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Suppose $f:R^n\to R$ is both convex and concave, how to prove that $f$ is linear? or exactly speaking, $f$ is affine?

I thought for the whole day, but I cannot figure it out.

When I was working on this problem, I met another problem, are all the convex function continuous?If not, is there any counter example?

Actually, I can prove for one dimensional case, in which $f:R\to R$. However, I cannot generalize it into n dimensional cases.

By the way, I use definition for convex(concave) like this: $$f(t\vec{x}+(1-t)\vec{y})\leq(or \geq)tf(\vec{x})+(1-t)f(\vec{y}), \forall t\in[0,1].$$

Thank you so much!

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3 Answers 3

Whether convexity implies continuity depends on the domain (are you on a bounded domain or all of $\mathbb{R}^n$?). Consider, e.g., a smiling frog; the graph of the function is the smile, but at the end points of the smile, it is the eyes. That's a convex but not continuous function.

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First off.... When I was working on this problem, I met another problem, are all the convex function continuous?If not, is there any counter example?

It is a requirement that all convex functions MUST be continuous, this is because the epigraph of a convex function is a convex set; but a discontinuity (of any type) would obviously preclude the epigraph of a function from being convex and hence the function which induces the epigraph could not be convex. I realize "obviously" is something people don't generally want to hear, but I feel like considering the epigraph isn't obvious but once you do the required implications are obvious.

Secondly...

Concerning that $f(x)$ is affine is equally trivial if you consider the epigraph and a simple geometric argument. $f(x)$ is concave iff $-f(x)$ is convex iff the epigraph of $-f(x)$ is convex. But then the epigraph of $f(x)$ must be convex AND the subgraph of $f(x)$ must be convex, hence the graph must be traced by an affine function.

Note: The epigraph is the set of points on a plot above the graph and the subgraph is the set of points on a plot below the graph.

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Convex functions are continuous. For a proof see Rudin Real and Complex chapter three. Whoops - sorry, didn't read past the first line. PS it is obvious but is tricky to write down. –  AreaMan Nov 28 '13 at 8:46
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Not sure how this comment is constructive? –  Squirtle Nov 29 '13 at 5:01
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I'm not sure how your comment is constructive, but you can have some points anyway. If it is necessary to defend the sanctity of my comment I can say that have provided a reference where the curious can find another proof of the implication convex -> continuous. –  AreaMan Nov 29 '13 at 5:06

Let $g(x) = f(x) - f(0)$. It suffices to show that $g$ (which is also both convex and concave, and satisfies $g(0)=0$) is linear. Next, note that for $t > 1$, $x = (1/t) (tx) + (1 - 1/t) (0)$.

That should give you a good start...

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Thank you so much! That helps a lot! –  breezeintopl Feb 2 '12 at 5:25
    
I'm done! Thank you sooooooo much! –  breezeintopl Feb 2 '12 at 6:05
    
@breezeintopl, Remember to accept Robert Israel's answer as the solution. –  Samuel Reid Feb 2 '12 at 6:36
    
@breezeintopl Click the green checkmark near the voting controls. –  Glen Wheeler Mar 23 '12 at 9:06

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