Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the inequality

$f''(x)x + f'(x) \leq 0$

Also, $f''(x)<0$ and $f'(x)>0$ and $x \in R^+$. And I need to figure out when it is true. I know it is a fairly general question, but I couldn't find any information in several textbooks I have skimmed. Also, I am not sure if integrating would require a sign reversal or not, so I cant go ahead and try to manipulate it my self.

Any help or mention of a helpful source would be much appreciated.

edit: forgot to mention $f(x)\geq 0$ for every $x \in R^+$

share|improve this question
add comment

2 Answers 2

$$0\geq f''(x)x+f'(x)=(f'(x)x)'$$

that is why the function $$f'(x)x$$ decreases for positive reals. Then the maximum should be at $0$, so $$f'(x)x\leq 0,\quad x \in R^+$$ which is a contradiction.

share|improve this answer
add comment

No.

It is not true that if $f(x) \geq 0$, $f'(x) >0$ and $f''(x) < 0$ for all $x \in \mathbb{R}^+$, then $x f''(x) + f'(x) \leq 0$.

Below are a class of counter examples.

Consider $f(x) = 1-\exp(-x) > 0$. We then have that $f'(x) = \exp(-x) > 0$ and $f''(x) = -\exp(-x) < 0$. However, $$xf''(x) + f'(x) = -x\exp(-x) + \exp(-x) = (1-x) \exp(-x)$$ which is negative only when $x>1$.

In general, you can consider $f(x) = 1 - \exp(-\alpha x) > 0$, where $\alpha > 0$.

We then have that $f'(x) = \alpha \exp(-\alpha x) > 0$ and $f''(x) = -\alpha^2 \exp(-\alpha x) < 0$.

Hence, we get that $$x f''(x) + f'(x) = -\alpha^2 x\exp(-\alpha x) + \alpha \exp(-\alpha x) = \alpha \exp(-\alpha x) (1 - \alpha x)$$

The above is negative only when $x > \dfrac1{\alpha}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.