Timeline for Hilbert representation of a bilinear form
Current License: CC BY-SA 4.0
13 events
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| Sep 23, 2018 at 20:42 | comment | added | YCor | @erz thanks for the remarks! Indeed in my answer I make an unnecessary complication by considering a basis (of some subspace of) the space of functions, which is indeed artificial. So indeed one sets $\sigma(f,g)=0$, $\sigma(f,e_n)=f(e_n)$ and $\sigma(e_n,e_m)=\delta_{m,n}$, or denoting this with integrals as you put it. | |
| Sep 23, 2018 at 20:15 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Fixes 2 typos (basis elements notation)
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| Sep 23, 2018 at 7:02 | comment | added | erz | Now for the sequence $v_n=0\oplus \delta_n$ the necessary condition does not hold, since we can take $v=f\oplus 0$, for $f$ growing arbitrary fast. | |
| Sep 23, 2018 at 7:00 | comment | added | erz | @YCor may I suggest a couple of improvements to simplify your answer? For the proof of the countable case we don't need to envelop a subspace into a non-degenerate one, we just need a usual Gram-Schmidt procedure, if I am not missing anything. As for you counterexample I propose the following simplification: take $V=\mathbb{R}^{\mathbb{N}}\oplus span\{ \delta_n, n\in \mathbb{N}\}$, where $\delta_n$ is viewed as a measure. Then $\sigma$ is defined by $\sigma(f\oplus \mu, g\oplus \nu) =\int f d\nu+\int g d\mu + \int\int\delta d\mu d\nu $, where $\delta$ is viewed as a function on $\mathbb{N}^2$. | |
| Sep 23, 2018 at 6:51 | comment | added | erz | @YCor this condition is necessary even for a weaker condition of existing maps $S:V\to E$ and $T:V\to E^*$, where $E$ is normed, such that $\sigma(u,v)=\left<Tu,Sv\right>$. Hence, the answer is "no" for the question about "Banach representation of bilinear forms" as well. | |
| Sep 23, 2018 at 6:44 | vote | accept | erz | ||
| Sep 22, 2018 at 14:26 | comment | added | YCor | So, well, a necessary condition for the existence of a representation can be written as: for each sequence $(v_n)$ in $V$, there exists a positive real sequence $(t_n)$ such that for every $v\in V$, one has $|\sigma(v_n,v)|=O(t_n)$. I don't know if it's sufficient. | |
| Sep 22, 2018 at 14:17 | comment | added | YCor | @NateEldredge oh thanks, so simple :) | |
| Sep 22, 2018 at 14:06 | comment | added | Nate Eldredge | If $\sigma > 0$ then $(V,\sigma)$ is an inner product space and we can simply take $H$ to be its completion, with $S$ the inclusion map and $T=I$. | |
| Sep 22, 2018 at 8:54 | comment | added | YCor | PPS At the end of the post I used the fact that if $(V,\sigma)$ has an orthogonal basis, then it has a Hilbert representation. The converse is false. Indeed, consider an infinite dimensional separable Hilbert space itself $H$, with orthonormal Hilbert basis $(e_n)$. If $H$ had an (algebraic) orthogonal basis $(v_i)_{i\in I}$, then $I$ has to be uncountable. Write $J_n=\{i:\langle e_n,v_i\rangle\neq 0\}$. Then $J_n$ is a finite subset of $I$, so $J=\bigcup_nJ_n$ is countable. Pick $i\notin J$: then $\langle v_i,e_n\rangle=0$ for all $n$, which forces $v_i=0$, a contradiction. | |
| Sep 22, 2018 at 8:48 | comment | added | YCor | PS at the moment I don't know if there's always a Hilbert representation when $\sigma>0$ (i.e., $\sigma(v,v)>0$ for all $v\neq 0$). | |
| Sep 22, 2018 at 7:38 | history | edited | YCor | CC BY-SA 4.0 |
added countable dimension case
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| Sep 22, 2018 at 6:36 | history | answered | YCor | CC BY-SA 4.0 |