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Define a pair of operators $S\colon\ell^2 \rightarrow \ell^2$ and $M\colon\ell^2 \to\ell^2$ as follows: $$S(x_1,x_2,x_3,x_4,\ldots)=(0,x_1,x_2,x_3,\ldots) $$ and $$M(x_1,x_2,x_3,x_4,\ldots)=(x_1,\frac{x_2}{2},\frac{x_3}{3},\frac{x_4}{4},\ldots).$$ Let $T=M \circ S$.

Then how to prove that $T$ has no eigenvalues? Is $T$ a compact operator? And self-adjoint? Also, if let $T^m$ denote the composition of $T$ with itself $m$ times. Then how to show directly from the formula for $T$ that $\lim_{m\rightarrow \infty} \|T^m\|^{1/m}=0$.

I think I need to use the Stirling's formula, but I don't know what to do. Please help me. Thank you.

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    I would start by computing $T$ and a few powers of it in order to see what $T^m$ does. Then show that $M$ is compact, so $T = MS$ is compact. Compute the adjoint of $T$. Does it equal $T$? Also, read [this page](http://en.wikipedia.org/wiki/Spectral_radius) attentively, in particular look at what is called Gelfand's formula there.2012-05-03
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    I changed the formatting a little to work around [this bug](http://meta.math.stackexchange.com/questions/3966/comment-with-line-filling-equation-gets-garbled-with-signature-on-top-of-text).2012-05-03

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