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$(X,||.||)$ a Normed space, prove that when we got the property $\sum x_n$ converges in $X$ if $\sum||x_n||$ converges in $\mathbb{R}$, that X is a Banach Space.

So far i have this:

Let $\{x_n\}$ be a cauchy sequence, then we can construct a sequence such that for $n,m>N_k$:

$||x_n - x_m|| < \frac{1}{2^k}$.

Choose a subsequence $y_k = \{x_s\}_{s \geq N_k}$.

Then

$\sum ||y_k|| \leq \sum \frac{1}{2^k} + ||x_m|| \leq \infty$

So we know that $\sum y_k$ is convergent in X.

I know that if a sub sequence of a Cauchy sequence is convergent, the Cauchy sequence is convergent as well.

But how does one now conclude that $y_k$ is a convergent sequence if its sum is convergent...?

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    $\begingroup$ The subsequence is not well defined. And what is $x_m$ thereafter? $\endgroup$ – Friedrich Philipp Mar 11 '16 at 17:34
  • $\begingroup$ hmmmmm, i am trying to say that we take a subsequence of x_n wherefor the inequalty holds but i don't know how to note that right... $\endgroup$ – Kees Til Mar 11 '16 at 17:40
  • $\begingroup$ maybe this is the right notation: $\{y_n\} = \{x_n : ||x_n - x_m|| < 0.5^k\}$ $\endgroup$ – Kees Til Mar 11 '16 at 17:48
  • $\begingroup$ No, this is nonsense. The question remains: what is $m$ here? $\endgroup$ – Friedrich Philipp Mar 12 '16 at 9:48
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Hint: Find a subsequence $x_{n_k}$ such that $\|x_{n_{k+1}} - x_{n_k}\| < 1/2^k, k \in \mathbb N.$ Then recall the standard way of turning a sequence into a series.

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  • $\begingroup$ Thanks, i used this to construct an own answer :) $\endgroup$ – Kees Til Mar 11 '16 at 20:57
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Your notation is a bit messy but the idea is correct. Construct a subsequence $\{ x_{n_k}\}$ so that $$\lvert \lvert x_{n_{k+1}} - x_{n_k}\rvert \rvert < \frac 1 {2^k}.$$ Then define $y_K$ = $x_{n_0} + \sum^K_{k=0} (x_{n_{k+1}} - x_{n_k})$. You'll find that $\{y_K\}$ is absolutely summable and hence $$\lim_{K\to \infty} \left[ x_{n_0} + \sum^K_{k=1} (x_{n_{k+1}} - x_{n_k}) \right] = x \text{ for some } x\in X \,\,\,\, \text{(i.e. the limit converges)}.$$ But what does the expression in the brackets simplify to?

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with User8128 and Zhw's tip:

construct $y_k = \{x_{n_{o}}, x_{n_{1}} - x_{n_{0}},x_{n_{2}} - x_{n_{1}}, \dots ,x_{n_{k}} - x_{n_{k-1}}\} $

Then we see that: $\sum_{k=0}^\infty ||y_k|| = \sum_{k=0}^\infty ||y_k + x_{n_{k}} - x_{n_{k}}|| \leq \sum_{k=0}^\infty \frac{1}{2^k} + ||x_{n_{k}}|| < \infty$.

So $\sum_{k=0}^\infty y_k$ converges in $X$.

Note now that the partial sums converge as well. So we always get:

$\sum_{k=0}^M y_k = \{x_{n_{m}}\}$

and since this subsequence of a cauchy sequence converges, we are finished $\quad$

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  • $\begingroup$ So, $y_k$ is a set? $\endgroup$ – Friedrich Philipp Mar 12 '16 at 9:45
  • $\begingroup$ $y_k$ is a sequence. $\endgroup$ – Kees Til Mar 12 '16 at 13:45
  • $\begingroup$ No, it isn't. At least not due to your definition. You define $y_k$ to be a set consisting of (at most) $k+1$ vectors. $\endgroup$ – Friedrich Philipp Mar 12 '16 at 23:46

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