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Assume $f$ is a function defi ned over real numbers for which $f(x)-f(y) \leq (x-y)^2$ for all $x,y \in R$. Prove that $f$ is a constant function.

Attempt

We have that $f(x)-f(0) \leq x^2$ and $f(0)-f(y) \leq y^2$ and thus $f(x)-f(y) \leq x^2+y^2$. But we also have that $f(x)-f(y) \leq (x-y)^2 = x^2+y^2 -2xy$. So that seems to imply that $xy$ is positive. I am not sure how much that helps, though.

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1 Answer 1

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Fix any $x\in\mathbb R$ and $h\in\mathbb R\setminus\{0\}$. Then, letting $y\equiv x+h$ and dividing by $|x-y|=|h|$, one has that $$\left|\frac{f(x+h)-f(x)}{h}\right|\leq|h|.$$ Letting $h\to 0$ implies that $f$ is differentiable at $x$ and $f'(x)=0$.

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  • $\begingroup$ What if $x = y$? $\endgroup$ Commented Feb 20, 2016 at 2:27
  • $\begingroup$ @user19405892 That’s why $h\neq 0$ needs to be assumed. $\endgroup$
    – triple_sec
    Commented Feb 20, 2016 at 2:37
  • $\begingroup$ I'm pretty sure if $x=y$ then $f(x)=f(y)$. What are you asking? @user19405892 $\endgroup$ Commented Feb 20, 2016 at 2:37

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