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I know the following result to be true:

Let $\mathbb K \in \{\mathbb R , \mathbb C\}$, let $(H, <\cdot,\cdot>_{H}, ||\cdot||_{H})$-be a $\mathbb K$-Hilbert space. Let $A : D(A) (\subseteq H) \to H$ be a diagonal linear operator with $\sigma_{P}(A) \subseteq (0,\infty)$ and $\mbox{inf} (\sigma_{P}(A)) > 0$. Then the triple $(D(A), <A(\cdot),A(\cdot))>, ||A(\cdot)||_{H}$ is a $\mathbb K$-Hilbert Space.

Here, I have followed the definition of diagonal linear operator, mentioned here:

Looking for a NON-Symmetric Diagonal Linear Operator!!

Now, my question is: suppose I want to have the assumption on $\sigma_{P}(A)$ to be weaker, i.e. suppose it is NOT given that $\mbox{inf}(\sigma_{P}(A)) > 0$ (Only assumption is $\sigma_{P}(A) \subseteq (0,\infty)$). In this situation is the result is true if we take $\mathbb K = \mathbb R$ ??

Rewriting the question clearly,

Prove or disprove (providing counterexample) that: Let $(H, <\cdot,\cdot>_{H}, ||\cdot||_{H})$-be a $\mathbb R$-Hilbert space. Let $A : D(A) (\subseteq H) \to H$ be a diagonal linear operator with $\sigma_{P}(A) \subseteq (0,\infty)$. Then the triple $(D(A), <A(\cdot),A(\cdot))>, ||A(\cdot)||_{H}$ is a $\mathbb R$-Hilbert Space

P.S. :- I know that Inverse of Laplace Operator (call it: A), with Dirichlet boundary condition is a diagonal linear operator with $\mbox{inf}(\sigma_{P}(A)) = 0$. But I am clueless how to show the domain to be a $\mathbb R$-Hilbert Space or disprove with some other example.

Thanking you,

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user92360
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1 Answers1

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Suppose you have a real Hilbert space $\mathcal{H}$ with an orthonormal basis $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ and real numbers $\{ r_{\alpha} \}$. Define $$ Ax = \sum_{\alpha} r_{\alpha}(x,e_{\alpha})e_{\alpha} $$ on the domain $$ \mathcal{D}(A) = \{ x \in \mathcal{H} : \sum_{\alpha} r_{\alpha}^2(x,e_{\alpha})^2 < \infty \}. $$ Then $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a densely-defined symmetric linear operator. If $r_{\alpha} \ge \rho$ for some real $\rho$ and for all $\alpha$, then $A$ is selfadjoint because $(A-\mu I)$ is invertible for all $\mu < \rho$.

So, suppose $r_{\alpha} > 0$ for all $\alpha$. Then $A$ is selfadjoint, which, in particular, means the graph of $A$ is a closed subspace of $H\times H$, making it a Hilbert space with respect to the inner product $$ (x,y)_A = (x,y)_H+(Ax,Ay)_H,\;\; x\in\mathcal{D}(A). $$ Your question is about whether or not $[x,y]_A=(Ax,Ay)_{H}$ is an inner product on $\mathcal{D}(A)$ that turns $\mathcal{D}(A)$ into a Hilbert space. Suppose that it is a Hilbert space. Then the natural injection $$ \mathcal{i} : (\mathcal{D}(A),(.,.)_A)\rightarrow(\mathcal{D}(A),[.,.]_A) $$ is continuous and surjective, making it continuously invertible, leading to the existence of a constant $C$ such that $$ \|x\|_H^2+\|Ax\|_H^2 \le C\|Ax\|_H^2,\;\;\; x\in\mathcal{D}(A). $$ The constant $C$ must satisfy $C > 1$. Therefore, $$ \|x\|_{H} \le \sqrt{C-1}\|Ax\|_{H},\;\; x\in\mathcal{D}(A), $$ which forces $r_{\alpha} \ge 1/\sqrt{C-1}$ for all $\alpha$.

So, $[.,.]_A$ generates a Hilbert space iff $r_{\alpha} \ge \rho > 0$ for some $\rho$ and all $\alpha$.

Disintegrating By Parts
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    Thanks a ton!! Just a small query ... my question is: $[.,.,.]$ generates "$\mathbb R$"-Hilbert Space or not!!?? In your proof, I am not sure, but I think I have missed the "REAL"-Hilbert Space part. Can you please explain this ?? – user92360 Mar 30 '17 at 22:35
  • Well, I think the assumption $\mathcal H$ is a real Hilbert Space has something to do with it. Although I am also not too sure!!So, yuup, +1 to the question. – user86511 Mar 30 '17 at 22:47
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    @user92360 : All of the spaces in my post are assumed real. – Disintegrating By Parts Mar 30 '17 at 23:07
  • @user86511 : All of the spaces in my post are assumed real. – Disintegrating By Parts Mar 30 '17 at 23:08
  • @TrialAndError, sorry to trouble you again, but in your proof , in the second paragraph you have written "If $r_{\alpha} \ge \rho$ for some real $\rho$ and for all $\alpha$, then $A$ is selfadjoint because $(A+\mu I)$ is invertible for all $\mu > -\rho$." Here you are allowing $\mu = 0$, which is not the case I guess. My whole concern is if the same statement would hold without positive infimum criteria, the statement of the known theorem could have been relaxed. Isn't it??? That's why I think omitting that criterion, there must be counterexample. – user92360 Mar 31 '17 at 10:26
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    @user92360 : I make a typo in that statement. The point is that you have semibounded, which you have because you have assumed $\sigma(A) \subseteq (0,\infty)$. – Disintegrating By Parts Mar 31 '17 at 19:11
  • @TrialAndError, what is the definition or meaning of "semibounded" you just said ?? – user92360 Apr 01 '17 at 08:19
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    @user92360 : A selfadjoint operator $T$ is semibounded if there exists a $\rho\in\mathcal{R}$ such that either (a) $(Tf,f) \ge \rho(f,f)$ for all $f\in\mathcal{D}(T)$ or (b) $(Tf,f) \le \rho(f,f)$ for all $f\in\mathcal{D}(T)$. The first condition is equivalent to $\sigma(T) \subseteq [\rho,\infty)$ and the second is equivalent to $\sigma(T) \subseteq (-\infty,\rho]$. The first condition is often stated as "$T$ is semibounded below" and the second as "$T$ is semibounded above." – Disintegrating By Parts Apr 01 '17 at 08:26
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    @user92360 : I edited the answer to correct the misstated condition. – Disintegrating By Parts Apr 01 '17 at 08:34
  • Thank you. I posted the same problem @ MO also. But, now I think I have to conclude there is no such example. (Y) – user92360 Apr 01 '17 at 08:51