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We consider the following Operator: $H_a=\frac{d^2}{dx^2}+a^2x^2$ were $a\in R^{*}$.

Let be $b\in R$. I want to construct a sequence $w_n$ (Which depends on $b$)such that:

1-$||w_n||_2=1$.

2-$w_n$ Converges weakly to $0$.

3- $ ||(H_a - b)w_n||_2\to 0$

Thank you in adavence

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    $\begingroup$ Typo: in order that $H_a\phi=ia\phi$ I guess you want $\phi(x)=e^{iax^2/2}$? $\endgroup$
    – Francois Ziegler
    Commented Dec 29, 2016 at 17:34
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    $\begingroup$ Typo: I believe "$\forall b\in\mathbf R$" should be moved to immediately before "I want". $\endgroup$
    – Francois Ziegler
    Commented Dec 29, 2016 at 22:08
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    $\begingroup$ The existence of such a sequence (for a fixed $b$, your formulation in (3) is misleading) is equivalent to $b$ being in the essential spectrum (Weyl's criterion). In the answers to your previous question, you already got more detailed information on the spectrum of $H_a$. $\endgroup$
    – Christian Remling
    Commented Dec 30, 2016 at 0:22
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    $\begingroup$ I'm voting to close this question as off-topic because you have already been given references to the relevant literature. Go study them. $\endgroup$
    – Michael Renardy
    Commented Dec 30, 2016 at 14:49
  • $\begingroup$ @Michael Renardy. I did not find my question about the references cited $\endgroup$
    – Fadil Kikawi
    Commented Dec 30, 2016 at 20:01

1 Answer 1

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There is no such ('Weyl') sequence, even satisfying just 1 and 3 (before your edit, i.e. with $b=ia$). In fact, replies to your previous question (assumed accepted...) show that $H_a$ has spectrum $\mathbf R$. So $ia$ is not in it, i.e. $H_a-ia$ has bounded inverse. So 3 implies $$ \|w_n\|=\|(H_a-ia)^{-1}(H_a-ia)w_n\|\leqslant\|(H_a-ia)^{-1}\|\|(H_a-ia)w_n\|\to0 $$ which contradicts 1.

Later. After your edit: existence of such $w_n$ is precisely Weyl's criterion that $b\in\sigma_{\mathrm{ess}}(H_a)=\mathbf R$. So you might just use its known proof. (Sketch: for $b$ in the spectrum, $H_a-b$ has unbounded inverse. So there are $v_n$ such that $\|v_n\|=1$ and $\|(H_a-b)^{-1}v_n\|\to\infty$. Then one checks that $$ w_n:=\frac{(H_a-b)^{-1}v_n}{\|(H_a-b)^{-1}v_n\|} $$ does the trick.)

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  • $\begingroup$ Yes, $\phi(x)=e^{i\frac{a}{2}x^2}$,It can also be seen by $||(H_a - ia)w_n||^2=||(H_a w_n||^2+||ia w_n||^2=$ so, if $||(H_a - ia)w_n||^2\to 0$ we get $ ||ia w_n||\to 0$ which contradicts 1. $\endgroup$
    – Fadil Kikawi
    Commented Dec 29, 2016 at 19:18
  • $\begingroup$ I corrected the statement $\endgroup$
    – Fadil Kikawi
    Commented Dec 29, 2016 at 19:26
  • $\begingroup$ Francois Ziegler@ i want to construct such sequence without use $\sigma_{ess}(H_a)=R$. In fact I want to show $\sigma_{ess}(H_a)=R$ $\endgroup$
    – Fadil Kikawi
    Commented Dec 30, 2016 at 10:47

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