Help required to prove the following inequality...

Question

Our professor wrote this problem on board and left the room stating, "Try it if you want. You'll succeed only if have properly gone through the concepts that were taught last week."

Let $f :[0,1]\rightarrow \mathbb{R}$ be continuous on $[0,1]$ and differentiable in $(0,1)$. If $f(0)=0$ and $f'(x) \geq f(x)$ $\forall$ $x \in [0,1],$ prove that $f(x) \geq 0$ $\forall$ $x \in [0,1].$

I tried solving the above problem using mean value theorems, but unfortunately it didn't work. Some suggestions or solutions of this problem will be really helpful.

@Mark, I went that way thinking the problem can be easily tackled... but alas. May be we sometimes cannot see things that are too close to our eyes. Can you help a bit? — Priyadarshi Mukherjee, Feb 20 '19 at 18:57
@PriyadarshiMukherjee Read Gabriel's comment. Try that, reorganize the inequality, and recognize a derivative. — Clement C., Feb 20 '19 at 18:57
Sorry, I thought for a moment that you have $f'(x)\geq 0$. But it is actually $f'(x)\geq f(x)$ so MVT will not be so easy. Anyway, Gabriel Romon already gave a hint. — Mark, Feb 20 '19 at 18:58
I don’t know who down voted. But doing it without an explanation seems very unproductive to me. — mathcounterexamples.net, Feb 20 '19 at 19:00
@mathcounterexamples.net Because literally this question shows no personal work, which is a clear thing in the website policy (which I assume you know?) — Clement C., Feb 20 '19 at 19:01
Possible duplicate of $f:\mathbb R \to \mathbb R$ be differentiable such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is $f(x)>0,\forall x>0$? — Martin R, Feb 20 '19 at 19:02
@ClementC. Yes I know. However I wouldn’t down vote. Recalling the policy looks enough to me. A question of feeling at the end! — mathcounterexamples.net, Feb 20 '19 at 19:03
Well, I voted to close, not downvoted. In the end, it's indeed a matter of feeling. — Clement C., Feb 20 '19 at 19:05

Robert Lewis · Answer 1 · 2019-02-20T20:17:34.980

3

I think it is only fair to note that our OP's assertion that $f'(x) \ge f(x), \; \forall x \in [0, 1]$ tacitly implies $f(x)$ is in fact differentiable on $[0, 1]$. Hence:

For all $x \in [0, 1]$ we have

$f'(x) \ge f(x), \tag 0$

whence

$f'(x) - f(x) \ge 0, \; \forall x \in [0, 1]; \tag 1$

now since $e^{-x} > 0$,

$e^{-x}f'(x) - e^{-x}f(x) \ge 0; \tag 2$

that is,

$(e^{-x}f(x))' \ge 0; \tag 3$

thus, with $f(0) = 0$,

$e^{-x}f(x) = e^{-x}f(x) - e^0 f(0) = \displaystyle \int_0^x (e^{-s}f(s))' \; ds \ge 0; \tag 4$

finally, multiply by $e^x$:

$f(x) \ge 0. \tag 5$

edited Feb 20 '19 at 20:17

answered Feb 20 '19 at 19:55

Robert Lewis

72,871

Thank you very much @Robert – Priyadarshi Mukherjee Feb 21 '19 at 07:09
@PriyadarshiMukherjee: my pleasure good sir! – Robert Lewis Feb 21 '19 at 07:10

Help required to prove the following inequality...

1 Answers1