I have the operator $T: \mathscr{C} \left[ 0, 1 \right] \rightarrow \mathscr{C} \left[ 0, 1 \right]$, given by $Tf = gf$, where $g \in \mathscr{C} \left[ 0, 1 \right]$ is fixed. I want to find the spectrum of $T$. First, we see that $T$ is bounded (with respect to $\| \cdot \|_{\infty}$ on $\mathscr{C} \left[ 0, 1 \right]$). To see this, we see that
$$\left| \left( Tf \right) \left( x \right) \right| = \left| g \left( x \right) \right| \left| f \left( x \right) \right| \leq \| g \|_{\infty} \| f \|_{\infty}.$$
So, $\| T \| \leq \| g \|_{\infty}$ and we can now say that the spectrum is contained in the disc of radius $\| g \|_{\infty}$. However, I would like to explicitly find the set! For constant $g$, at least we can say that $g$ is an eigenvalue and hence the spectrum contains at least $g$. However, how can we conclude about the other elements of the disc? It is the case that the spectrum is either empty (when $g$ is non-constant) and it is singleton and contains only $g$ (when $g$ is constant)?
Any help will be appreciated!
Edit:
As per the comment, I am trying to prove that the spectrum is $g \left[ 0, 1 \right]$. To do this, I first take $\lambda \notin g \left( \left[ 0, 1 \right] \right)$. Then, for all $x \in \left[ 0, 1 \right]$, $g \left( x \right) - \lambda \neq 0$. Therefore, if I look at $T_{\lambda} = T - \lambda I$, then I have $T_{\lambda} f = \left( g - \lambda \right) f = 0$ if and only if $f = 0$. That is, $T_{\lambda}$ is invertible. Also, the image of $T_{\lambda}$ is whole of $\mathscr{C} \left[ 0, 1 \right]$, so it is dense in $\mathscr{C} \left[ 0, 1 \right]$.
Finally, it is also easy to see that $T_{\lambda}^{-1} f = \dfrac{f}{g - \lambda}$ which is well-defined and bounded. In fact, $\| T_{\lambda}^{-1} f \| \leq M$, where $\left| g \left( x \right) - \lambda \right| \geq \dfrac{1}{M} > 0$. Such an $M$ exists because $g \left( \left[ 0, 1 \right] \right)$ is compact and $\lambda \notin g \left( \left[ 0, 1 \right] \right)$.
This proves one part: the spectrum is contained in $g \left( \left[ 0, 1 \right] \right)$. However, while proving the other way inclusion, we take $\lambda \in g \left( \left[ 0, 1 \right] \right)$. Then, either $T_{\lambda}$ is not invertible in which case $\lambda$ is a spectral value (in fact, and eigenvalue) or $T_{\lambda}$ is invertible. However, if $T_{\lambda}$ is invertible, I am getting that $T_{\lambda}^{-1}f = \dfrac{f}{g - \lambda}$ (as before), which is not possible since $\dfrac{f}{g - \lambda}$ is not well-defined, since $\lambda \in g \left( \left[ 0, 1 \right] \right)$.
So, can we say from here that the case that $T_{\lambda}$ is invertible cannot occur and all the elements of $g \left( \left[ 0, 1 \right] \right)$ are in fact eigenvalues?