Spectrum of Multiplication Operator $T$ in $C[0,1]$ "Let $X=C[0,1]$ and $v \in X$ be a fixed function. Let $T$ be the multiplication operator by $v$, i.e. $Tx(t)=v(t)x(t)$. Find the spectrum of $T$."
This is an exercise from a PDF of notes I found online. I'm trying to better understand Spectral Theory.
So $\lambda$ is a regular value if $(T-\lambda I)^{-1}$ exists, is bounded, and dense (I think there is a lemma which lets us not worry about the dense part). The set of all regular values is $\rho(T)$ and the spectrum is $\sigma(T)=\mathbb{C}\setminus \rho(T)$.
It seems to me that $(T-\lambda I)^{-1}$ maps some $y(t)$ to $\frac{y(t) + \lambda}{v(t)}$. However, this would be problematic if $v(t)=0$ for some values of of $t \in [0,1]$
I don't have much of an understanding of all these definitions so if someone could give a solution to this example, I think that would help clear up some of the ideas for me. 
Thanks in advance.

The map $(T-\lambda I)^{-1}$ is defined as the solution mapping of the equation
$$
(T-\lambda I) y = z,
$$
i.e. $ y= (T-\lambda I)^{-1}z$. In case  of the multiplication with $v$, this is equivalent to
$$
(v(t)-\lambda) y(t) = z(t).
$$
First consider the case that $\lambda\ne v(t)$ for all $t\in [0,1]$. Then the above equation has a unique solution, moreover, $(T-\lambda I)^{-1}$ can be shown to be bounded.
Second, suppose there is $t_0$ such that $\lambda = v(t_0)$. Then $((T-\lambda I)y)(t_0)=0$ for all $y\in X$, hence the range of $(T-\lambda I)$ and its closure are not equal to $X$. Thus, the spectrum of $T$ is the set of all function values of $v$. 
