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Let $f$ be a continuous function , with the intial conditions: $f(0)=1$, $\frac{df}{dx} \geq f$.

How does one show that $f(x) \geq e^x$ for every $x \geq 0$?

I tried using Taylor series

Thank you.

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    @Fabia$n$: which is what I did couple of minutes before your comment :)2011-03-01

3 Answers 3

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$\frac{d(e^{-x}f(x))}{dx} = e^{-x} \frac{df}{dx} - e^{-x} f = e^{-x} \left( \frac{df}{dx} - f \right) \geq 0$ Hence, $e^{-x} f(x)$ is an increasing function and therefore $e^{-x}f(x) \geq e^{-0}f(0)$.

Hence, $f(x) \geq f(0) e^x = e^x$

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Define $\psi(x)=\log f(x)$. Then $\psi(0)=0$ and \psi'(x) = f'(x)/f(x) \geq 1. Hence $\psi(x) \geq x$, that is $f(x) \geq e^x$.

Remark: Note that $\psi(x)$ is well-defined (since $f(x) > 0$ for all $x > 0$; consider the mean value theorem).

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    @Sivaram: Then use the mean value theorem, it gives you immediately that $\psi(x) \geq x$ (using $\psi(0)=0$ and $\psi' \geq 1$).2011-03-01
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Try using the definition of derivative: f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h} for which you know that f'(a) \geq f(a).

Try using a helper function $h(x) = f(x) - e^x$ and prove that it's positive or zero.

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    Thank you for the comment.2011-03-01