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My question is strictly related to this question.

Frechet Differentiability versus Strict Differentiability

The author said it can be proved that $f(x) - f(y) \le 3x(x - y)$ for all $x > y > 0$. After several attempts, I have not found how this could be proved.

I did find out how the author built his function though. The piece-wise linear function $f$ was built from $x^2$ by using the points on the graph at $x = 2^j$ and $x = -2^j$, $j \in \mathbb{Z}$ and connect them all together by the increasing order of $j$.

Even with this discovery, I can not prove the prementioned inequality. I have gone through the properties of a convex function to, but this also came to no avail.

Please give me a hint regarding this problem. Thank you.

Robert Z
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ElementX
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1 Answers1

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Note that $f$ is a convex increasing function in $(0,+\infty)$ and therefore for $y>0$, $$x\to \frac{f(x)-f(y)}{x-y}$$ is increasing in $(y,+\infty)$.

Let $i\in\mathbb{N}$ such that $2^{i-1}< x\leq 2^{i}$ then $$\frac{f(x)-f(y)}{x-y}\leq \frac{f(2^{i})-f(2^{i-1})}{2^{i}-2^{i-1}}= \frac{4^{i}-4^{i-1}}{2^{i}-2^{i-1}}=(4-1)\cdot 2^{i-1}<3x.$$ Note that if $2^{i-1}\leq y< x\leq 2^{i}$ the first inequality is actually an equality.

Robert Z
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