$f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$ and $f(0) \geq 0$ and $f^\prime(x) \geq f(x)$ to show $f(x)\geq 0 \forall x \in (0,\infty)$

my answer: if $\exists x_0 \in (0,\infty)$ s.t. $f(x_0)<0$ then $\frac{f(x_0)-f(0)}{x_0} = f^\prime(a_1) \leq 0$ now $a_1 \in (0,x_0) $

$\frac{f(a_1)-f(0)}{a_1}=f^\prime(a_2) \leq 0$

continuing this way we get a sequence $(a_i) \rightarrow 0$ s.t.$f^\prime(a_i)\leq 0 \leq f(a_i)$ which is a contradiction

is this correct or not?

  • 1
    $\begingroup$ 1 - How do you prove that $a_i\rightarrow 0$? 2 - Why is $f'(a_i)\leq 0\leq f(a_i)$? $\endgroup$ – Tomás Jan 23 '13 at 18:11
  • $\begingroup$ +1 I'm having a lot more trouble with this question than I expected. I can prove it for $f$ analytic; but there must be some easy trick. $\endgroup$ – user7530 Jan 23 '13 at 18:47

$f$ satisfies the ODE $$u'(x) = u(x) + g(x)$$ with initial conditions $u(0) = f(0)$, and where $g(x) = f'(x)-f(x) \geq 0.$

This ODE has a solution $$u(x) = f(0)e^x + e^x\int_0^x e^{-y}g(y)\,dy$$ which is clearly positive everywhere. Uniqueness isn't immediate since $g(x)$ is not necessarily continuous, but since $0$ is the unique solution to the ODE $$(f-u)' = (f-u)$$ with initial conditions $(f-u)(0) = 0$, it follows that $f=u$.

  • $\begingroup$ Looks fine to me. Also, I think it might be possible to show $g(x)$ is actually continuous since otherwise there would need to be an essential discontinuity for $f'$, which means that it would be oscillating a lot and this violates $f'(x) \geq f(x)$ immediately. $\endgroup$ – Marek Jan 24 '13 at 10:00
  • $\begingroup$ Also, as you said, it is not hard to prove this for analytic $f$ (since $f$ behaves like $x^k$ and $f'$ like $x^{k-1}$ near a zero of $f$). I think this generalizes to arbitrary differentiable functions but I can't really come up with the proof. Any ideas? $\endgroup$ – Marek Jan 24 '13 at 10:16
  • $\begingroup$ I'm not sure... even a $C^{\infty}$ function can have a Taylor series with radius of convergence 0 at its roots, in which case the series can't be used to extend the positive-ness of $f$ past the root. $\endgroup$ – user7530 Jan 24 '13 at 16:18

$$f^\prime(x) \geq f(x) \Longleftrightarrow e^{-x}(f^\prime(x) - f(x))\ge 0 \Longleftrightarrow \frac{d}{dx}[e^{-x}f(x)]\ge 0$$

This implies after integration that: $$f(x)-f(0)= \int_0^x \frac{d}{dt}[e^{-t}f(t)]dt \ge 0$$ Hence $$f(x)\ge f(0)\ge 0$$

  • $\begingroup$ Is it true that $f(x)\geq 0 \iff f'(x)\geq 0$ ? What about $f(x)=sinx ,\ x \in [0,\pi] $? Somehow we can extend this function suitable way in $[0,\infty)$ with positive value. $\endgroup$ – nurun nesha May 14 '18 at 11:22
  • $\begingroup$ @mathiu_lady Of course that is not true. But check it again I have not made use of that. the last equivalence is equality $\endgroup$ – Guy Fsone May 28 '18 at 16:42

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