1

Let

  • $(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space
  • $Q$ be a bounded, linear, nonnegative and symmetric operator on $U$
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$ with $$Qe_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\lambda_n)_{n\ge 0}\subseteq[0,\infty)$
  • $U_0:=Q^{\frac 12}(U)$ and $$\langle u,v\rangle_0:=\langle Q^{-\frac 12}u,Q^{-\frac 12}v\rangle\;\;\;\text{for }u,v\in U_0$$ where $Q^{-\frac 12}$ is the pseudo inverse of $Q^{\frac 12}$

We can prove that $(U_0,\langle\;\cdot\;,\;\cdot\;\rangle_0)$ is a separable Hilbert space. Let $$e^{(0)}_n:=Q^{\frac 12}e_n\;\;\;\text{for }n\in\mathbb N\;.$$ How can we prove that $\left(e^{(0)}_n\right)_{n\in\mathbb N}$ is an orthonormal basis of $U_0$?

I fail even to prove that $\left(e^{(0)}_n\right)_{n\in\mathbb N}$ is an orthonormal system, cause I don't know how I need to deal with $Q^{-\frac 12}$.

0xbadf00d
  • 13,422
  • If $\lambda_n=0$ could $e^{(0)}_n$ be non zero? Hint: you might want to consider the direct sum of the $\ker Q$ and its orthogonal complement. – user251257 Feb 17 '16 at 15:30
  • reason is unsound... But, not all $\lambda_n$ needs to be zero, some may be. So ${ e^{(0)}_n \mid n\in\mathbb N }$ might not be a ONB, but the subset of non-zero elements is. – user251257 Feb 17 '16 at 17:21
  • could you prove it, if $\lambda_n > 0$ for every $n\in\mathbb N$? – user251257 Feb 17 '16 at 17:28
  • @user251257 Could you be more explicit in what you got in mind? I could imagine that I need to use the result I've asked for in the following question: http://math.stackexchange.com/questions/1661302/if-q-is-an-operator-on-a-hilbert-space-with-qe-n-%CE%BB-ne-n-for-all-n-then-q. Without that, I don't know how I need to deal with $Q^{-\frac 12}$ in $$\displaystyle \delta_{ij}\stackrel !=\langle e^{(0)}_i,e^{(0)}_j\rangle_0=\langle Q^{-\frac 12}Q^{\frac 12}e_i,Q^{-\frac 12}Q^{\frac 12}e_j\rangle;,$$ when $Q^{\frac 12}$ is not injective. – 0xbadf00d Feb 18 '16 at 12:53
  • @user251257 By definition of $Q^{-\frac 12}$, $$Q^{\frac 12}Q^{-\frac 12}v=v;;;\text{for all }v\in Q^{\frac 12}(U);.$$ However, in order to take benefit from that, I would need that $Q^{\frac 12}$ and $Q^{-\frac 12}$ commute, i.e. that $$Q^{-\frac 12}Q^{\frac 12}=Q^{\frac 12}Q^{-\frac 12};,$$ but I don't know whether or not we can prove that and I don't see a way to prove the desired statement without that. – 0xbadf00d Feb 18 '16 at 13:17
  • I am irritated. $Q$ is obviously defined on a ONB. So $Q$ is densely defined. Further it is bounded. So it is self adjoint. If every eigenvalues are positive, then $Q$ is injective. If not, just divide out the kernel of $Q$. – user251257 Feb 19 '16 at 03:46
  • @user251257 Since $Q$ is injective on ${\ker Q}^\perp$ and $Q({\ker Q}^\perp)=Q(U)$, we can define $$Q^{-1}:=\left(\left.Q\right|{{\ker Q}^\perp}\right)^\perp:Q(U)\to{\ker Q}^\perp;,\tag 1$$ which is one of two equivalent definitions for $Q^{-1}$. Since $U={\ker Q}^\perp\oplus\ker Q$, $$e_n=v_n+w_n;;;\text{for all }n\in\mathbb N$$ for some unique $(v_n){n\in\mathbb N}\subseteq{\ker Q}^\perp$ and $(w_n)_{n\in\mathbb N}\subseteq\ker Q$. – 0xbadf00d Feb 20 '16 at 13:03
  • @user251257 Since $Q^{\frac 12}$ is a nonnegative and symmetric element of $\mathfrak L(U)$, the stuff above remains true if we replace $Q$ by $Q^{\frac 12}$. Especially, $$Q^{\frac 12}e_n=Q^{\frac 12}v_n$$ and hence $$Q^{-\frac 12}Q^{\frac 12}e_n=v_n$$ for all $n\in\mathbb N$ by $(1)$. So, all we need to do is proving that $(v_n)_{n\in\mathbb N}$ is an orthonormal system of $U$. How can we do that? – 0xbadf00d Feb 20 '16 at 13:04
  • Ask yourself how $e_n$ relays to $v_n$ – user251257 Feb 20 '16 at 13:09
  • @user251257 I did and I've obtained $$\langle v_i,v_j\rangle=\langle e_i-w_i,e_j-w_j\rangle=\langle e_i,e_j\rangle-\underbrace{\langle e_i,w_j\rangle}{=-\langle w_i,w_j\rangle}-\underbrace{\langle w_i,e_j\rangle}{=-\langle w_i,w_j\rangle}+\langle w_i,w_j\rangle=\delta_{ij}+3\langle w_i,w_j\rangle;,$$ since $$\langle v_n,w\rangle=0;;;\text{for all }n\in\mathbb N\text{ and }w\in\ker Q^{\frac 12}$$ So, I guess we need to have $\langle w_i,w_j\rangle=0$, but again, I don't know how I can show that. – 0xbadf00d Feb 20 '16 at 13:22
  • hmm $e_n$ is an eigen basis. So $v_n=e_n$ if $\lamba_n > 0$. – user251257 Feb 20 '16 at 13:24
  • @user251257 Let me try to understand your argument:$$\lambda_ne_n=Qe_n=Q^{\frac 12}Q^{\frac 12}e_n=Q^{\frac 12}Q^{\frac 12}v_n=Qv_n;.$$ Now, $v_n\in\left(\ker Q^{\frac 12}\right)^\perp$ and $Q^{\frac 12}:\left(\ker Q^{\frac 12}\right)^\perp\to Q^{\frac 12}(U)$ is a bijection. But I still got trouble to conclude. – 0xbadf00d Feb 20 '16 at 13:58
  • Sorry. I probably got something wrong. I will think about it again. – user251257 Feb 20 '16 at 15:10
  • @user251257 Thanks. Please tell me, when you found a solution. – 0xbadf00d Feb 20 '16 at 15:13
  • I posted an answer. hope it helps. – user251257 Feb 20 '16 at 16:14

