Let $f$ be a differentiable function such that $f(0)=0$

Question

Let $f$ be a differentiable on $(0,\infty)$function such that $f(0)=0$

$0 \le f'(x) \le x.f(x) \forall x \in (0,\infty)$ then show that $f(x)=0$

My attempt:

Solving the equation I got

$A_1 \le f(x) \le Ae^{x^2}$

Now $f(0)=0$ then $A_1\le0$ and $A \ge 0$.

I don't think we can claim that $f(x)=0$.

Is the question wrong? Or else where am I making a mistake?

Am I missing something?
$f(x)=\frac{x^2}{2}$ vanishes at $x=0$ and $f'(x)\le x$ $\forall x\in (0,\infty)$. — Umesh Shankar, Feb 01 '23 at 11:02
I think it's supposed to read $f'(x) \leqslant xf(x)$, I'm not sure.. — AlvinL, Feb 01 '23 at 11:04
This is Gronwall's inequality. Just repeat the steps of the proof. — Hyperbolic PDE friend, Feb 01 '23 at 11:18
Possibly related: https://math.stackexchange.com/q/648706/9911 (there the inequality is $f'(x) \le f(x)$) — doetoe, Feb 01 '23 at 12:05

geetha290krm · Answer 1 · 2023-02-01T12:08:25.717

2

Assuming continuity on $[0,\infty)$ we get $(e^{-x^{2}/2}f(x))'\leq 0$ so $e^{-x^{2}/2}f(x)$ is decreasing on $[0,\infty)$ and $0\leq e^{-x^{2}/2}f(x) \leq e^{0}f(0)=0$. [$f(x) \geq 0$ because $f$ is increasing and $f(0)=0$].

Without continuity at $0$ the assumption that $f(0)=0$ is of no use and the result is false.

edited Feb 01 '23 at 12:08

answered Feb 01 '23 at 11:22

geetha290krm

36,632

$e^{-x^2/2}f(x)$ is a decreasing function and it is less than $0$. How would that ensure $f(x)=0$? – Antimony Feb 01 '23 at 11:41
Is it because of the fact that $f'(x) \ge 0$ and $f(x)$ is an increasing function? – Antimony Feb 01 '23 at 11:53
What if it is of the form $0 \le f'(x) \le sinxf(x)$ ? – Antimony Feb 01 '23 at 11:53
@Antimony I have added some details. – geetha290krm Feb 01 '23 at 11:57
@Antimony $\sin x \leq x$ for all $x \geq 0$ so $0\leq f'(x)\leq \sin x f(x)$ also has the same answer. – geetha290krm Feb 01 '23 at 12:07

Let $f$ be a differentiable function such that $f(0)=0$

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