# Prove that the min and max of 2 continuous function are continuous

Prove that if $f$ and $g$ are continuous functions the so are min⁡{f(x),g(x)} and max⁡{f(x),g(x)}

I know this is true when $f$ and $g$ are not intersect each other, then I can compare them. However, I don't know how to prove it's true when they are intersect.

• You write this is in calculus/proof-writing. Do you know $\epsilon, \delta$ limits and continuity? Is this how you are expecting to write this up? – davidlowryduda Oct 17 '13 at 19:15
• You can prove that $\min,\,\max \colon \mathbb{R}^2 \to \mathbb{R}$ are continuous. Then the continuity of compositions of continuous functions does the rest. – Daniel Fischer Oct 17 '13 at 19:16
• A different hint: $\max(f,g)=\frac12(f+g+|f-g|)$. – dls Oct 17 '13 at 19:20
• @mixedmath yes, I'm expecting a ϵ-δ proof. – Diane Vanderwaif Oct 17 '13 at 19:21
• @DanielFischer, I don't think I have learn that technique yet – Diane Vanderwaif Oct 17 '13 at 19:22

## 1 Answer

Let $h(x) = \min\{f(x),g(x)\}$. Suppose $x_0$ is such that $f(x_0) = g(x_0)$. We want to show $h$ is continuous at $x_0$.

Take $\epsilon > 0$, then there is a $\delta_f$ so that $|f(x) - f(x_0)| < \epsilon$ for $|x-x_0| < \delta_f$, and similarly for $g$ and some $\delta_g$ (with the same $\epsilon$).

Use this, and the fact that $h(x_0) = f(x_0) = g(x_0)$ to show that $|h(x) - h(x_0)| < \epsilon$ whether $h(x) = f(x)$ or $h(x) = g(x)$ as long as $|x-x_0| < \delta$ for some $\delta$.

• and you can do the same to max{f(x),g(x)}? – Diane Vanderwaif Oct 17 '13 at 19:47
• Yep, the exact same argument will work. – BaronVT Oct 17 '13 at 20:44
• This is true for finitely many continuous functions, correct? How can you show that this does not hold for infinitely many continuous functions? @BaronVT – kathystehl Jun 15 '15 at 13:39
• Also, what is your reasoning for setting $x_0$ s.t. $f(x_0)=g(x_0)$? Does this prove the statement generally, or only for the case in which we have an $x_0$ with this property? – kathystehl Jun 15 '15 at 13:42
• @baronVT 'and the fsct that $h(x_0)=f(x_0)=g(x_0)$'? I don't see how you may say this – Bahbi May 17 '16 at 12:06