Norm and eigenvalues of a multiplication operator

Question

Let $M$ be a multiplication operator from $L^2(\mathbb{R}, \mathbb{R})$ to itself, written as the following: $Mf(x)= (-3\chi\left\{x \in [-5,1]\right\} + 2\chi\left\{x \in [-1,2]\right\})f(x)$.

Problem

1. Compute the norm of $M$.
2. Find all the eigenvalues of $Nf(x)= (-3\chi\left\{x \in (-5,1)\right\} + 2\chi\left\{x \in (-1,2)\right\})f(x)$.

Attempt at a solution:

1. I wrote $M = -3P_1 + 2P_2$, where $P_1$ and $P_2$ are the orthogonal projections from $L^2$ to the subspace of functions defined on $[-5,1]$ and $[-1,2]$ respectively. Then, I wrote $||M|| \leq 3 + 2 = 5$ , because $P_1$ and $P_2$ have both norm 1, since they are orthogonal projections. My claim is that $||M||= 5$: is that correct? How do I prove this?

2. To find the eigenvalues, I wrote $Nf(x)= \lambda f(x)$ . Now, if $\lambda = 0$, $f(x)$ must be orthogonal to 1, so $\lambda = 0$ is the first eigenvalue. How do I find the entire point spectrum?

Answer 1

About the eigenvalues. Since $Mf(x)=g(x)f(x)$ where $g(x)=-3$ on $(-5,-1)$, $g(x)=2$ on $(1,2)$, $g(x)=-3+2=-1$ on $(-1,1)$ and $g(x)=0$ on $(-\infty,-5)\cup(2,\infty)$, then point spectrum is the set $\{-3,-1,0,2\}$. Because with these values of $\lambda$ the equation $(g(x)-\lambda)f(x)=0$ has a nontrivial solution of the form $f(x)=0$ outside the interval corresponding to the value of $\lambda$, and $f$ arbitrary non-zero in this interval. Since $g$ does not take other essential values, we found all points of the spectrum.

Answer 2

As JustDroppdIn wrote in his comment:

$$||M||= ||g||_{\infty}= \sup \{|g(x)|: x \in \mathbb R\}.$$

Since $|g(x)| \le 3$ for all $x$ and $|g(0)|=3,$ we get

$$||M||=3.$$