Problem of the Week #286 - Mar 05, 2019

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Euge

MHB Global Moderator
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Here is this week's POTW:

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1. Show that if $G$ is an abelian group and $p$ is a prime such that $px = 0$ for all $x\in G$, then $G$ has the structure of a vector space over $\Bbb Z/p\Bbb Z$.

2. If $S$ is a bounded linear operator on a Banach space $X$, show that the spectral radius of $S$ is the infimum of $\|S^n\|^{1/n}$, as $n$ ranges over the positive integers.
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Euge

MHB Global Moderator
Staff member
Here is a hint for problem 1: -- I already posted a solution to this somewhere on this site. Find it!

Euge

MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solutions below.

1. This is answered on https://mathhelpboards.com/linear-a...ds-small-order-11387-post53377.html#post53377

2. If $c_n = \log \|S^n\|$, then $c_{n + m} \le c_n + c_m$ for all $n$ and $m$. Fix a postive integer $k$. We may write $n = kq + r$ where $q$ and $r$ are positive integers with $0 \le r < k$. So by the inequality satisfied by the sequence, $$\frac{c_n}{n} \le \frac{qc_k + c_r}{n} \le \frac{qc_k}{qk} + \frac{c_r}{n} = \frac{c_k}{k} + \frac{c_r}{n}$$ Hence
$$\limsup_{n\to \infty} \frac{c_n}{n} \le \frac{c_k}{k}$$ As $k$ was arbitrary, $$\limsup_{n\to \infty} \frac{c_n}{n} \le \inf_k \frac{c_k}{k}$$ On the other hand, from $\dfrac{c_n}{n} \le \dfrac{c_k}{k} + \dfrac{c_r}{n}$, we find $\inf_m \dfrac{c_m}{m} \le \dfrac{c_k}{k} + \dfrac{c_r}{n}$. In the limit as $n \to \infty$, then in the inferior limit as $k \to \infty$, we obtain $$\inf_{m} \frac{c_m}{m} \le \liminf_{k \to \infty} \frac{c_k}{k}$$ Consequently, $\lim_n \dfrac{c_n}{n} = \inf_n \dfrac{c_n}{n}$, and the result follows.

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