Show that F is uniformly continuous function where $F(x_{1},x_{2},\dots, x_{n})= \max\{ |x_{1}|,|x_{2}|,\dots, |x_{n}| \}$ .

Question

$F\colon \mathbb R^n \to \mathbb R$ defined as $F(x_{1},x_{2},\dots, x_{n})= \max\{ |x_{1}|,|x_{2}|,\dots, |x_{n}| \}$ Show that F is uniformly continuous function. So we have to show $\forall \varepsilon >0,\exists \delta > 0\text{ s.t } x,y \in \mathbb R^n \text{ and } |x-y| < \delta \Rightarrow \left|F(x)-F(y)\right|<\varepsilon.$ Now let $x:=(x_{1},x_{2},\dots, x_{n})$ and $y=(y_{1},y_{2},\dots, y_{n})$ so $|F(x)-F(y)|=|\{\max\{ |x_{1}|,|x_{2}|,\dots, |x_{n}| \}- \{\max\{ |y_{1}|,|y_{2}|,\dots, |y_{n}| \}|$

After this I don't understand how to choose $\varepsilon,\delta$ such that if $|x-y| < \delta \Rightarrow \left|F(x)-F(y)\right|<\varepsilon.$

Answer 1

$\delta = \epsilon$ will work, if we use the $l^\infty$ norm for $\mathbb R ^n$.

I'm having trouble generating the equations here on my phone.

Answer 2

Let $\varepsilon>0$. Define $\delta=\varepsilon$.

Let $|x_i|=\max\{|x_1|,...,|x_n|\}, y_j=\max\{|y_1|,...,|y_n|\}$.

Notice that $|x-y|<\delta=\varepsilon$ implies $|x_i-y_i|<\varepsilon$ (easy to prove using the definition of norm).

From that we get$$|x_i|\le|x_i-y_i|+|y_i|<|y_i|+\varepsilon$$ That is$$|x_i|-|y_i|<\varepsilon$$ Now, from the definition of $j$ we get $-|y_j|\le-|y_i|$, and thus$$|x_i|-|y_j|<\varepsilon$$ By simetry we have also$$|y_j|-|x_i|<\varepsilon$$ which is$$-\varepsilon<|x_i|-|y_j|$$And therefor$$||x_i|-|y_j||<\varepsilon$$Which is$$|F(x)-F(y)|<\varepsilon$$ EOP