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 $ .)
Form of the inner product in $\ l_2$
0
$\begingroup$
functional-analysis
hilbert-spaces
inner-product-space
-
0I don't see any reason for this to be true of an inner product which is not _continuous._ – 2011-08-16
1 Answers
4
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 $\langle x,y \rangle_{A} = \langle Ax, Ay \rangle$.
-
2And the latter scalar product is of the desired form iff $A^*A$ is a "diagonal matrix". – 2011-02-22