f' bounded by f gives relation to e

I found the following exercise:

Given that $f : [0,\infty] \to \mathbb{R}$ is a differentiable function with $$k \cdot f < f' < K \cdot f$$ for some constants $k,K$. Show that

$$f(0) e^{kx} \leq f(x) \leq f(0) e^{Kx}$$

for $x\geq 0$.

How do you solve that? I noticed that this statement is very easy to solve when assuming that f is of the form $f(x) = e^{Lx}$. But otherwise, I don't see how to solve it.

The inequality above implies that $f(x)\neq 0$, as it will have $0<0$. Let's assume that $f(x)>0$. If it's $f(x)<0$, a similar proof holds.
Then, you can write $$k<\frac{f'}{f} < K \ \Rightarrow \ \int_0^x k \ dt \leq \int_0^x \frac{f'(t)}{f(t)} \ dt \leq \int_0^x K \ dt \ \Rightarrow \ kx\leq \ln(f(x))-\ln(f(0)) \leq Kx$$
which leads to $$kx\leq \ln\left(\frac{f(x)}{f(0}\right) \leq Kx \ \Rightarrow \ f(0)e^{kx}\leq f(x) \leq f(0)e^{Kx}$$
• And of course we could never have $f(x) = 0$ because then the inequality $$kf < f' < Kf$$ would yield $0 < 0$. – User8128 Feb 20 '17 at 23:39
• Strictly speaking, you could have $f(x)<0$, which would reverse the inequalities everywhere. (in the last step the inequalities would be reversed back because $f(0)<0$) – Evangelos Bampas Feb 21 '17 at 0:11