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Let $f: R \rightarrow R$. Consider the following properties:

$(1)$ - There are positive constants $a$ and $r$ such that $\forall x, y$ $$|f(x)-f(y)|\leq a(1 + |x|^r+|y|^r)|x-y|.$$

$(2)$ - There is a positive constants $b$ such that $\forall x, y$ $$(f(x)-f(y))(x-y)\leq b(x-y)^2.$$

$(3)$ - $f(0)=0$, (or $f(0)$ is limited).

$(4)$ - There is a positive constant $c$ such that $$f(x)x \leq c(1 + x^2).$$

The questions are:

$(1) + (3) + (4) \Rightarrow (2)$ ????


There is $f$ that satisfies $(1) + (3) + (4)$ but, not satisfies $(2)$????

REMARK: Is easy to see that $(1) + (2) + (3)$ implies $(4)$, or only, $(2) + (3)$ implies $(4)$.

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For $(1)$, When $x < y$, the right hand side is certainly negative, but the left hand side is non negative. Is there a typo? – Kenneth Hung Aug 12 '11 at 16:40
(1) can only hold for $a=0$ and $f$ constant. Did you mean to have $|x-y|$ on the right-hand side? Are there any other absolute values missing in (2) or (4)? And what does “$f(0)$ is limited” mean? – Emil Jeřábek Aug 12 '11 at 16:43
Thanks, i had forgotten, in (1) is |x-y| and not (x-y). – de Araujo Aug 12 '11 at 16:56
As given, $f(x)=\begin{cases}x^{42}(1+\sin x),&x<0,\\0,&x\ge0\end{cases}$ is a counterexample. Are you really sure you don’t want absolute values on the left-hand side of (4)? – Emil Jeřábek Aug 12 '11 at 17:11
I don't find it easy to see that $(1)+(2)+(3)$ implies $(4)$ considering $f(x)=-x^3$ appears to be a counterexample. I think Emil is right and you want absolute values in $(2)$ and $(4)$. – Aubrey da Cunha Aug 12 '11 at 18:43

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