# Form of the inner product in $\ l_2$

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 $\ l_2$ .)

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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
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

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 $\langle x,y \rangle_{A} = \langle Ax, Ay \rangle$.
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