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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.2012-12-26
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    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?2012-12-26
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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 ;)2012-12-27
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    Dear Johan, thank you for your answer! : ) I wonder if he takes them from a book. [Lax](http://www.amazon.co.uk/Functional-Analysis-Pure-Applied-Mathematics/dp/0471556041/ref=sr_1_1?ie=UTF8&qid=1356598725&sr=8-1) 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.2012-12-27

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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 $X$.