Let $M: C([0,1]) \rightarrow C([0,1])$ be defined by $ Mf(x) = f(x/2), \;\; x\in[0,1]$ Show that $M$ is bounded and that its spectrum is containd in the closed unit disc $\{ \lambda \in \mathbb{C} |\lambda| \leq 1 \}$
my try: $ \|M\| = \sup_{\|f\| \leq 1} \|Mf\| = \sup_{\|f\| \leq 1} \|f(x/2) \| \leq \|f\|$
since $\lim_{k\rightarrow \infty} M^kf(x) = f(0)$ and the spectral radius $\sigma (M) = \lim_{k\rightarrow \infty} |M^k|^{1/k} = |f(0)|^{1/k} = 1$
Is this correct?