Show that the given function is a uniformly continuous function.

Question

Let $F : \mathbb{R}^{n} → \mathbb{R}$ be defined by $F(x_1, x_2, . . . , x_n) = \max\{|x_1|, |x_2|, . . . , |x_n|\}$. Show that $F$ is a uniformly continuous function.

I really have nothing to show as to what I have tried.

Thanks.

Answer 1

Hint. The triangle inequality on $\def\R{\mathbf R}\R$ gives $|x_i + y_i| \le |x_i| + |y_i|$. Hence, $$ F(x + y) \le F(x) + F(y) $$ So $$ F(x) \le F(x-y) + F(y), \text{ and } F(y) \le F(x-y) + F(x) $$ giving $$ |F(x) - F(y)| \le F(x-y) $$

Answer 2

$||(x_1,...,x_n)||_{max} = \mathop{max}_{i: 1,...,n} |x_i|$ is a norm on $\mathbb{R}^{n}$. So by the second triangle inequality: $$\left| \:||(x_1,...,x_n)||_{max} -||(y_1,...,y_n)||_{max} \right| \leq ||(x_1,...x_n)-(y_1,...y_n)||_{max}$$ Hence $F$ is Lipschitz continuous (with respect to the $||.||_{max}$-norm) and from that is is uniformly continuous.