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.

this might be a dumb question but is the following true? $$<\sum_{n=1}^\infty x_n, y>=\sum_{n=1}^\infty <x_n, y>$$

share|improve this question
2  
In a vector space, infinite sums do not make sense. If your vector space has a topology on it, and if your inner product is continuous with respect to that topology, then the answer is: yes. –  Mariano Suárez-Alvarez Feb 28 '11 at 2:23
    
This does not make sense because $n$ is free on the left hand side, but not on the right hand side. –  lhf Feb 28 '11 at 2:24
    
sorry im still not understanding...why dont direct sums make sense? –  jack Feb 28 '11 at 2:24
    
@lhf, fixed it. –  jack Feb 28 '11 at 2:26
3  
@jack: Here's a small LaTeX tip: Use \langle ($\langle$) and \rangle ($\rangle$) instead of $\lt$ and $\gt$ when writing inner products. It looks better plus they can be autosized with amsmath's \left and \right commands: $$ \left\langle \sum_{n=1}^\infty x_n, y_n \right\rangle.$$ Right click the math above to see the source –  kahen Feb 28 '11 at 2:47

1 Answer 1

up vote 6 down vote accepted

As Mariano points out, you need to have a topological vector space - otherwise there is no such thing as an infinite sum, because an "infinite sum" is really defined to be a limit of partial sums, and no topology means no limits. However, if there is a topology, and the inner product is continuous with respect to it, we have

$$\langle\sum_{n=1}^\infty x_n,y\rangle=\langle\lim_{N\rightarrow\infty}\sum_{n=1}^N x_n, y\rangle=\lim_{N\rightarrow\infty}\langle\sum_{n=1}^N x_n,y\rangle=\lim_{N\rightarrow\infty}\sum_{n=1}^N\langle x_n,y\rangle=\sum_{n=1}^\infty \langle x_n,y\rangle$$

where the second equality is justified by the assumption that the inner product is continuous and hence preserves limits, and the third equality is justified by the bilinearity of the inner product.

share|improve this answer

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.