $j \geq k$, and the thesis follows.

Answer 2

[I answer here because this is too long to be a comment].

Ok, maybe I did not express myself correctly (it was late). The proof works but it's a bit messy. Of course this is normal at the beginning. The easiest way to do less computations (hence less mistakes) is to apply theorems: if you already know that $\ell^\infty$ is complete, then you can avoid to show again that $v^{(m)}$ converges to $v$ uniformly (2nd and 3rd inequalities). Also, the not important mistake is that by taking the supremum, strict inequalities like $< \epsilon$ transform in inequalities like $\leq \epsilon$. Finally, the last line of math is completely useless! The proof was completed with the 4th line of math, but you have to clarify that $N$ is the one given by the uniform convergence $v^{(m)} \to v$, otherwise you will have that $N$ depends on $i$ and $j$. Thus the first and the last addenda are bounded by $\epsilon$ (you take $m=N$), and as $v^{(N)}_j$ is Cauchy then all the expression is bounded by $3 \epsilon$ for every $i, j \geq J$.