Domain of Linear operator forming a Hilbert Space 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,
 A: 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 $\alpha$.
