6

Let $I:=[a,b]$ and let $g: I \to \Bbb R$ be continuous on $I$. Suppose that there exists $K > 0$ such that $$|g(x)| \leq K \int_a^x|g| \ \ \forall x \in I.$$ Then $g(x) = 0\ \ \forall x \in I $.

I am stuck with the problem please help!

Martin Argerami
  • 205,756
  • Deleted my earlier comment. This follows right from Gronwells, since you can set any $C$ in Gronwells, you have your function is bounded by $ce^{x-a}$ for all $c>0$, taking limits, you get that it must be 0 – Alan Jul 26 '15 at 04:57
  • See this answer. – leo Jul 26 '15 at 05:06

1 Answers1

1

We directly apply Gronwall's inequality in its integral form to

$\vert g(x) \vert \leq K \int_a^x \vert g \vert dx = \int_a^x K\vert g \vert dx; \tag{1}$

as far as the present purposes are concerned, Gronwall's may be taken to state that continuous $w(x)$ on $I = [a, b]$ satisfying

$w(x) \le \beta + \int_a^x cw(t) dt, \tag{2}$

$\beta$ constant, also satisfy

$w(x) \le \beta e^{\int_a^x cdt} = \beta e^{c(x - a)}; \tag{3}$

since in the present case $\beta = 0$, we thus find that

$0 \le \vert g(x) \le 0 e^{\int_a^x Kdt} = 0; \tag{4}$

thus

$\vert g(x) \vert = 0, \tag{5}$

so

$g(x) = 0, x \in I. \tag{6}$

QED!

Robert Lewis
  • 71,180