Here's a homework question I'm struggling with:

Let $f,g$ two convex functions. Prove that $h(x)=\max\{f(x),g(x)\}$ is also convex

I don't know where to begin. The only thing I had in mind was was to try proving that if a function is convex on two sets $A$ and $B$, it is also convex on their union. That does not seem right though, for example if I glue together $f(x)=x^2, g(x)=\frac{x^2}{1000}$ where $f$ is defined on $[0,1]$ and $g$ on $(1,2]$.

Anyway, that was the only thing I thought about. Any better ideas? thanks!

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    $\begingroup$ @yotamoo: While typesetting, use \max instead of max (similarly use \sin instead of sin etc) $\endgroup$ – user17762 May 20 '12 at 17:54
  • $\begingroup$ @Didier Can you add that as an answer adding, probably another line or two, so that this question gets an answer? $\endgroup$ – user17762 May 20 '12 at 18:04

Hint: Use the characterization that $h$ is convex if and only if, for every $t$ in $[0,1]$ and every $(x,y)$, $h(tx+(1-t)y)\leqslant th(x)+(1-t)h(y)$.

Second hint: One wants to prove that $h(z)\leqslant th(x)+(1-t)h(y)$ where $z=tx+(1-t)y$. Since $h=\max\{f,g\}$, this is equivalent to the two inequalities $$ f(z)\leqslant th(x)+(1-t)h(y),\qquad g(z)\leqslant th(x)+(1-t)h(y). $$ Consider the first inequality. By convexity of $f$, one knows that $f(z)\leqslant tf(x)+(1-t)f(y)$. Furthermore, $f(x)\leqslant$ $____$ and $f(y)\leqslant$ $____$, hence...

  • $\begingroup$ Okay I can assume that $h=g$ at $tx_1+(1-t)x_2$, but the behavior of $h$ at $x_1$ and $x_2$ has four different options, depending on what function is greater at these points. How can I come to a conclusion? $\endgroup$ – yotamoo May 21 '12 at 7:08
  • $\begingroup$ Nevermind I saw it, thanks! $\endgroup$ – yotamoo May 21 '12 at 7:31
  • $\begingroup$ What about the case $ \max_{i} {x}_{i} - \min_{i} {x}_{i} $? $\endgroup$ – Royi May 30 '17 at 14:33
  • $\begingroup$ @Royi If you have a new problem (which frankly I did not get), please post a new question. $\endgroup$ – Did May 30 '17 at 14:34
  • $\begingroup$ @Did, I fixed an error I had there. $\endgroup$ – Royi May 30 '17 at 14:38

The hint of @Did solves the problem, but there is another proof, which is more intuitive I think.

A function is convex if and only if the area above its graph is convex. But then, the region above $h(x) = \max\{f(x),g(x)\}$ is the intersection of the area above $f$ and the region above $g$. Moreover, intersection of convex sets is convex, and that concludes the proof.

  • $\begingroup$ Does this apply similarly for the maximum of two concave functions? $\endgroup$ – bright-star Apr 18 '14 at 5:02
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    $\begingroup$ @TrevorAlexander Not really, take for example $f(x) = -(x-2)(x-4)$ and $g(x) = -(x+2)(x+4)$, then $h(-4) = h(-2) = h(2) = h(4) = 0$, but $h(-3) = h(3) = 1$, hence $h$ is not concave. $\endgroup$ – dtldarek Apr 18 '14 at 7:27
  • $\begingroup$ @dtldarek is it equivalent to saying that minimum of two concave sets is concave? $\endgroup$ – kaka Oct 9 '16 at 2:45
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    $\begingroup$ @kaka Yes, you can just use $h(x) = -\max\{-f(x),-g(x)\}$. $\endgroup$ – dtldarek Oct 9 '16 at 6:34
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    $\begingroup$ @kaka If $f$ is concave, then $-f$ is convex. $\endgroup$ – dtldarek Oct 9 '16 at 18:44

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