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Let $X$ separable Hilbert space and $B \in \mathcal{L}(X)$ (i.e. a bounded linear operator) be strictly positive and self-adjoint. Then I can define the fractional power $B^{1/2}$ and if I set $$|x|_{-1}=| B^{1/2}x|$$ then $|\cdot|_{-1}$ is a norm on $X$.

Now $(X,|\cdot|_{-1})$ is not a Hilbert space because it is not complete but I can define $X_{-1}$ to be the completion of $X$ under $|\cdot|_{-1}$, i.e. see [Fabbri G, Gozzi F, Swiech A. Stochastic optimal control in infinite dimension. Probability and Stochastic Modelling. Springer. 2017.] p.172.

Now my question is: does $X_{-1}$ admit an orthonormal basis made of elements of $X$ as it is claimed in [Fabbri G, Gozzi F, Swiech A. Stochastic optimal control in infinite dimension. Probability and Stochastic Modelling. Springer. 2017.] p.189?

It could be useful to note that $B^{1/2}$ can be extended to $B^{1/2} \in \mathcal{L}(X_{-1},X)$ and it is an isometry, i.e. for $x \in X_{-1}$ we have: $$|B^{1/2}x|=|x|_{-1}$$ Morally I would like do something like this: let $\{ e_k\}$ be orthoormal basis for $X$. Then for $x \in X_{-1}$ there exist $x_n=\sum_{k=1}^{\infty}x_{nk}e_k$ in $X$ such that $|x_n-x|_{-1}\to 0$, i.e. \begin{align} x=\lim_{n}\sum_{k=1}^{\infty}x_{nk}e_k =\sum_{k=1}^{\infty} \lim_{n} x_{nk}e_k=\sum_{k=1}^{\infty} x_{\infty k}e_k \end{align} where the limit is in the $X_{-1}$-topology. This of course can't be done.

Any suggestion?

carlos85
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1 Answers1

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It is certainly convenient that $X$ is separable, but I think it's not essential. A/the more general situation would just need a stronger equivalent of Axiom of Choice. Let $D=\{x_n: n=1,2,3,...\}$ be an enumeration of a dense subset of $X$. Then apply Gram-Schmidt to $D$ in $X_{-1}$ to obtain a Hilbert-space basis as desired.

Yes, in many specific situations ($L^2$ Sobolev spaces on nice spaces, such as the circle) there is an orthogonal basis in $H^{+\infty}$ for all the spaces $H^s$. One abstraction of this is that the operator $B$ in the question should be compact (and self-adjoint). Then there is an orthogonal basis of $X$ that is an orthogonal basis of $X_{-1}$.

paul garrett
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  • this answer claims that separability is essential. – Martin Argerami Nov 13 '21 at 00:42
  • @MartinArgerami, that would not amaze me... but/and I myself am not too intensely interested in the non-separable case, so I'd completely defer to anyone else who is interested. Thanks for the apparent fact! :) – paul garrett Nov 13 '21 at 00:47
  • Yes in my case I can assume $X$ is separable (since I'm actually working with $X=\mathbb{R}\times L^2[0,T]$) – carlos85 Nov 13 '21 at 16:57
  • When you say: “applying GS to D you obtain the orth. Basis”. How does GS work in Hilbert spces? Do you have any reference of GS in Hilbert spaces? – carlos85 Nov 13 '21 at 19:33
  • At least with a countable set of vectors, ordered by the positive integers, it works much like finite-dimensional Gram-Schmidt: given non-zero vectors $v_1,v_2,v_3,\ldots$, adjust $v_1$ by a scalar so that it's length $1$. Adjust $v_2$ by a multiple of $v_1$ so that it's orthogonal to $v_1$, and then adjust its length to be $1$. Adjust $v_3$ by a linear combination of $v_1,v_2$ so that it's orthogonal to both, then make its length $1$. Continue. Apparently (thanks, @MartinArgerami) there is a fatal difficulty with non-separable Hilbert spaces, though. – paul garrett Nov 13 '21 at 19:46
  • Yes but at the end of the process you obtain a set $(e_n)$ countable and dense in $X_{-1}$ made of orthonormal elements. But how do you show that it is a basis? – carlos85 Nov 13 '21 at 20:23
  • @carlos85, first, show that at each ("finite") stage the Hilbert-span of the set of vectors doesn't change (not just algebraic span). This is almost as in the finite-dimensional case. That the "infinite process" does the same takes a little more finesse. Unfortunately, I do not immediately know a reference. Sources that treat Hilbert spaces in great detail would surely do such things. – paul garrett Nov 13 '21 at 20:39
  • Thanks. For conpleteness the answer to my question can be found here: https://math.stackexchange.com/questions/2709501/every-separable-hilbert-space-has-an-orthonormal-basis – carlos85 Nov 13 '21 at 21:30