What does "$\max_{x>0}$" mean in $\max_{x > 0} \min_{1 \leq i \leq n} \frac 1{x_i} \sum_{j=1}^n a_{ij}x_j$?

Question

Corrolary 8.1.31. Let $A = [a_{ij}] \in M_n$ be nonnegative. If $A$ has a positive eigenvector, then $$ \rho(A) = \max_{x > 0} \min_{1 \leq i \leq n} \frac 1{x_i} \sum_{j=1}^n a_{ij}x_j = \min_{x > 0} \max_{1 \leq i \leq n} \frac 1{x_i} \sum_{j=1}^n a_{ij} x_j. $$

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Can someone please help me read this notationally?

What does the "$\max_{x>0}$" mean?

Answer 1

In this context, $x > 0$ means that the vector $x \in \Bbb C^n$ is such that each of its entries is greater than $0$. That is, $x = (x_1,\dots,x_n)$ with $x_j > 0$ for $j = 1,\dots,n$.

With that established, we can rewrite $$ \max_{x > 0}\min_{1 \leq i \leq n} \frac 1{x_i} \sum_{j=1}^n a_{ij}x_j = \max \left\{ \min\left\{\frac 1{x_i} \sum_{j=1}^n a_{ij}x_j : 1 \leq i \leq n \right\}:x = (x_1,\dots,x_n) > 0\right\}, $$ where $\max(S)$ denotes the maximum element of a set $S$.

Answer 2

Unpacking this statement, this says that $\rho(A)$ is equal to the largest value among all $x > 0$ of $f(x)$, where $f(x)$ is equal to the smallest value among all $1 \leq i \leq n$ of $g(x,i)$, where $g(x,i) = \frac{1}{x_i}\sum\limits_{j=1}^na_{ij}x_j$.

Specifically, $\max\limits_{x > 0}$ means "of all of the possible values that we could get out by putting different $x$ in there, the one that gives the biggest number".