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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??

Community
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user92360
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2 Answers2

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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 $\ell^2$.

s.harp
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  • Thanks!! So, finally we can say that the intuition was correct and the inf$\sigma_{P}(A) > 0$ is really strong condition and without that there exist examples such that $(D(A), <A(\cdot),A(\cdot))>, ||A(\cdot)||_{H})$ is NOT a $\mathbb R$-Hilbert Space. – user92360 Mar 31 '17 at 14:26
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    Oh, excuse me, I somehow forgot $0\notin\sigma(A)$. This example is wrong in this case. Indeed for bounded $A$ with $0\notin\sigma(A)$ and global domain $\mathrm{im}(A)$ is a Hilbert space. – s.harp Mar 31 '17 at 14:33
  • If $0\notin\sigma(A)$ then $A$ must be invertible and $\mathrm{im}(A)=H$ and this is a Hilbert space. – s.harp Mar 31 '17 at 14:37
  • I am following these two threads with curiosity. And I think the question posed here is wrong to the best of my knowledge. There is probably no such counterexample! – user86511 Mar 31 '17 at 17:17
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    @user92360 Although, I must admit I am optimistic about the existence of such an example for unbounded linear operator!! – user86511 Mar 31 '17 at 17:19
  • @user86511 The condition $0\notin\sigma(A)$ means $A:D(A)\to H$ is invertible, ie $1$-to-$1$ and $D(A)$ with norm $|A\cdot|$ becomes the same as $H$. – s.harp Mar 31 '17 at 17:48
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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 $A:\ell^2\to\ell^2$, $(Ax)_k = 2^kx_k$.

Dirk
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  • @ Dirk, are you asking to try with this operator $A:l^{2}(\mathbb N) \to l^{2}(\mathbb N)$, defined by $\sum_n^\infty x_ne_n\mapsto \sum_n^\infty 2^{n}x_n e_n$ ?? which is unbounded for sure. And, it maps any Cauchy Sequence to NOT-Cauchy sequence (being divergent!!!). Isn't it ?? – user92360 Apr 01 '17 at 09:23
  • This specific operator does make the space $(D(A),\langle A\cdot,A\cdot\rangle)$ into a Hilbert space. Indeed of $0\notin\sigma(A)$ you should always have that $(D(A),\langle A\cdot,A\cdot\rangle)$ is a Hilbert space. – s.harp Apr 01 '17 at 11:38
  • As far as I see, you do not need the operator to be bounded. Try to find out what $D(A)$ is. – Dirk Apr 02 '17 at 15:50