What is the point spectrum of the operator $(M_a u)(x) = |x|^2 u(x)$?

Question

Exercise

Consider on $L_2 (\mathbb{R}^n)$ the operator $M_a$ of multiplication by $$a(x) = |x|^2 := x_1^2 + ... + x_n^2, \quad x = (x_1, ..., x_n) \in \mathbb{R}^n$$ i.e., $$(M_a u)(x) = |x|^2 u(x), \quad x \in \mathbb{R}^n \ a.e.$$ The domain $D(M_a)$ is defined by $$\int_{\mathbb{R}^n} (1 + |x|^4) |u(x)|^2 dx < \infty$$ Show that this defined operator $M_a$ has purely continuous spectrum which fills $\mathbb{R}_+ = [0, \infty)$

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So we have to show that the point spectrum and the resiual spectrum are empty sets, whereas the continuous spectrum is equal to $\mathbb{R}_+$.

And here's my problem: How can I show it?

If I try to show that the point spectrum is empty, this is what I get:

$$(M_a - \lambda I)u = 0$$ $$|x|^2 u(x) - \lambda u(x) = 0$$

If for example we take $\lambda = 4$, then it's eigenvectors are all functions $u$ for $x = 2$ (because $|2|^2 u(2) - 4 u(2) = 4 u(2) - 4 u(2) = 0$).

Wouldn't that mean that the point spectrum here is actually equal to $\mathbb{R}_+$?

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I don't know whether I'm doing something wrong or not. I also don't really know how to find the continuous and residual spectrum. These are the definitions we use:

$$\sigma_p (M_a) = \left\{ \lambda \in \mathbb{C} \ | \ \ker (M_a - \lambda I) \neq 0 \right\}$$ $$\sigma_c (M_a) = \left\{ \lambda \in \mathbb{C} \ | \ R(M_a - \lambda I) \neq \overline{R (M_a - \lambda I)} \right\}$$ $$\sigma_r (M_a) = \left\{ \lambda \in \mathbb{C} \ | \ R(M_a - \lambda I) = \overline{R(M_a - \lambda I)} \neq X \right\}$$

Answer 1

Suppose $\lambda =a+b$ with $b\neq 0$. Then $||x|^{2}-\lambda|\geq |b|$. So the equation $M_au-\lambda u=v$ has the unique solution $v=\frac u {|x|^{2}-\lambda}$ and $\|v\| \leq |b|^{-1}\|u\|$. This proves that $M_a-\lambda $ has a bounded inverse and hence $\lambda \notin \sigma (M_a)$. A similar argument works if $b=0$ and $a<0$.

We have proved that the spectrum of $M_a$ is contained in $[0,\infty)$.

If $\lambda$ is an eigen value then $(M_a-\lambda )u=0$ has the non -trivial solution $u$. But this gives $u(x)=0$ except when $|x|^{2}=\lambda$ and the set of points $x$ with this property has Lebesgue measure $0$. This proves that there are no eigen values.

It remains to show that $M_a-\lambda$ (is injective and) has dense range properly contained in $L^{2}$ when $\lambda \in [0,\infty)$. For any $v \in L^{2}$ and any $n \geq 1$ let $u(x)=\frac {v(x)} {|x|^{2}-\lambda}$ if $|x|^{2}-\lambda \geq \frac 1 n$ and $0$ otherwise. Then $M_au=v\chi_\{{||x|^{2}-\lambda| \geq \frac 1 n}\}$. Conclude that $v$ is in the closure of the range of $M_a-\lambda$.

I will let you finish by checking that $M_a-\lambda$ is not surjective whne $\lambda \in [0,\infty)$.

Answer 2

The function $u \in L^2(\mathbb R^n)$ is a solution of $(M_a - \lambda I) = 0$ if $(|x|^2 - \lambda)u(x)$ holds for all $x$.

This implies that for $x$ such that $|x|^2 \neq \lambda$, $u(x) = 0$. Since the set of points $x$ such that $|x|^2 = \lambda$ has measure zero, this means that $u =0$ in $L^2$.

This shows that the point spectrum is empty.