6
$\begingroup$

Let $\,\,f,g\in C^2([0,1])$ such that $f'(0)g''(0)-f''(0)g'(0)\neq 0$ and $g'(x)\neq 0$ for all $x\in (0,1)$. Let $\theta(x)$ be a real number such that

$$\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(\theta(x))}{g'(\theta (x))}.$$

What can I say about $\lim_{x\rightarrow 0^+} \frac{\theta(x)}{x}$?

$\endgroup$
3
  • 1
    $\begingroup$ i would conjecture that it is $1/2.$ $\endgroup$
    – abel
    Dec 31, 2014 at 3:52
  • $\begingroup$ Why you say that? $\endgroup$ Dec 31, 2014 at 13:34
  • $\begingroup$ @FeliceIandoli: it is easy to guess by taking $g(x)=x$, for instance. $\endgroup$ Jan 1, 2015 at 14:47

1 Answer 1

4
$\begingroup$

We have: $$ f(x) = f(0)+f'(0) x +\int_{0}^{x}f''(t)(x-t)\,dt $$ $$ f'(x) = f'(0)+\int_{0}^{x}f''(t)\,dt $$ hence: $$\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(0)+\int_{0}^{x}f''(t)(1-t/x)\,dt}{g'(0)+\int_{0}^{x}g''(t)(1-t/x)\,dt}=\frac{f'(0)+\int_{0}^{\theta(x)}f''(t)\,dt}{g'(0)+\int_{0}^{\theta(x)}g''(t)\,dt}.\tag{1}$$ Assuming that over the interval $I_x=[0,x]$ both $f''$ and $g''$ have small variation ($\leq\varepsilon$), since $\theta(x)\in I_x$ we have that $\int_{0}^{\theta(x)}f''(t)$ behaves like $\theta(x)\,f''(0)$ while $\int_{0}^{x}f''(t)(1-t/x)\,dt$ behaves like $\frac{x}{2}f''(0)$, hence

$$\frac{f'(0)+\frac{x}{2}f''(0)}{g'(0)+\frac{x}{2}g''(0)} =\frac{f'(0)+\int_{0}^{x}f''(t)(1-t/x)\,dt}{g'(0)+\frac{x}{2}g''(0)}=$$$$ =\frac{f'(0)+\int_{0}^{\theta(x)}f''(t)\,dt}{g'(0)+\frac{x}{2}g''(0)} = \frac{f'(0)+\theta(x)\,f''(0)}{g'(0)+\frac{x}{2}g''(0)}.\tag{2}$$

Collecting...

$$\frac{f'(0)+\frac{x}{2}f''(0)}{g'(0)+\frac{x}{2}g''(0)} = \frac{f'(0)+\theta(x)\,f''(0)}{g'(0)+\frac{x}{2}g''(0)} \Rightarrow \frac{x}{2}f''(0) = \theta(x)\,f''(0) .\tag{3}$$

And taking limit we get that $$\lim_{x\to 0^+}\frac{\theta(x)}{x}=\frac{1}{2}.$$

$\endgroup$
2
  • $\begingroup$ Aren't you using "not enough" hypotesis? Looks too strong to me. $\endgroup$ Jan 2, 2015 at 12:29
  • $\begingroup$ The first hypothesis is enough to grant that we do not divide by zero in our manipulations, and the $C^2$ condition is enough to ensure we can use the Taylor approximation with the integral remainder, just as done. Moreover, since we are interested only in the behaviour in a right neighbourhood of zero, the $C^2$ condition can be lifted to a $C^k\;(k>2)$ condition by convolution. $\endgroup$ Jan 2, 2015 at 12:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .