Let $X:=C([0,1],\mathbb{C})$ be equipped with $\Vert f\Vert_{\infty}:=sup_{t\in [0,1]}\vert f(t)\vert$ and define $M:X\to X$ by $(Mf)(t):=tf(t)$. Prove that the spectrum of $M$ is $\sigma(M)=[0,1]$
I solved this exercise but my solution is completely different than the one we got in the class. Is mine still correct?
My solution:
Assume there is some $\tilde{\lambda}\in [0,1]$ such that $\tilde{\lambda}\mathbb{1}-M$ is bijective. Then since $X$ is a Banach space by the inverse operator theorem $(\tilde{\lambda}\mathbb{1}-M)^{-1}$ is bounded. But this is not possible since $(\tilde{\lambda}\mathbb{1}-M)f(t)=(\tilde{\lambda}-t)f(t)$ and so its inverse diverges at $t=\tilde{\lambda}$.
Solution from the class: Assume $\lambda\in \mathbb{C}\backslash [0,1]$ and define $R_{\lambda}f(t):=\frac{1}{\lambda-t}f(t)$. By assumption $inf_{t\in[0,1]}\vert \lambda - t\vert >0$. Hence $C:=\sup_{t\in[0,1]}\frac{1}{\lambda-t}<\infty$ is bounded. So $\Vert R_{\lambda}f\Vert_{\infty}=sup_t \vert \frac{1}{\lambda-t}f(t)\vert\leq C \Vert f\Vert_{\infty}$ $\Longrightarrow$ $\Vert R_{\lambda}\Vert=\sup_f \frac{\Vert R_{\lambda}f\Vert_{\infty}}{\Vert f\Vert_{\infty}}\leq C$ hence $R_{\lambda}$ is bounded. Obviously $(\lambda \mathbb{1}-M)R_{\lambda}=R_{\lambda}(\lambda \mathbb{1}-M)=\mathbb{1}$ so $\forall \lambda \in \mathbb{C}\backslash [0,1]$ $\lambda\mathbb{1}-M$ has an inverse.