Definition talk:Hilbert Space

Different definition of Hilbert space
A work under study defines a Hilbert space as $H = \left({\R^\infty, d}\right)$ where:
 * $\displaystyle \R^\infty := \prod_{n \mathop = 1}^\infty \R$
 * $\displaystyle d \left({u, v}\right) := \left({\sum_{n \mathop = 1}^\infty \left({u_i - v_i}\right)^2}\right)^{1/2}$

where $u = \left({u_1, u_2, \ldots}\right), v = \left({u_1, u_2, \ldots}\right)$ for $u_1, u_2, \ldots, v_1, v_2 \ldots \in \R$.

Can this be reconciled with the definition as given in this page? Clearly Mendelson's definition is a specific instance of what is given here, in the limited context of a generalization of a Euclidean vector space, but I am having difficulty finding sources to back up Mendelson's definition.

Is anyone able to help? In the meantime I am skipping this chapter in Mendelson until I get an idea of where this is going to need to go. --prime mover (talk) 12:59, 4 January 2015 (UTC)


 * The space defined above is intended to be the Hilbert sequence space $\ell^2(\R)$. However, it omits the square-summability condition which is necessary to make $d$ a metric. Does that provide you with a start for further inquiry? &mdash; Lord_Farin (talk) 13:54, 4 January 2015 (UTC)


 * Thanks -- yes I wondered about that as well ... but as you say, it doesn't have that extra bit.


 * I may merely add it as an extension of the Euclidean vector space and call it an "infinite-dimensional Euclidean vector space" and take it from there, adding an Also known as. --prime mover (talk) 18:00, 4 January 2015 (UTC)

I think Hilbert space is $\left({V, d}\right)$, not $V$. So we need modification.--Bltzmnn.k (talk) 04:04, 26 January 2018 (EST)
 * I am afraid I don't understand your remark. Could you please clarify? &mdash; Lord_Farin (talk) 12:15, 28 January 2018 (EST)