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How can we show that a bounded and convex function on $\mathbb R$ is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all.

Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.

Asaf Karagila
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DatFaceTo
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4 Answers4

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Suppose that $f$ is a convex function that is not constant. Then there must be $a$ and $b$ with $f(a)<f(b)$. Without loss of generality, assume that $a<b$.

Then for any $c>b$ we must have, due to convexity, $$ f(c) \ge f(a) + (c-a)\frac{f(b)-f(a)}{b-a} $$ which grows without bounds as $c\to \infty$. So $f$ cannot be bounded.

hmakholm left over Monica
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    Not quite following your notation: In which interval does f lie below a straight line? Is c some x outside of the interval in which f lies above the line? Regarding the logic: You showed the implication "f convex AND f non-constant => f not bounded". How is that equivalent to "f convex AND f bounded => f constant"? – DatFaceTo Oct 07 '13 at 21:36
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    The inequality is a consequence of the convexity inequality for $a$, $b$, $c$ -- that is $f(b)$ must be below or at the line between the points at $a$ and $c$. This is the same as saying that $f(c)$ must be above the line defined by the values at $a$ and $b$ (which is easiest to see geometrically, but you can also throw algebra at it). – hmakholm left over Monica Oct 07 '13 at 21:40
  • @Dat, as for logic: Yes, these are equivalent; both implications are equivalent to the statement "it is not true that $f$ is convex and non-constant and bounded", by the rule that $P\Rightarrow Q$ is the same as $\neg(P\land\neg Q)$. If in doubt, draw up a truth table. – hmakholm left over Monica Oct 07 '13 at 21:42
  • Ok I understand the origin of the expression. However I am not entirely sure of the proof: How does that "force" the function to be constant? We assumed it wasn't constant and were to able to show that it will not be bounded due to its convexity. So just one last thing before I get it: What exactly forced it to be a constant function? It would be nice to algebraically show that f gets locked in between two values which must equal each other (e.g. A <= f(b) <= B and A = B, thus f being constant). Thanks a lot! – DatFaceTo Oct 07 '13 at 22:10
  • @Dat: You can also view it as an indirect proof: Let $f$ be bounded and convex. Suppose for a contradiction that it is not constant. Then (insert above argument here) it is unbounded. But this contradicts our knowledge that it is bounded. Therefore it is constant. – hmakholm left over Monica Oct 07 '13 at 22:13
  • I think I get it now. By the way, does it not have to contradict all premises including the one about convexity? – DatFaceTo Oct 07 '13 at 22:19
  • @Dat: When you do an indirect proof you just need to reach some contradiction -- that is, any two things that cannot possibly both be true at the same time. Here we have both "bounded" and "unbounded" for $f$, which contradict each other no matter that we also know other things. – hmakholm left over Monica Oct 07 '13 at 22:49
  • Thank you very much! I am attempting an other approach than contradiction. I'll start a new thread to avoid further extension of the discussion. – DatFaceTo Oct 07 '13 at 23:15
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Hint: Without loss of generality, let $f(y)<f(z)$ for $y<z$. Then $$\frac{z-y}{z-x}f(x)+\frac{y-x}{z-x}f(z) \le f(y)$$ for all $x < y$. Now let $x \to -\infty$ and use boundedness of $f$.

njguliyev
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    How'd you arrive at that expression? Convexity states that: f((1-t)a + tb) <= af(1-t) + tf(b) where a<b and 0<=t<=1. The right-hand side is a straightline. How did you go from the definition to that expression? – DatFaceTo Oct 07 '13 at 21:39
  • Your "definition" is not quite correct. It should be $$f((1-t)x+tz) \ge (1-t)f(x)+tf(z).$$ Now put $t = \frac{y-x}{z-x}$. – njguliyev Oct 07 '13 at 21:44
  • But your expression says the function lies above the line, which is not the case with convex functions, right? And that choice for t, any rationale behind that? – DatFaceTo Oct 07 '13 at 21:59
  • Sorry, when I said "not quite correct" I meant $f(1-t)$ in your expression and I didn't notice that my proof is for convex upwards functions and you are asking about convex downwards ones. But the principle is the same. To understand the rationale behind that just imagine two points with different values on the graph of $f$ and try find out what can you say about infinitely distant point. – njguliyev Oct 07 '13 at 22:16
  • Ok thanks, will do that! – DatFaceTo Oct 07 '13 at 22:24
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Consider a secant of the graph of the function. Inside the segment the graph is below the secant, outside the segment the graph is above the secant. If the secant were not horizontal, the graph would rise faster than the secant, so the function would be not be bounded above, contradiction. So all of the secant are horizonta, and the function is constant.

orangeskid
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I don't think that any of this is true: consider the function f(x) = \exp(-x) on [0,\inf). This function is strictly convex and bounded but not constant.

user155993
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