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

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 $g(x)$

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

• 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? – MC989 Nov 21 '15 at 22:37
• @MC989 yes, since we assume $f'\geq f$, it may be possible that $g'(x)=0$ in the case that $f'(x)=f(x)$ – charlestoncrabb Nov 21 '15 at 22:43

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. ($$f'(x) \geq f(x) \geq e^x , x \geq 0)$$