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.

Is it true that every inner product in $\ l_2 $ is of the form $\langle x,y\rangle_a =\sum_{n=1} ^ {\infty} {a_n x_n y_n}$ ? (Of course $\ x=(x_n) , y=(y_n) $ are in $\ l_2 $ .)

share|improve this question
3  
A note on notation: please don't use $\lt\,\cdot\,,\,\cdot\,\gt$ for a scalar product. The symbols $\lt$ \lt and $\gt$ \gt are interpreted as relations and result in awkward spacing. Use $\langle\,\cdot\,,\,\cdot\,\rangle$ instead, which you get by using $\langle$ \langle and $\rangle$ \rangle. –  t.b. Feb 22 '11 at 18:37
1  
Not even the standard inner product is of this form: –  Nate Eldredge Feb 22 '11 at 18:57
    
Thanks for your comments.I edited the question.It was obviously wrong! –  t.k Mar 1 '11 at 18:39
    
I don't see any reason for this to be true of an inner product which is not continuous. –  Qiaochu Yuan Aug 16 '11 at 22:02
add comment

1 Answer 1

up vote 4 down vote accepted

I'm not sure I understand the question. First of all, every bounded sequence $a = (a_{n})$ with $0 \lt a_{n}$ for all $n$ will give a scalar product on $\ell^{2}$ by $\langle x,y \rangle_{a} = \sum_{n = 1}^{\infty} a_{n} x_{n} y_{n}$. Not every bounded sequence is square-summable. For instance, the usual scalar product is not of the form you're asking about.

Moreover, for every bounded and injective operator $A : \ell^{2} \to \ell^{2}$ you get a scalar product by setting $\langle x,y \rangle_{A} = \langle Ax, Ay \rangle$.

share|improve this answer
2  
And the latter scalar product is of the desired form iff $A^*A$ is a "diagonal matrix". –  Nate Eldredge Feb 22 '11 at 21:20
add comment

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.