0
$\begingroup$

Let's say I have a Banach space X . How can I prove that X is Banach if and only if every infinite series $\sum_{n=1}^{\infty }y_{n}$ with $y_{n}\in$ X and $\left \| y_{n} \right \|\leq \frac{1}{2^{n}}$ is convergent?

I know it's trivial in the first direction when X is a Banach space, but I have no idea how to prove it when beginning from the convergence argument. Please help.

  • 1
    You started by **assuming** $X$ is a Banach space, so there's nothing to prove. Perhaps you meant: "I have a **normed linear** space"...2017-01-29

1 Answers 1

2

tAs I commented, I suppose you mean: suppose $X$ is a normed linear space in which every series $\sum_{n=1}^\infty y_n$ with $\|y_n\| \le 2^{-n}$ converges. You want to show that $X$ is a Banach space, i.e. is complete.

Consider a Cauchy sequence $c_n$ in $X$. You want to prove that it converges. Take a subsequence $c_{n_j}$ such that $\|c_{n_j} - c_k\| < 2^{-j}$ for all $k > n_j$. Consider the series $\sum_{j=1}^\infty {y_j}$ where $y_j = c_{n_j} - c_{n_{j+1}}$. Then $\|y_j\| < 2^{-j}$, so this converges, say to $s$. But note that $\sum_{j=1}^k y_j = c_{n_1} - c_{n_{k+1}}$, so the subsequence $c_{n_j}$ converges to $c_{n_1} - s$. And a Cauchy sequence with a convergent subsequence converges.

  • 0
    Yes, a normed space is what I meant but forgot to specify. Thank you so much for your help!2017-01-29