Proving that the maximum of two convex functions is also convex

Question

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!

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

score 33 · Answer 1 · edited Oct 15 '14 at 08:15

33

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.

edited Oct 15 '14 at 08:15

Did

284,245

answered May 20 '12 at 18:23

dtldarek

37,969

Does this apply similarly for the maximum of two concave functions? – bright-star Apr 18 '14 at 05:02
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@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. – dtldarek Apr 18 '14 at 07:27
@dtldarek is it equivalent to saying that minimum of two concave sets is concave? – kaka Oct 09 '16 at 02:45
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@kaka Yes, you can just use $h(x) = -\max{-f(x),-g(x)}$. – dtldarek Oct 09 '16 at 06:34
@dtldarek so $f(x)$ and $g(x)$ are considered concave and not convex in $h(x)=-\operatorname{max}{ -f(x),-g(x)}$ ? – kaka Oct 09 '16 at 14:53
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@kaka If $f$ is concave, then $-f$ is convex. – dtldarek Oct 09 '16 at 18:44
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Just a short note: the area above the graph is called "epigraph" - https://en.wikipedia.org/wiki/Epigraph_(mathematics) – noamgot Jun 24 '19 at 19:19

Did · Accepted Answer · 2012-05-21T07:31:55.097

19

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...

edited May 21 '12 at 07:31

answered May 20 '12 at 19:47

Did

284,245

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? – yotamoo May 21 '12 at 07:08
Nevermind I saw it, thanks! – yotamoo May 21 '12 at 07:31
What about the case $ \max_{i} {x}{i} - \min{i} {x}_{i} $? – Royi May 30 '17 at 14:33
@Royi If you have a new problem (which frankly I did not get), please post a new question. – Did May 30 '17 at 14:34
@Did, I fixed an error I had there. – Royi May 30 '17 at 14:38

score 4 · Answer 3 · answered Nov 14 '24 at 02:32

A function $ f(x) $ defined for $ x \geq 0 $ is said to be convex if, for every $ 0 \leq x_1 \leq x_2 $ and every $ 0 \leq \lambda \leq 1 $, the inequality $$ f((1 - \lambda)x_1 + \lambda x_2) \leq (1 - \lambda)f(x_1) + \lambda f(x_2) $$ holds.

Now, suppose $ f(x) $ and $ g(x) $ are convex functions defined for $ x \geq 0 $. Define $$ h(x) = \max\{f(x), g(x)\}. $$

Question: Show that $ h(x) $ is also convex.

To show that $ h(x) = \max\{f(x), g(x)\} $ is convex, given that $ f(x) $ and $ g(x) $ are convex, we will proceed as follows:

Proof:

Let $ x_1, x_2 \geq 0 $ and $ 0 \leq \lambda \leq 1 $. We need to show that $$ h((1 - \lambda)x_1 + \lambda x_2) \leq (1 - \lambda)h(x_1) + \lambda h(x_2). $$

By definition of $ h(x) $, we have: $$ h((1 - \lambda)x_1 + \lambda x_2) = \max\{f((1 - \lambda)x_1 + \lambda x_2), g((1 - \lambda)x_1 + \lambda x_2)\}. $$

Since $ f(x) $ and $ g(x) $ are convex, we know that: $$ f((1 - \lambda)x_1 + \lambda x_2) \leq (1 - \lambda)f(x_1) + \lambda f(x_2), $$ $$ g((1 - \lambda)x_1 + \lambda x_2) \leq (1 - \lambda)g(x_1) + \lambda g(x_2). $$

Taking the maximum of both sides, we obtain: $$ h((1 - \lambda)x_1 + \lambda x_2) = \max\{f((1 - \lambda)x_1 + \lambda x_2), g((1 - \lambda)x_1 + \lambda x_2)\} \\ \leq \max\{(1 - \lambda)f(x_1) + \lambda f(x_2), (1 - \lambda)g(x_1) + \lambda g(x_2)\}. $$

Note that: $$ \max\{(1 - \lambda)f(x_1) + \lambda f(x_2), (1 - \lambda)g(x_1) + \lambda g(x_2)\} \\ \leq (1 - \lambda) \max\{f(x_1), g(x_1)\} + \lambda \max\{f(x_2), g(x_2)\}. $$

Thus, we have shown that: $$ h((1 - \lambda)x_1 + \lambda x_2) \leq (1 - \lambda)h(x_1) + \lambda h(x_2). $$

This completes the proof that $ h(x) $ is convex.

Proving that the maximum of two convex functions is also convex

3 Answers3

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