Problem Of The Week # 342 - Mar 05, 2019

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Euge

MHB Global Moderator
Staff member
Hi all,

I was sick for some time, so I had not posted any new problems for either the uni POTW or the grad POTW for a couple weeks. Just this time, there will be a special of two problems posted today for both the university and graduate levels! Here is this week's two POTW:

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1. Suppose $f$ is a continuous, complex-valued function on the complex plane $\Bbb C$ such that $\lim\limits_{\lvert z\rvert \to \infty} \lvert f(z)\rvert = 0$. Prove that $f$ has maximum modulus in $\Bbb C$.

2. If $X$ and $Y$ are $n\times n$ matrices over a field $F$, show that the trace of $X\otimes Y$ is the product of the traces of $X$ and $Y$.

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You may submit a solution one of the two problems or submit solutions to both of the problems. Remember to read the POTW submission guidelines to find out how to submit your answers!

Euge

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

1. If $f = 0$ there is nothing to prove, so assume $f$ is not identically zero. Since $\lim_{\lvert z\rvert \to \infty} f(z) = 0$, $f$ is bounded, so that $\sup_{z\in \Bbb C} \lvert f(z)\rvert$ exists, call it $\alpha$. As $f$ is not identically zero, $\alpha > 0$. Set $S_k := \{z\in \Bbb C : \lvert f(z)\rvert \ge (1 - 1/k)\alpha\}$ for $k\in \Bbb N$. Each of the sets $S_k$ is nonempty; indeed, given $k\in \Bbb N$, $\alpha/k > 0$ so there must be a $z\in \Bbb C$ with $\lvert f(z)\rvert > \alpha - \alpha/k = (1 - 1/k)\alpha$. By continuity of $f$, each $S_k$ is closed. Furthermore, the $S_k$ are bounded. For as $f(z) \to 0$ as $\lvert z\rvert \to \infty$, there is an $R_k > 0$ such that $\lvert f(z)\rvert < (1 - 1/k)\alpha$ for all $\lvert z\rvert > R$. Therefore $S_k$ is contained in the closed disk of radius $R_k$ centered at the origin. We deduce from the Heine-Borel theorem that each $S_k$ is compact. Since $S_1 \supset S_2 \supset S_3 \supset \cdots$, Cantor's nested intersection theorem yields an element $z_0 \in \bigcap S_k$. Then $\lvert f(z_0)\rvert \ge (1 - 1/k)\alpha$ for all $k$. Taking limits as $k \to \infty$, $\lvert f(z_0)\rvert \ge \alpha$. This forces $\lvert f(z_0)\rvert = \alpha$. Hence, $\lvert f\rvert$ achieves its maximum at $z_0$.

2. The diagonal elements of $X\otimes Y$ come from the diagonal elements of $X_{ii}Y$ for $1\le i \le n$. So the trace of $X\otimes Y$ is $\sum_{i,j} X_{ii} Y_{jj} = \sum_i X_{ii} \sum_j Y_{jj} = \operatorname{trace}(X) \operatorname{trace}(Y)$.

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