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.

  • $\begingroup$ 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? $\endgroup$
    – davidlowryduda
    Commented Oct 17, 2013 at 19:15
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    $\begingroup$ 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. $\endgroup$ Commented Oct 17, 2013 at 19:16
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    $\begingroup$ A different hint: $\max(f,g)=\frac12(f+g+|f-g|)$. $\endgroup$
    – dls
    Commented Oct 17, 2013 at 19:20
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    $\begingroup$ @mixedmath yes, I'm expecting a ϵ-δ proof. $\endgroup$ Commented Oct 17, 2013 at 19:21
  • $\begingroup$ @DanielFischer, I don't think I have learn that technique yet $\endgroup$ Commented Oct 17, 2013 at 19:22

1 Answer 1


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 $\delta$.

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    $\begingroup$ and you can do the same to max{f(x),g(x)}? $\endgroup$ Commented Oct 17, 2013 at 19:47
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    $\begingroup$ Yep, the exact same argument will work. $\endgroup$
    – BaronVT
    Commented Oct 17, 2013 at 20:44
  • $\begingroup$ This is true for finitely many continuous functions, correct? How can you show that this does not hold for infinitely many continuous functions? @BaronVT $\endgroup$
    – kathystehl
    Commented Jun 15, 2015 at 13:39
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    $\begingroup$ 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? $\endgroup$
    – kathystehl
    Commented Jun 15, 2015 at 13:42
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    $\begingroup$ @baronVT 'and the fsct that $h(x_0)=f(x_0)=g(x_0)$'? I don't see how you may say this $\endgroup$
    – Bahbi
    Commented May 17, 2016 at 12:06

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