# Problem of the Week #286 - Mar 05, 2019

Status
Not open for further replies.

#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

-----
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.
-----

#### 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.

Status
Not open for further replies.