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I'm a high school student from Bonn, Germany and I have to solve the following problem: If $g:R \rightarrow R $ is a function with the property $g(ab)-ag(b)\leq bg(a)$, for all real numbers a and b, then: I. $g(-a)$ is always smaller or equal $-g(a)$ for all real numbers $a$($g(a)+g(-a)\leq0$)? II.The following identity holds: $ag(\frac{1}{a})+\frac{1}{a}g(a)=0$ for all $a>0$. Thank you!

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  • $\begingroup$ I presume $\;f\;$ is another real functions...? $\endgroup$ – DonAntonio Feb 22 '14 at 20:07
  • $\begingroup$ Sorry! It's all about the function g. My mistake! $\endgroup$ – Mark S. Feb 22 '14 at 20:32
  • $\begingroup$ You still left a little $\;f\;$ on the right side of the inequality, @Mark ... $\endgroup$ – DonAntonio Feb 22 '14 at 20:42
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Since, $g(1)\le2g(1)$ and $g(-1)\le -g(1)+g(-1) \implies g(1)=0$. Also $g(1)\le -2g(-1)\implies g(-1)\le0$.

So, you have $g(-1.a)\le ag(-1)-g(a)$ and $g(-1.-a)\le -ag(-1)-g(-a)$, together adding implies $g(a) + g(-a)\le 0$.(I)

For the second part,$\frac{1}{a}g(a)\le-\frac{1}{a}g(-a)$ and $ag(\frac{1}{a})\le-ag(-\frac{1}{a})$ (using I and $a>0$),by adding we have $0=g(a.\frac{1}{a})\le ag(\frac{1}{a})+\frac{1}{a}g(a)\le -ag(-\frac{1}{a})-\frac{1}{a}g(-a)$ ... (t).

After that I think $g(-1)=0$, is essential to the problem to show that $ag(\frac{1}{a})+\frac{1}{a}g(a)=0$. Because, to show that $ag(\frac{1}{a})+\frac{1}{a}g(a) \le 0$, we need that condition.

Adding, $ag(-\frac{1}{a})-\frac{1}{a}g(a)\ge g(-1)$ and (t), we get $-\frac{2}{a}g(a)\ge g(-1)$. Implying $-2ag(\frac{1}{a})\ge g(-1)$, as well.

That is on adding the last two expressions, $ag(\frac{1}{a})+\frac{1}{a}g(a) \le -g(-1)$, is the best I can get without $g(-1)=0$.

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