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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.

Souza
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Diane Vanderwaif
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1 Answers1

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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 $\delta$.

BaronVT
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    and you can do the same to max{f(x),g(x)}? – Diane Vanderwaif Oct 17 '13 at 19:47
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    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
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    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
  • In the case there is no such $x_0$, you can show that $f(x) < g(x)$ for all $x$ (or vice-versa) and so $\max(f,g) = g$ (or $f$), which is known to be continuous a priori. (note that in OP's question, they note that they already know how to handle this case, so I omitted it from my original answer)

    For finitely many functions, you can write $\max(f,g,h)$ etc. as a composition of continuous functions $\max, f, g, h,$ etc.

    In the case of infinitely many functions, more things can go wrong: for instance, let $f_n(x) = nx$, then $\max_n(f_n) = \infty$ for most values of $x$.

     – BaronVT Jun 15 '15 at 18:13
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    @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
  • This is by assumption, if it is not true, the proof is trivial (do you see why?) – BaronVT May 19 '16 at 19:53