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The problem is, Let $F:\Bbb R^n\to \Bbb R$ be defined as $F(x_1,x_2...x_n)$=max{$|x_1|,|x_2|...|x_n|$}. Show that $F$ is a uniformly continuous function.

I thought to proceed using mathematical induction. After simple computing, I was able to show that $F$ is uniformly continuous for $n=1,2$. (In both the cases, I used Lipschitz condition). I assume this holds for any $n=m$. But cannot prove that it is also true for $n=m+1$. Please help.

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Hint: Show $F(x)$ is a norm. Hence $|F(y)-F(x)|\le F(y-x)$ by the triangle inequality. Also note $F(x) \le \|x\|_e,$ where $ \|\,\|_e$ is the euclidean norm.

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