If (x_j) is a sequence in an inner product space X, such that the series
\|x_1\| + \|x_2\| + \cdots converges.
Show that (s_n) is a Cauchy sequence , where s_n = x_1 + \cdots + x_n.
Hello,
Let $\displaystyle S$ be the limit of $\displaystyle T_n=\sum_{j=1}^n \|x_j\|$
Hence, $\displaystyle \forall \epsilon>0,\exists N\in\mathbb{N},\forall k\geq N, \|T_k-S\|<\frac\epsilon 2$
In particular, we have :
$\displaystyle \begin{array}{ll} \forall m\geq N ~:~ \|T_m-S\|<\frac\epsilon 2 \\ \forall n\geq N ~:~ \|T_n-S\|<\frac\epsilon 2\end{array}$
So $\displaystyle \|T_n-S\|+\|T_m-S\|<\epsilon$
Which is equivalent to $\displaystyle \|T_n-S\|+\|S-T_m\|<\epsilon$
Since $\displaystyle \|.\|$ is a norm, it satisfies the triangle inequality :
$\displaystyle \|T_n-T_m\|=\|T_n-S+S-T_m\|\leq \|T_n-S\|+\|S-T_m\|<\epsilon$
But (assuming n>m) $\displaystyle \|T_n-T_m\|=\left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=a \left(\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right)$, where a is the norm of the unit vector.
Again, by the triangle inequality, we know that $\displaystyle \|s_n-s_m\|=\|x_n+x_{n-1}+\dots+x_{m+1}\|\leq \|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|$
So finally, we have (letting $\displaystyle \epsilon'=\epsilon/a$)
$\displaystyle \forall \epsilon'>0,\exists N\in\mathbb{N},\forall n,m\geq N, \|s_n-s_m\|<\epsilon'$
And this proves that $\displaystyle (s_n)$ is a Cauchy sequence
Hi Moo.
Thank you very much for your elegant answer.
I want to check something in your solution I do not understand.
Its were you state an equality (where assumed n>m) and then you have:
a = the norm of the unit vector
I do not understand this equality can you explain where this whole line comes from please??
Kind Regards!
Hi
No problem. This proves that you're following what I've been doing...
...wrong$\displaystyle \left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=a \left(\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right)$
I'm sorry, I guess I was confusing myself... The norm of a number is its absolute value.
Here, $\displaystyle \|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|$ is a number, a positive number, since a norm is always positive.
So $\displaystyle \left\|\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|\right\|=\|x_n\|+\|x_{n-1}\|+\dots+\|x_{m+1}\|$
And in the final steps, you juste have to keep $\displaystyle \epsilon$ instead of $\displaystyle \epsilon'$
Again, excuse me for my mistake