Spectrum of a Multiplication Operator I have a question involving the spectrum of a multiplication operator.
We are in the space of square integrable functions over $\mathbb{R}$, $L^2(\mathbb{R})$, and we define $$(T\psi)(x)=f(x)\psi(x)$$ where
$$ f(x) = \left\{
     \begin{array}{lr}
       1 &  x \geq 0\\
       0 &  x<0
     \end{array}
   \right. $$
I'm looking for the spectrum of $T$. It is easy to find that the eigenvalues of this function are 0 and 1; My argument is that the spectrum is $\{0,1\}$. If if $z\neq0,1$, $$(T-z)\psi=0$$ implies that $\psi=0$, thus $T-z$ is injective, and any $(f(x)-z)\psi(x)$ in the image of $T-z$ can simply map back to $\psi(x)$, thus $T-z$ is onto, and the spectrum only consists of the eigenvalues. Is this correct? This implies that the continuous spectrum is the empty set, however, and I was under the impression that the image of $T-z$ was dense. 
EDIT: I think I see my mistake, my proof for surjectivity is wrong. If I want to find $g$ such that $(Tg)=y$ for some $y\in L^2(\mathbb{R})$, I actually never can, because I only get a piecewise version of it, so $T-z$ is actually not surjective for any $z$, and the spectrum is all of $\mathbb{C}$.
 A: Firstly a preliminary remark: the spectrum of a bounded operator is always compact so it cannot possibly be all of $\Bbb C$.
If $\lambda\neq 0,1$, then $(T-\lambda)f = 0$ implies that $f = \lambda f$ for $x > 0$ but this definitely is not the case unless $f=0$, giving that $T-\lambda$ is injective as you noted. Thus we need to see when $T-\lambda$ is surjective, that is given some $\lambda$ (we don't know what it can be yet) we want to solve for $f$ given $g\in L^2(\Bbb R)$:
$$(T-\lambda)f = g.$$
Consider $x < 0$, then
$$-\lambda f(x) = g(x).$$
Hence $$f(x) = -\frac{1}{\lambda} g(x).$$
If instead $x\ge 0$, we have that
$$(T-\lambda)f = f-\lambda f = (1-\lambda)f.$$
Thus
$$f(x) = \frac{1}{1-\lambda}g(x)$$
So we define
$$f(x) = \begin{cases} \dfrac{1}{1-\lambda}g(x) & x\ge 0 \\ -\dfrac{1}{\lambda} g(x) & x < 0\end{cases}.$$
The only thing left to see is that $f$ is in $L^2(\Bbb R)$ but that is pretty easy to see by splitting your interval up. Notice that the above works for any $\lambda\neq 0,1$ and the fact that $0,1$ do not work appeared very naturally in our function $f$ as these are the poles.

Here is a cleaner approach, I think. Note that $T^2 = T$ and thus $T$ is a projection operator.$^{[1]}$ In general, projection operators tend to have nice "spectral" inverses. Consider $(T+\mu I)(T-\lambda I)$. This is nothing more than
$$(T+\mu I)(T-\lambda I) = T^2 + \mu T-\lambda T - \mu\lambda I = (\mu+1-\lambda)T-\mu\lambda I.$$
If $\mu = \lambda-1$, then we get $-(\lambda-1)\lambda I$. So unless $\lambda = 0$ or $\lambda = 1$, $T-\lambda I$ has an inverse (and the inverse is $T+(\lambda-1)I$). Furthermore, $0$ and $1$ are realized as spectral values by considering functions with support on $(-\infty,0)$ and $(0,\infty)$, respectively.
$^{[1]}$ $T$ acts as a projection onto the subspace $V$ of functions with support on $[0,\infty)$; the orthogonal complement of $V$ is nothing more than the subspace of functions with support on $(-\infty,0)$ and it's not hard to see that these two span the whole space.
