# Continuous spectrum of multiplication operator in $L_2$

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 $$\sigma_c=\mathbb{R}$$

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

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 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. Commented May 16, 2020 at 15:38
• So if $Ran(T)$ is orthogonal only to $g=0$ then $Ran(T)$ is dense? Why is that? Commented May 16, 2020 at 16:17
• 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. Commented May 16, 2020 at 16:20
• 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. Commented May 16, 2020 at 16:50