Given that $f: \mathbb{R}^n \rightarrow{} \mathbb{R},\ f(\mathbf{x}) = \max\{|x_1|,…,|x_n|\}$ is continuous, show that $g: \{\mathbf{x} = (x_1, x_2) \in \mathbb{R}^2 | x_1, x_2 \geq 1\} \rightarrow{} \mathbb{R}$ where $g(\mathbf{x}) = \min\{|x_1 + x_2|, |2x_1 + 3x_2|\}$ is continuous.
So $g(\mathbf{x})$ is continuous if $\forall a \in g, \forall \epsilon > 0, \exists \delta > 0$ such that $||x-a|| < \delta \implies |g(x) – g(a)| < \epsilon$.
I tried to tackle this by breaking down the last bit first, $ |f(x) – f(a)|$. Taking $\mathbf{x}, \mathbf{y}$, we want to show that $||\mathbf{x}-\mathbf{y}|| < \delta \implies |g(\mathbf{x}) – g(\mathbf{a})| < \epsilon$.
It is :
$$|g(\mathbf{x}) – g(\mathbf{a})| = |\min\{|x_1 + x_2|, |2x_1 + 3x_2|\} – \min\{|a_1 + a_2|, |2a_1 + 3a_2|\}\ \leq \sqrt{(\min\{|x_1 + x_2|, |2x_1 + 3x_2|\} – \min\{|a_1 + a_2|, |2a_1 + 3a_2\})^2}$$
I'm guessing I need to relate this to the function given somehow and use the fact that it's continuous but not sure how exactly. Would appreciate tips/help.
Best Answer
Let $f_1(x_1,x_2) =\min \{|x_1|,|x_2|\}$. $f_1$ is continuous because $f_1 (x_1,x_2)=|x_1|+|x_2| -f(x_1,x_2)$. Define $h(x_1,x_2)=(x_1+x_2,2x_1+3x_2)$. Then $h$ is a continuous function and $g=f_1\circ h$ (where $f$ is as defined by you with $n=2$). Composition of two continuous functions is always continuous.