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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.

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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 $$kx\leq \ln\left(\frac{f(x)}{f(0}\right) \leq Kx \ \Rightarrow \ f(0)e^{kx}\leq f(x) \leq f(0)e^{Kx}$$

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    $\begingroup$ And of course we could never have $f(x) = 0$ because then the inequality $$kf < f' < Kf$$ would yield $0 < 0$. $\endgroup$ – User8128 Feb 20 '17 at 23:39
  • $\begingroup$ Thank you very much, I didn't realize that detail. $\endgroup$ – A. Salguero-Alarcón Feb 20 '17 at 23:40
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    $\begingroup$ No problem. Great proof! $\endgroup$ – User8128 Feb 20 '17 at 23:41
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    $\begingroup$ 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 $f(0)<0$) $\endgroup$ – Evangelos Bampas Feb 21 '17 at 0:11
  • $\begingroup$ You're right, I erased that part. Thank you @EvangelosBampas $\endgroup$ – A. Salguero-Alarcón Feb 21 '17 at 0:12

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