4
$\begingroup$

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?

  • 0
    Dear Johan, thank you $f$or your answer! : ) I wonder if $h$e 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

1 Answers 1

2

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 $X$.