If $f:[0,\infty) \to \mathbb{R}$ be a differentiable function with $f(0)=1$ and $f'(x)\geq f(x)$ for all $x>0$. Show that $f'(x)\geq e^x$ for all $x>0$.

Hint: Consider $g(x)=e^{-x}f(x)$.

My main concern is what does $g(x)$ have to do with anything. I feel if I understood its relevance, then I could solve the problem. I'm just not sure how I can declare it's relevance to the problem at hand.

Edit: charlestoncrabb is correct on the proof, I'm just not certain of the relevance of $g(x)$

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    $\begingroup$ What is the sign of $g'(x)$ for $x > 0$? $\endgroup$ – r9m Nov 21 '15 at 8:42
  • $\begingroup$ I assume it must be negative given the definition of g. $\endgroup$ – MC989 Nov 21 '15 at 8:48
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    $\begingroup$ Rather than assuming, you might want to apply the product rule and compute it... $\endgroup$ – Eric Towers Nov 21 '15 at 8:55
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    $\begingroup$ "I have no clue where to begin" how about you begin with the hint you are given? $\endgroup$ – Najib Idrissi Nov 21 '15 at 14:34

Here is an answer for why it must be that $f(x)\geq e^x$:

This is in some sense a "reverse Grönwall inequality". Compute $g'(x)$, as suggested, using the product rule: $$g'(x)=e^{-x}(f'(x)-f(x))\geq 0.$$ Since $f(0)=1$, we have $g(0)=1$ as well, thus $g$ is a non-decreasing function with $g(0)=1$, implying that $g(x)\geq1$, so that $e^{-x}f(x)\geq1\Longrightarrow f(x)\geq e^x$. Thus $f'(x)\geq e^x$ as well.

(Edit: Remember that we assumed $f'(x)\geq f(x)$ to begin with, so $f(x)\geq e^x\Longrightarrow f'(x)\geq e^x$)

(2nd Edit: Also notice the $f(0)=1$ assumption is essential, as throwing it out yields the counterexample $f\equiv0$)

(3rd Edit: The relevance of $g$ is that we have shown $g(x)\geq 1$ for all $x$, which is equivalent the desired inequality)

  • $\begingroup$ I now actually understand the relevance of $g(x)$, but I have a question about your use of "non-decreasing function". Is this because $g'(x)$ can be 0, and thus we can't assume it is strictly increasing, so it is only non-decreasing, not increasing? $\endgroup$ – MC989 Nov 21 '15 at 22:37
  • $\begingroup$ @MC989 yes, since we assume $f'\geq f$, it may be possible that $g'(x)=0$ in the case that $f'(x)=f(x)$ $\endgroup$ – charlestoncrabb Nov 21 '15 at 22:43

Here is another answer.

Suppose $A:=\{x \geq 0:f(x) \leq0 \}$ is non - empty set and take $c := \inf A$

Obviously, $f(c) \leq 0$ (Since $f$ is conti)

By Mean value theorem, we can take $a \in (0,c)$ such that $f'(a) \leq 0$.

However, since $f'(x) \geq f(x)$, $f'(a) \geq f(a)$ so $a \in A$.

This is contradiction to minimality of $c$.

So, $A$ is empty and this implies that $f$ is positive function.

Now, let's get this over with.

$$f'(x) \geq f(x) \Leftrightarrow \frac{f'(x)}{f(x)} \geq 1$$ $$\Leftrightarrow \int_0^x \frac{f'(t)}{f(t)} dt \geq x \Leftrightarrow \ln f(x) \geq x \Leftrightarrow f(x) \geq e^x $$

Since $f'(x) \geq f(x)$, so we're done. ($f'(x) \geq f(x) \geq e^x , x \geq 0)$


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