Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is the definition of a normed vector space my book uses:

link

And here is a remark I do not understand:

link2

I do not understand that a sequence can converge to a vector in one norm, and not the other. For instance: Lets say $s_n$ converges to $u$ with the $\|\|_1$-norm. From definition 4.5.2 (i) we must have that $s_n$ becomes closer and closer to $u$. Why is it that it could fail in the other norm, when it can become as close as we want in the first norm?Are there any simple examples of this phenomenon?

PS:I know that they say we will see examples of this later in the book, but what comes later is too hard for me to udnerstand now.

share|improve this question
1  
It should be mentioned that this never happens in finite dimensions. In finite dimensions, every norm is "equivalent" meaning that if a sequence converges in one norm, in converges in the other (which implies a series converges in one norm if it converges in the other). This should give you an idea of how bizarre things get when you have an infinite dimensional vector space. –  Cameron Williams Mar 24 at 0:24

2 Answers 2

up vote 4 down vote accepted

Consider the following sequence of elements in the space $V$ of finite sequences: $$ u_1=(1,0,0,\ldots),\ \ u_2=(0,\frac12,0,\ldots),\ \ u_3=(0,0,\frac13,0,\ldots) $$ Then $$ \sum_{k=1}^nu_k=(1,\frac12,\frac13,\ldots,\frac1n,0,\ldots) $$ Now consider these two norms on $u=(a_1,a_2,\ldots)$: $$ \|u\|_1=\sum_{k=1}^\infty|a_k|,\ \ \ \|u\|_2=\left(\sum_{k=1}^\infty|a_k|^2\right)^{1/2} $$ Then, for the $u_n$ defined above, $$ \left\|\sum_{k=M}^nu_k\right\|_1=\sum_{k=M}^N\frac1k,\ \ \ \left\|\sum_{k=M}^Nu_k\right\|_2=\sum_{k=M}^N\frac1{k^2} $$ So, in $\|\cdot\|_1$, the tails of the series $\sum_{k=1}^\infty u_k$ are unbounded, which means that the series diverges; while in $\|\cdot\|_2$, the tails go to zero, so the series converges.

share|improve this answer
1  
Thank you, so I guess in the first case we have that the distance between |(1,1/2,1/3...,1/n,0,0,...)-(1/1,1/2,1/3,1/4,1/5,1/6.......)|=|(0,0,0,0,0,0,0,1/‌​(n+1),1/(n+1),1/(n+3),.....| will never be as small as we want, but it will be that in the second case? Thank you for your help! –  user119615 Mar 24 at 1:26
    
Yes indeed. The point, as your book states, is that the notion of series depends on the choice of a topology, as the series is a limit of the sequence of partial sums. –  Martin Argerami Mar 24 at 4:16

You can get in trouble if the convergence is not absolute, i.e. $$\sum_n \|x_n\| = +\infty$$ but $$\lim_{N\to\infty} \sum_{n\le N} x_n $$ exists.

share|improve this answer
1  
This would be appropriate as a comment. –  Sanath Mar 24 at 0:14
    
What is norm 1 and what is norm 2 in your example? –  user119615 Mar 24 at 0:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.