Showing $X$ is a Banach space using a sequence of elements $\{x_n\}$ that satisfies two conditions?

Question

Let $X$ be a normed space. I want to show that $X$ is a Banach space if and only if for all sequences $\{x_n\}\in X$ such that $\sum_n||x_n|| < \infty$ there exists an element $y\in X$ such that $$ \lim_{n\to \infty} ||y - (x_1+\dots+x_n)||=0. $$

1. $=>$ First showing that $X$ is a Banach space. Let $\{x_n\}$ be a Cauchy sequence in $X$ such that $\sum_n ||x_n|| < \infty$ and choose $X \ni x = y - (x_1+ \dots + x_n)$ as a candidate limit for the Cauchy sequence $\{x_n\}$. Then we have $$ \begin{align} \lim_{n\to \infty}||x - x_n|| & = \lim_{n\to \infty}||y - (x_1+ \dots + x_n) - x_n|| \\ & \le \lim_{n\to \infty}||y - (x_1+ \dots + x_n)|| + \lim_{n\to \infty}||x_n|| \\ & \le 0 + 0 = 0, \end{align} $$ where $\lim_{n\to \infty}||x_n|| = 0$ as $\sum_n ||x_n|| < \infty$. But it seems I missed something here as to show $X$ is complete we need to be able to find a limit $x\in X$ for all Cauchy sequences, not just ones satisfying $\sum_n ||x_n|| < \infty$. So how can I fix this proof?
2. $<=$ Now to show the other direction. Let $X$ be a Banach space and let $\{x_n\}$ be a sequence in $X$ such that $\sum_n||x_n|| < \infty$. This means that $\lim_{n\to\infty}||x_n|| = 0$. Set $y = x_1 + \dots + x_{n-1}$. Then we have $$ \begin{align} \lim_{n\to\infty}||y-(x_1+\dots+x_n)|| & = \lim_{n\to\infty}||(x_1 + \dots + x_{n-1}) - (x_1 + \dots + x_n)|| \\ & = \lim_{n\to\infty}||x_n|| = 0. \end{align} $$ I didn't use the completeness of the Banach space here so I'm not sure if I missed something. Does this look correct?

Answer 1

Let $X$ be a Banach space and $(x_n)\subseteq X$ a sequence such that $\sum_n||x_n||<+\infty$. Then the sequence $$y_n:=x_1+...+x_n$$ is in $X$ for all $n$ and satisfies for $n\geqslant m$: $$||y_m-y_n||=\Big|\Big|\sum_{m\leqslant k\leqslant n}x_k\Big|\Big|\leqslant \sum_{m\leqslant k\leqslant n}||x_k||\to 0\hspace{0.2cm}\text{as}\hspace{0.2cm}m,n\to+\infty$$ Therefore $(y_n)\subseteq X$ is a Cauchy sequence. Because $X$ is complete (by definition a Banach space is a complete normed linear space) then there exists some element $y\in X$ such that $||y-y_n||\to 0$ as $n\to\infty$. This shows one direction. For the other direction. Let $(x_n)\subseteq X$ be a sequence such that $\sum_n||x_n||<+\infty$ and that there exists some $y\in X$ such that $||y-y_n||\to 0$ as $n\to+\infty$ with $y_n$ as defined previously. Let $(z_n)\subseteq X$ be a Cauchy sequence. Then by definition $\forall \varepsilon>0$ there exists some $N\in\mathbb{N}$ such that $$||z_m-z_n||<\varepsilon\hspace{0.2cm}\text{whenever}\hspace{0.2cm}m,n\geqslant N$$ This implies in particular that there is some subsequence $(z_{n_k})$ such that $$||z_{n_k}-z_{n_{k-1}}||<\frac{1}{2^{k-1}}$$ for $k\geqslant 1$. Define $w_k:=z_{n_{k-1}}-z_{n_k}$. Then we get $$\sum_k||w_k||= \sum_k||z_{n_{k-1}}-z_{n_k}||<\sum_k\frac{1}{2^{k-1}}=2<+\infty$$ Therefore by our hypothesis there is some element $w\in X$ such that $$\lim_m||w-(w_1+w_2+..+w_m)||=0$$ But $w_1+...+w_m=z_{n_1}-z_{n_m}$ thus we have $$\lim_m||w-(z_{n_1}-z_{n_m})||=0\Rightarrow\lim_m||\tilde{w}-z_{n_m}||=0$$ with $\tilde{w}:=z_{n_1}-w$. This then gives in turn $$||z_n-\tilde{w}||\leqslant||z_n-z_{n_m}||+||z_{n_m}-\tilde{w}||\to 0$$ as $n,m\to +\infty$. The first term on the right vanishes since $(z_n)$ is Cauchy, the second vanishes from the previous line. Therefore $X$ is complete.

Answer 2

Left to right is not correct. The $y$ is not fixed. Anyway, it suffices to prove that the sequence $\{s_{n}\}$ is Cauchy, where $s_{n}=x_{1}+\cdots+x_{n}$, $n=1,2,...$, then for $n> m$ \begin{align*} \left\|s_{n}-s_{m}\right\|&=\|x_{m+1}+\cdots+x_{n}\|\\ &\leq\|x_{m+1}\|+\cdots+\|x_{n}\|, \end{align*} but $\displaystyle\sum_{n}\|x_{n}\|<\infty$, we have $\displaystyle\sum_{n\geq m}\|x_{m}\|\rightarrow 0$ as $m\rightarrow\infty$, so $\|s_{n}-s_{m}\|\rightarrow 0$ as $n,m\rightarrow\infty$, as expected.

Right to left. Let $\{x_{n}\}$ be a Cauchy sequence but it has no limit. Now choose $n_{1}<n_{2}<\cdots$ such that \begin{align*} \|x_{n_{k+1}}-x_{n_{k}}\|<\dfrac{1}{2^{k}},~~~~k=1,2,..., \end{align*} then \begin{align*} \|x_{n_{k+1}}-x_{n_{1}}\|=\left\|\sum_{j=1}^{k}(x_{n_{j+1}}-x_{n_{j}})\right\|\leq\sum_{j=1}^{k}\|x_{n_{j+1}}-x_{n_{j}}\|<\sum_{j=1}^{k}\dfrac{1}{2^{j}}<1<\infty, \end{align*} so $x_{n_{k+1}}-x_{n_{1}}-y\rightarrow 0$ for some $y\in X$, but then $x_{n}\rightarrow x_{n_{1}}+y$, a contradiction.