1 Answers1

0

I am little bit rusty at operator theory, so bear with me.

  1. $Q: D(Q)\to U$ is bounded, $E:=(e_n)_{n\in\mathbb N}$ is an ONB, and $E\subseteq D(Q)$. So, $D(Q) = U$
  2. Since $Q$ is symmetric and defined on $U$, $Q$ is self-adjoint.
  3. From $Qe_n = \lambda_n e_n$ for every $n\in\mathbb N$, it follows that $E$ is orthonormal eigenbasis. Since $Q$ is nonnegative, for $L:=\{l\in\mathbb N \mid \lambda_l > 0\}$ we have \begin{align} Q &= \sum_{l\in L} \lambda_l e_l e_l^*, & R:= Q^{\frac12} &= \sum_{l\in L} \lambda_l^{\frac12} e_l e_l^*, \end{align} where $e_l^*$ denotes linear functional $x\mapsto \langle e_l, x\rangle$.
  4. Notice that $$ U_0 = R(U) = \left\{ \sum_{l\in L} x_l e_l \mid \sum_{l\in L} \frac{x_l^2}{\lambda_l} < \infty \right\}. $$ Thus, on $U_0$ the pseudo inverse $R^+$ of $R$ is given by $$ R^+ = \sum_{l\in L} \lambda_l^{-\frac12} e_l e_l^*. $$ In particular, for $l\in L$ we have $$ R^+ R e_l = e_l. $$
user251257
  • 9,229
  • What is $D(Q)$? $Q$ in the question is a mapping $U\to U$. And why is $R$ equal to $\sum_{l\in L} \lambda_l^{\frac12} e_l e_l^*$ and $U_0$ equal to $\left{ \sum_{l\in L} x_l e_l \mid \sum_{l\in L} \frac{x_l^2}{\lambda_l} < \infty \right}$? I suppose you've used that $$Q^{\frac 12}e_n=\sqrt{\lambda_n}e_n;,$$ right? – 0xbadf00d Feb 20 '16 at 18:56
  • @0xbadf00d the domain of $Q$. Since you called $Q$ symmetric instead of self adjoint, I assume $Q$ may be partially defined. Just ignore item 1. – user251257 Feb 20 '16 at 19:01
  • What about the other questions? – 0xbadf00d Feb 20 '16 at 19:14
  • @0xbadf00d $R^2 = Q$ isn't it? What is your definition of $Q^{1/2}$ otherwise? – user251257 Feb 20 '16 at 19:17
  • Yes, but why is $U_0$ equal to $\left{ \sum_{l\in L} x_l e_l \mid \sum_{l\in L} \frac{x_l^2}{\lambda_l} < \infty \right}$? Especially, what is $x_l$? – 0xbadf00d Feb 20 '16 at 20:11
  • @0xbadf00d $x_l$ is the scalar coefficient to the basis vector $e_l$. For $u\in U$ we have $u = \sum_{n\in\mathbb N} u_n e_n$ and $R(u) = \sum_{n\in\mathbb N} \sqrt{\lambda_n} u_n e_n = \sum_{l\in L} \sqrt{\lambda_l} u_l e_l$. Thus $\sum_{l\in L} (\sqrt{\lambda_l} u_l)^2 / \lambda_l = \sum_{l\in L} u_l^2 \le |u|^2 < \infty$. On other hand, if you have a $x = \sum_{l\in L} x_l e_l$ with $\sum_{l\in L} x_l^2 / \lambda_l < \infty$, it follows $v = \sum_{l\in L} x_l / \sqrt{\lambda_l} e_l \in U$ as $\sum_{l\in L} (x_l / \sqrt{\lambda_l})^2 < \infty$ by assumption. – user251257 Feb 20 '16 at 20:23
  • This shows that $\left(e^{(0)}l\right){l\in L}$ is an orthonormal system in $U_0$. How can we conclude that is even a basis? – 0xbadf00d Feb 21 '16 at 08:54
  • @0xbadf00d ${ R e_l \mid l \in L }^\bot = 0$ in $U_0$. – user251257 Feb 21 '16 at 12:22