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Find all functions $ F : R \rightarrow R $ having the property that for any $x_1$ and $x_2$ the following inequality holds:

$ F(x_1) - F(x_2) \le (x_1 - x_2)^2 $

My attempt:

Observe that

$ -(x_1 - x_2)^2 \le F(x_1) - F(x_2) \le (x_1 - x_2)^2 $

Assume WLOG that $ x_1 - x_2 >0 $, then we have

$ -(x_1 - x_2) \le (F(x_1) - F(x_2))/(x_1 - x_2) \le x_1 - x_2 $

As $ x_1 $ tends to $x_2$, we have

$ 0 \le dF/dx \le 0 $

hence

$ dF/dx = 0 $

Therefore $ F(x) = $ constant

Is this solution correct? (probably not because the problem didn't say nothing about the function be differentiable) Any tip will be great, thanks.

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    $\begingroup$ Actually you just proved that F is differentiable $\endgroup$ – Aymane Gr Jan 8 '18 at 13:51
  • $\begingroup$ I think that this solution is correct and it is the most natural one. $\endgroup$ – Giuseppe Negro Jan 8 '18 at 14:15
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I like your approach. My only remark is that you should give some more details on this inequality that you used: $$\tag{1} |F(x_1)-F(x_2)|\le (x_1-x_2)^2.$$ This is a consequence of the one given in the text, namely $$\tag{2} F(x_1)-F(x_2)\le (x_1-x_2)^2,$$ because in this inequality the right-hand side is invariant under permutation of $x_1, x_2$ while the left-hand side is not. So we can upgrade (2) to $$\tag{3} \max\{ F(x_1)-F(x_2), F(x_2)-F(x_1)\} \le (x_1-x_2)^2, $$ and this is exactly (1).

P.S. The upgrading from (2) to (3) is an example of the amplification technique, in the words of Terry Tao.

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  • $\begingroup$ How do you proceed to find F(x) from (1) ? $\endgroup$ – Rick Jan 8 '18 at 14:21
  • $\begingroup$ My bad, i should have been clearer. $\endgroup$ – Vinicius L. Deloi Jan 8 '18 at 14:26
  • $\begingroup$ @Rick: see the original post of Vinicius. I am just adding details on the first inequality he used. $\endgroup$ – Giuseppe Negro Jan 8 '18 at 14:27
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We have $$|F(x+nh)-F(x)| \le \sum_{k=1}^n|F(x+kh)-F(x+(k-1)h)|\le n h^2.$$ Let $x=x_1$ and $h=(x_2-x_1)/n$ so $|F(x_2)-F(x_1)|\le (x_2-x_1)^2/n$. Now let $n\to\infty$.

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  • $\begingroup$ That last inequality should be $|F(x_2)-F(x_1)|\le (x_2-x_1)^2/n^3$, but it still works. $\endgroup$ – Paul Sinclair Jan 8 '18 at 16:52
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You are correct, and as a point of interest this problem is one of the coffin problems. Problem 2 on This link

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I think you wanted to say abs(F(x1)-F(x2))<(x1-x2)^2 . Anyway, if you divide by abs(x1-x2) , and then tend x1 to x2 you can prove that F is differentiable in x1

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    $\begingroup$ The absolute value is implied by interchanging $x_1$ and $x_2$ $\endgroup$ – clark Jan 8 '18 at 14:14
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I agree with user Aymane Gr. An alternative solution is to notice that, for arbitrary $x$ $$F(x) - F(0) = \sum_{i=1}^n F\left(i\frac{x}{n}\right)-F\left((i-1)\frac{x}{n}\right) \leq n \left(\frac{x}{n}\right)^2 = \frac {x^2}{n} \quad \forall n,$$ so $F(x) = F(0)$.

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    $\begingroup$ The absolute value is implied by interchanging $x_1$ and $x_2$ $\endgroup$ – clark Jan 8 '18 at 14:14

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