# Does the multiplication operator $T_f$ given by $T_f\phi = f\phi$ have a discrete spectrum? Can a multiplication operator have continuous spectrum?

Consider the multiplication operator $T_f:L^2(\mathbb{R}) \to L^2(\mathbb{R})$ given by

$$\phi(x) \mapsto (T_f\phi)(x) = f(x)\phi(x),$$ where $$f(x) = \begin{cases} 1, \quad x \ge 0, \\ 0, \quad x < 0. \end{cases}$$

So I checked and found that the spectrum of this operator is $\sigma(T_f) = \sigma_p(T_f) = \{0,1\}$. That is, $(T_f-\lambda I)$ is not injective when $\lambda = 0,1$, and hence the spectrum is simply the point spectrum.

1. Now I have read that the point spectrum $\sigma_p$ has a subset $\sigma_d$ called the discrete spectrum. Are $0$ and $1$ in the discrete spectrum $\sigma_d$ or are they in $\sigma_p\setminus \sigma_d$? One of the conditions for $\lambda$ to be in $\sigma_d$ is that 'the root subspace $L_\lambda(A)$ corresponding to $\lambda$ is finite-dimensional. It seems to me that the eigenfunctions corresponding to, for example, $\lambda = 0$ form an infinite dimensional subspace. I.e. $\phi(x) \in \mathbb{R}$ for $x <0$ and $\phi(x) = 0$ for $x \ge 0$. So this means that $0$ and $1$ are not in the discrete spectrum?
2. Is it possible to construct a simple explicit example of a multiplication operator that has a continuous spectrum? In this case $(T_f-\lambda I)$ should be injective but only have a dense image (i.e. not the whole space). Can a multiplication operator have this type of spectrum?

Your first multiplication operator is a projection because $T_f^2 = T_f$. This operator is selfadjoint, which makes $T_f$ an orthogonal projection.
The operator $T_f$ where $f=\tan^{-1}x$ has continuous spectrum $[-\pi/2,\pi/2]$, and it has no discrete spectrum because there is no set $E$ of positive measure on which $f$ is constant.