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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?

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Ok for boundedness and the spectrum. An operator of norm $1$ has his spectrum contained in the unit disk. –  Davide Giraudo Dec 26 '12 at 15:47
    
Dear Johan, I have a curious question: I thought you were reading Lax's Functional Analysis. But Lax doesn't seem to contain any exercises (at least a quick search on the preview on amazon seems to suggest so). Have you switched to a different book? –  Matt N. Dec 26 '12 at 18:46
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@MattN. I am reading that book, and yes it contains exercises, without solutions. But most of the exercises I work on are not from there. It is from my teacher but I don't know where he finds them ;) –  Johan Dec 27 '12 at 8:43
    
Dear Johan, thank you for your answer! : ) I wonder if he takes them from a book. Lax only seems to contain exercises of the form "prove this theorem". I wouldn't count this as real exercise. A book should also contain exercises with computations. –  Matt N. Dec 27 '12 at 8:58
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up vote 2 down vote accepted

Correct. Moreover, you don't need to compute spectral radius, because
$$ \sigma(T)\subset \{z\in\mathbb{C}:|z|\leq\Vert T\Vert\} $$ for any bounded opearator $T:X\to X$ on arbitrary Banach space $X$.

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