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If $f(x)$ is a continuous function on $\mathbb R$, and $f$ is doesn't vanish on $\mathbb R$, does this imply that the function $\frac{f\,'(x)}{f(x)}$ be bounded on $\mathbb R$?

The function $1/f(x)$ will be bounded because $f$ doesn't vanish, and I guess that the derivative will reduce the growth of the function $f$, so that the ration will be bounded, is this explanation correct!?

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2 Answers 2

up vote 2 down vote accepted

Take $f(x)=e^{x^2}$ then $$\frac{f'(x)}{f(x)}= 2x$$ which is not bounded on $\mathbb{R}$

Also if $f(x)$ does not vanish it does not mean that $\frac{1}{f(x)}$ is bounded for example take $e^{-x}$.

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In fact for any positive continuous function $g(x)$ on $\mathbb R$ and any $y_0 \ne 0$, the differential equation $\dfrac{dy}{dx} = g(x) y$ with initial condition $y(0) = y_0$ has a unique solution $y = f(x) = y_0 \exp\left(\int_0^x g(t)\ dt\right)$, and this is a nonvanishing continuous function with $f'(x)/f(x) = g(x)$. So $f'/f$ can do anything a continuous function can.

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