# $g(x) = \frac{f'(x)}{f(x)}$ not bounded

Let $$f : ]0,1[ \rightarrow ]0,+\infty[$$ a differentiable function such that

$$\lim_{x\to0+} f(x) = 0$$

Show that the function $$g:]0,1[ \rightarrow \mathbb{R}$$ defined by

$$g(x) = \dfrac{f'(x)}{f(x)}$$

is not bounded.

I tried a lot of things: I tried the extension by continuity on 0, tried to pass by Mean value theorem and tried to proof by absurd by bounding $$g(x)$$ but it doesn't lead me anywhere... Thanks for help in advance

Hint: observe that $$g(x) = h'(x)$$, with $$h(x) := \log f(x)$$, and that, by assumption, $$h(x) \to -\infty$$ as $$x\to 0^+$$. Assume by contradiction that $$h'$$ is bounded. Can $$h(x)\to -\infty$$ as $$x\to 0^+$$?

• +1, beat me by one minute ... Jan 5, 2021 at 14:13
• even me also @MichaelHoppe Jan 5, 2021 at 14:14

By the mean value theorem, for any $$0 < x < y < 1$$, there exists $$\xi(x,y) \in (x,y)$$ such that

$$\frac{\log f(y) - \log f(x)}{y-x} = \frac{f'(\xi(x,y))}{f(\xi(x,y))}= g(\xi(x,y))$$

Thus,

$$\lim_{x \to 0+}g(\xi(x,y)) = \lim_{x \to 0+}\frac{\log f(y) - \log f(x)}{y-x}= \frac{\log f(y) - \lim_{x \to 0+}\log f(x)}{y} = +\infty$$

Hence, for any $$n \in \mathbb{N}$$ there exists $$\delta_n \in (0,y)$$ such that $$g(\xi(\delta_n,y))> n$$.

As the mean value theorem is non-constructive it is not possible to determine if $$\xi(\delta_n,y) \to 0$$ with $$y$$ fixed as $$n \to \infty$$. However, since $$y$$ can be chosen arbitrarily close to $$0$$, this shows that $$g$$ is unbounded in every neighborhood of $$0$$.