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Now I have seen a lot of answers around here which seem to be good enough. Problem is, our teacher asked us to prove it his way.

Suppose we know that

$$|u(x)−u(y)|≤(x−y)^2$$ Prove, by adding and subtracting $$u((y+x)/2)$$

that u is a constant function.

Now I made use of the triangular inequality and got to

$$||u(x)-u((x+y)/2)|−|u(y)-u((x+y)/2)||≤(x−y)^2$$

I know that I should show that if between any two numbers the midpoint also returns the same constant as with the other two numbers, function is constant. But I cannot seem to get there.

Thanks.

ABC
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Danny
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  • It seems easier to note the derivative is zero – Alex Youcis Apr 13 '13 at 05:19
  • I don't know what you mean by, "I have seen a lot of answers around here which seem to be good enough." Do you mean you have posted this question here before? – Gerry Myerson Apr 13 '13 at 05:23
  • No, Garry, not at all. I just looked around for a possible answer here before asking this, and unfortunately none contained this specific idea. – Danny Apr 13 '13 at 05:30

2 Answers2

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I think the idea you're supposed to use if $$|u(x)-u(y)|\le|u(x)-u((x+y)/2)|+|u((x+y)/2)-u(y)|\le(1/2)(x-y)^2$$ Notice this gives you a better bound than the one you start with. So repeat the procedure. Forever.

Gerry Myerson
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  • But how did you get there? Excuse my lack of math skills, that is. – Danny Apr 13 '13 at 05:33
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    The assumption is that $|u(s)-u(t)|\le (s-t)^2$ for all $s$ and $t$. Let $s=x$ and $t=(x+y)/2$. Then $s-t=(x-y)/2$, and its square is $(x-y)^2/4$. There is a similar term from the second part, giving a total of $(x-y)^2/2$. – André Nicolas Apr 13 '13 at 05:45
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another method is,let consider $x-y=\Delta x$, then $u(x)-u(y)=\Delta u$, and we get:

$|\Delta u| \leq \Delta x^2$, ie.$ |\dfrac{\Delta u}{\Delta x}| \leq |\Delta x|$, $\Delta x \to 0, |u'| \leq 0$, that is $u'=0$, which means $u=C$.

Edit: I misunderstand the question. the above is right one.

chenbai
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