# 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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