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Consider the multiplication operator by the idependent variable on the space $L_2(R)$. It is defined by

$(Mu)(t) = tu(t), t \in R$ a.e.

$dom(M)$ consists of all measurable functions $u$ which satisfy

$$ \int_{R}(1+t^2)|u(t)|^2dt < \infty $$ Show that $\sigma _c =\mathbb{R}$.

Let $t \neq t_0$ and $$g_{\varepsilon} = \left\{ \begin{array}{ll} 1 & \textrm{gdy $|t-t_0|< \varepsilon $}\\ 0 & \textrm{gdy $|t-t_0| \ge \varepsilon$} \end{array} \right.$$ $$\int_{\mathbb{R}}(t-\lambda)fg_{\varepsilon}dx=\int_{|t-t_0|<\varepsilon}(t-\lambda)fdx \xrightarrow{\varepsilon \to 0}0$$ It follows that $g_\varepsilon=0$, because $ran(t-\lambda)$ is dense but I don't know how it's imply that $\sigma_c=\mathbb{R}$

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  • $\begingroup$ A bounded operator has bounded spectrum.. $\endgroup$ Commented May 16, 2020 at 13:59
  • $\begingroup$ My operator isn't bounded, because $t$ isn't in $L_{\infty}$, my bad. So what can i use to prove this equality? $\endgroup$
    – KrzysztofB
    Commented May 16, 2020 at 14:20

1 Answer 1

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To show that the continuous spectrum is $\Bbb R$, you want to show that $M-\lambda I$ has dense range. Meaning that if $g$ is such that $\langle (M-\lambda I) f, g\rangle=0$ for all $f$, then $g\equiv 0$.

However, $\langle (M-\lambda I) f, g\rangle = \langle (t-\lambda) f, g\rangle$. Pick a delta sequence that is centered around $x_0$ for $x_0\neq \lambda$.

I think this should help.

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  • $\begingroup$ I should add that you should show the spectrum must be real by showing $M-\lambda I$ has a bounded inverse for complex $\lambda$. This is not too hard to do. $\endgroup$ Commented May 16, 2020 at 15:38
  • $\begingroup$ So if $Ran(T)$ is orthogonal only to $g=0$ then $Ran(T)$ is dense? Why is that? $\endgroup$
    – KrzysztofB
    Commented May 16, 2020 at 16:17
  • $\begingroup$ That's a general theorem of Hilbert spaces. If $X$ is dense, every $g$ can be approximated by elements in $X$. However if $g$ is orthogonal to everything in $X$, then it is orthogonal to every element of a sequence in $X$ converging to it. You can easily conclude $g$ must be $0$ from that. $\endgroup$ Commented May 16, 2020 at 16:20
  • $\begingroup$ That make sens. I will try to use your hint then. I used the sequence from the task I did before. So let $t_0 \in R: \ t-\lambda \neq 0$ a.e. and $g_{\varepsilon}(x)=f(x) for |t_0-\lambda | \ge \varepsilon$ and $0$ otherwise. It follows that $g_{\varepsilon}$ is convergent to $f$ so $Ran(t-\lambda)$ is dense. I could also prove this using orthogonal property as you wrote without sequence? The second part about boundedness I proved in other task. $\endgroup$
    – KrzysztofB
    Commented May 16, 2020 at 16:50
  • $\begingroup$ @Cameron-could you check it? $\endgroup$
    – KrzysztofB
    Commented May 16, 2020 at 19:05

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