Spectrum of a multiplication operator in L^2

Question

I have following excercice:

Let $X=\mathbb{L}^2(\mathbb{R})$ and $D=\{f \in \mathbb{L}^2(\mathbb{R}) : \int |xf(x)|^2 < \infty \}$

$A:D\subset X \rightarrow X$ $\; \; Af(x):=xf(x)$,
determine the spectrum of A

I have shown, that $\sigma (A) =\mathbb{R}$ and that $\sigma_p (A)=\emptyset$
Now Im trying to determine the continous spectrum.

Let $z \in \mathbb{R}$ and $f \in C_0^{\infty}(\mathbb{R})$, then $(A-z)f \in C_0^{\infty}(\mathbb{R})$ . Since these functions are dense in $\mathbb{L}^2(\mathbb{R}) $ it follows that Ran(A-z) is dense and therefore $\mathbb{R} \subset \sigma_c(A)$. Is this correct?

kind regards

Answer 1

Let $z\in \mathbb{R}$ and $B_n:= \{x \in \mathbb{R} : |x-z| > \frac{1}{n}\}$ and let g be an arbitrary test function. Now $g \in D(A)$ and hence $1_{B_n}g \in D(A)$ and $1_{B_n}g \in Ran(A-z)$ because $g_n:=1_{B_n}(x)\frac{1}{x-z}g(x) \in D(A)$ and $(A-z)g_n=g(x)1_{B_n}$ also $1_{B_n}g \rightarrow g $ pointwise almost surely and by dominated convergence and Riesz in $\mathbb{L}^2$ Since $\overline{Ran(A-z)}$ is closed it follows that g is in there and therefore the claim.

Answer 2

$A-\lambda I$ is invertible for all $\lambda\notin\mathbb{R}$. This is because $$ \frac{1}{x-\lambda }(A-\lambda I)f=f, \;\;\; f\in \mathcal{D}(A), $$ and because $\frac{1}{x-\lambda }f \in \mathcal{D}(A)$ for all such $\lambda$, with $$ (A-\lambda I)\frac{1}{x-\lambda}f=f. $$ Therefore $\sigma(A)\subseteq\mathbb{R}$, because $\frac{1}{x-\lambda}$ is a bounded multiplication operator on $L^2(\mathbb{R})$.

Every $\lambda\in\mathbb{R}$ is in the spectrum because $\mathcal{R}(A-\lambda I)\ne L^2$ for any such $\lambda$, which can be seen by observing that the following is not in $L^2$ for all $\lambda\in\mathbb{R}$: $$ e^{-x^2}\frac{1}{x-\lambda}. $$