Bound on first derivative $\max \left(\frac{|f'(x)|^2}{f(x)} \right) \le 2 \max |f''(x)|$

Question

I want to show that for a function $f \in C_c^2((a,b))$ non-negative, the inequality $$\sup \left(\frac{|f'(x)|^2}{f(x)} \right) \le 2 \sup |f''(x)|$$ holds.

I noticed that the left term is equal to $2 |\sqrt{f(x)}'|^2.$ So the question is equivalent to: Can I bound this term just by the second derivative? Currently, I don't see how this could work.

The problem I am also having with the exercise is that we get problems if $f(x)=0$ cause then we may divide zero by zero on the left-hand side and it is not immediate to me that the limit is finite.

If anything is unclear, please let me know.

Is this even correct? on $[1,2]$ with $f(x)=x$, $f'(x)=1$, $f''(x)=0$, your lhs is $2$ and your rhs is $0$… there exists a similar inequality: http://en.wikipedia.org/wiki/Landau%E2%80%93Kolmogorov_inequality — gniourf_gniourf, May 30 '15 at 14:57
if $[a,b]$ replace $R$ is also true. It is famous inequality $$M_{1}\le \sqrt{2M_{0}M_{2}},M_{i}=sup {|f^{(i)}(x)|}<+\infty$$ — math110, May 30 '15 at 15:02
@math110, your “famous” inequality is a particular case of the LK-inequalities. Here, OP's question is not exactly the same (yet looks similar or related). — gniourf_gniourf, May 30 '15 at 15:08
@math110 because here it's about the max of $|f'|^2/f$ and not about the max of $|f'|^2$ divided by the max of $f$. — gniourf_gniourf, May 30 '15 at 15:26
@JackD'Aurizio as you can see in the comment section, this questions is different from the one that you linked. — Roadrunner324, May 30 '15 at 17:37
@Roadrunner324 I left him a comment on the other question (not sure if it's a good thing to do, but it's the only way I could contact him)—let's hope the question will be reopened… — gniourf_gniourf, May 30 '15 at 18:11
possible duplicate of http://math.stackexchange.com/questions/883476/to-control-first-derivative-with-the-function-itself-fx2-leq-cfx-near-w — Jonas Meyer, May 30 '15 at 19:08

score 0 · Answer 1 · answered May 30 '15 at 15:47

0

I have use proof by contradiction;

Assmue there exsit $x_{0}\in [a,b]$, $$\dfrac{f'^2(x_{0})}{f(x_{0})}>2\max_{x\in[a,b]}|f''(x)|\Longrightarrow 1-\dfrac{2f(x_{0})f''(\theta)}{f'(x_{0})}<0,\forall \theta\in[a,b]$$ Use Taylor's Formula,we have

$$f(x)=f(x_{0})+f'(x_{0})(x-x_{0})+\dfrac{f''(\theta)}{2}(x-x_{0})^2,\theta\in [a,b]$$ Let $x-x_{0}=-\dfrac{2f(x_{0})}{f'(x_{0})}$

then we have \begin{align*}f(x)&=f(x_{0})-2f(x_{0})+\dfrac{f''(\theta)}{2}\cdot\dfrac{4f^2(x_{0})}{f'^2(x_{0})}\\ &=-f'(x_{0})\left(1-\dfrac{2f(x_{0})f''(\theta)}{f'(x_{0})}\right)\\ &<0 \end{align*} contradiction

answered May 30 '15 at 15:47

math110

93,304

2

your solution is full of errors. – Roadrunner324 May 30 '15 at 17:56
Yep, lots of errors. – gniourf_gniourf May 30 '15 at 18:10
which is wrong? – math110 Jun 01 '15 at 01:13
@math110 Why is $f'(x_0) > 0$? You also need to exercise more care at the boundary: to assert that $\theta \in [a,b]$ in Taylor's theorem you need that $x \in [a,b]$, which $x_0 - 2f(x_0)/f'(x_0)$ might not be. – Erick Wong Aug 19 '15 at 14:26

Bound on first derivative $\max \left(\frac{|f'(x)|^2}{f(x)} \right) \le 2 \max |f''(x)|$

1 Answers1

Linked