Let $f:\mathbb R\to\mathbb R$ be a twice differentiable function. Suppose that for all real numbers $x$ and $y$, the function $f$ satisfies $f'(x)-f'(y) \leq 3|x-y|$

(a) Show that for all $x$ and $y$, we must have $|f(x)-f(y)-f'(y)(x-y)|\leq 1.5(x-y)^{2}$.

(b) Find the largest and smallest possible values for $f''(x)$ under the given conditions.

I think (b) is trivial. The range is $\pm {3}$. I'm looking for example of that function. For (a) I started with substituting $t=x-y$ assuming $x>y$ but i have no idea about about range of $t$ and how to approach further.

Edit- As mentioned in comment one example of (b) is $f(x)=1.5x^{2}$.

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    $\begingroup$ For (b), if you suspect you can attain $f''(x) = 3$, try moving from that to $f(x)$. $\endgroup$ – BallBoy Jan 10 '18 at 15:22
  • $\begingroup$ @Y.Forman Ahh got it now. It should be $f(x)=1.5x^{2}$. I feel really stupid for missing this point. Are there any other examples? $\endgroup$ – ILiveInValhalla Jan 10 '18 at 15:25

(a) is false. Let $f(x) = 1.5x^2 + 1.$ Then $f'$ satisfies the condition. But if $x=y= 1,$ the left side of the alleged inequality is $|f(1)-f(1) -3f(0)|=3,$  while the right side is $0.$

I'm guessing the inequality in (b) should be

$$|f(x)-f(y)-f'(y)(x-y)|\leq 1.5(x-y)^{2}.$$

  • $\begingroup$ Ah you are right. It was typo. $\endgroup$ – ILiveInValhalla Jan 12 '18 at 13:50

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