Show that a normed vector space is complete

Question

I have been struggling on that exercise for some time. I am currently reading and practicing the material from Donald L. Cohn's book Measure Theory.

Show that a normed vector space $(V, ||·||)$ is complete if and only if for every sequence $(v_n)_{n=1}^\infty$ in $V$ with $\sum_{n=1}^\infty||v_n||\lt ∞$ there is $v ∈ V$ such that $v = \sum_{n=1}^\infty v_n$ in $V$ (meaning that $||\sum_{n=1}^k v_n − v||$ tends to $0$ as $k → ∞$).

We will consider the following :

• the metric on $V$ is given by $d(x,y) := ||x − y||$ for $x,y ∈ V$

• the metric space $(V,d)$ is complete by definition if for every Cauchy sequence $({x_n})_{n=1}^∞$ in it there is $v ∈ V$ with $lim_{n→∞} d(x_n,v) = 0$.)

Thank you in advance

Answer 1

The 'if' part: let $(v_n)$ be a Cauchy and choose $n_k$ increasing to $\infty$ such that $\|v_{n_{k+1}}-v_{n_k} \|<\frac 1 {k^{2}}$. By hypothesis $\sum (v_{n_{k+1}}-v_{n_k})$ is convergent. Conclude that $\lim v_{n_k}$ exists. In any metric space if a Cauchy sequence has a convergent subsequence then the whole sequence converges.

Converse is very easy.

Answer 2

Assume that the hypothesis on convergent norm series holds, and show that $X$ is complete: if $(x_n)$ is Cauchy then pick $n_1$ such that

$$\forall n,n \ge n_1: \|x_n - x_m\| < \frac{1}{2}$$

and then $n_2 >n_1$ such that

$$\forall n,n \ge n_2: \|x_n - x_m\| < \frac{1}{2^2}$$

and in general $n_{k+1} > n_k$ such that

$$\forall n,n \ge n_{k+1}: \|x_n - x_m\| < \frac{1}{2^{k+1}}$$

and so on.

Now consider the series of the elements $(x_{n_{m+1}}-x_{n_m})_m$, which is telescopic and has $m$-th partial sum $x_{n_{m+1}}-x_{n_1}$ and by all previous inequalities has $$\sum_m \| x_{n_{m+1}}-x_{n_m}\| = 1$$

and so by assumption $$\sum_{m=1}^\infty (x_{n_{m+1}} - x_{n_m}) = \lim_{m \to \infty} x_{n_{m+1}}-x_{n_1}$$

exists with some limit $p$. This implies that $\lim_{m \to \infty} x_{n_m}$ exists and so our original Cauchy sequence has a convergent subsequence and thus is convergent (to that same limit).

For the converse, if $X$ is complete and $\sum_m \|x_m\| < \infty$ then consider the partial sums $S(n)=\sum_{k=1}^n x_k$ and note that for $m_1 < m_2$ in $\Bbb N$ we have

$$\|S(m_2)-S(m_1)\| = \| \sum_{k=1}^{m_2} x_k - \sum_{k=1}^{m_1}\| = \sum_{k=m_1+1}^{m_2} x_k\| \le \sum_{k=m_1+1}^{m_2} \|x_k\| = \sum_{k=1}^{m_2}\|x_k\| - \sum_{k=1}^{m_1}\|x_k\|$$

which implies (as the last is the difference between partial sums of a convergent series) that $S(n)$ is Cauchy in $X$ and so converges by completeness and so $v=\sum_{k=1}^\infty x_k$ exists by definition.