Looking for a possible counterexample

Question

Continuing from the problem of this link:

Domain of Linear operator forming a Hilbert Space

If I try to work backwards, and pose the question that:

Give an example of $(H, <\cdot,\cdot>_{H}, ||\cdot||_{H})$ a $\mathbb R$-Hilbert space and $A : D(A) (\subseteq H) \to H$ be a diagonal linear operator with $\sigma_{P}(A) \subseteq (0,\infty)$ such that triple $(D(A), <A(\cdot),A(\cdot))>, ||A(\cdot)||_{H})$ is NOT a $\mathbb R$-Hilbert Space

Then what do I need?? Please let me know if I am wrong.

i) I need $D(A)$ to contain a Cauchy Sequence of real-valued functions which converge to some complex limit.

ii) This $D(A)$ has to be eligible to be a domain of a diagonal linear operator $A$

and

iii) A has to have all positive eigen values in its point-spectrum.

Isn't it??

Answer 1

You can either say you want a Cauchy sequence in $D(A)$ wrt the norm $x\mapsto \|Ax\|$ that does not converge or you can say you want a Cauchy sequence in $\mathrm{im}(A)$ that does not converge to a point in$\mathrm{im}(A)$.

It is not necessary to consider unbounded operators, for example look at the operator: $$A:\ell^2(\Bbb N)\to\ell^2(\Bbb N),\ \sum_n^\infty x_ne_n\mapsto \sum_n^\infty 4^{-n}x_n e_n$$ Now the finitely supported sequences lie in the image of $A$, but the element $\sum_n^\infty 2^{-n}e_n=\lim_N \sum_n^N 2^{-n}e_n$ does not lie in the image, since its preimage would necessarily be $\sum_n^\infty 2^n e_n$, which does not live in $\ell^2$.

Answer 2

To get some things straight: This has nothing to do with complex numbers. $\inf \sigma_P(A)>0$ is much stronger than $\sigma_P(A) \subset (0,\infty)$. You need unbounded operators (otherwise their domain is closed, hence Hilbert).

Try $A:\ell^2\to\ell^2$, $(Ax)_k = 2^kx_k$.