Given that function $g$ is positive and continuous on the interval $[0,1]$, function $f$ is increasing and differentiable on $[0,1]$, $f(0)=1$. We have

$$g(x) \leq 1+\int_0^{x}{\frac{f'(t)}{f(t)}g(t)dt}$$ Prove that $$g(x)\leq f(x), \forall x \in [0,1].$$

My idea is:

Let $H(x)=f(x)-1-\int_0^{x}{\frac{f'(t)}{f(t)}g(t)dt}$, $H(0)=0$ and I've tried to prove $H\geq 0$. We have $$H'(x)=f'(x)-\frac{f'(x)}{f(x)}g(x)+\frac{f'(0)}{f(0)}g(0)$$

As the condition $g(x) \leq 1+\int_0^{x}{\frac{f'(t)}{f(t)}g(t)dt}$, so $$H'(x)\geq f'(x)-\frac{f'(x)}{f(x)} -\frac{f'(x)}{f(x)}\int_0^{x}{\frac{f'(t)}{f(t)}g(t)dt}+\frac{f'(0)}{f(0)}g(0)$$

I know $f'(x)-\frac{f'(x)}{f(x)}\geq 0,\frac{f'(0)}{f(0)}g(0)\geq 0$. I'm trying to prove $H' \geq 0$, but I can't, help me.

Thank you so much.

  • 1
    $\begingroup$ Note that $H'(x)=f'(x)-\frac{f'(x)}{f(x)}g(x)$, without the $\frac{f'(0)}{f(0)}g(0)$ term. $\endgroup$ – Martin R Apr 6 '18 at 11:40

This is a special case of Grönwall's inequality in it's integral form (with $u = g, \alpha = 1, \beta = \frac{f'}{f}$).

Adapting the general proof to this special case would look as follows: Define $$ \tag{*} v(x) = \frac{\int_0^x \frac{f'(t)}{f(t)}g(t) dt}{f(x)} \, . $$ Then $v(0) = 0$ and $$ v'(x) = \frac{f'(x) g(x) - f'(x)\int_0^x \frac{f'(t)}{f(t)} g(t) dt } {f(x)^2} \le \frac{f'(x)}{f(x)^2} $$ which implies $$ v(x) \le \int_0^x \frac{f'(t)}{f(t)^2} dt = 1 - \frac{1}{f(x)} \, . $$ Substituting this back into $(*)$ gives $$ \int_0^x \frac{f'(t)}{f(t)}g(t) dt \le f(x) \left( 1 - \frac{1}{f(x)}\right) = f(x) - 1 $$ and therefore $$ g(x) \leq 1+\int_0^{x}{\frac{f'(t)}{f(t)}g(t)dt} \le f(x) \, . $$

So in order to compare $\int_0^x \frac{f'(t)}{f(t)}g(t) dt$ with $f(x)$, the “trick” is to compute the derivative of the quotient $(*)$ and not of the difference, as you tried with $H(x)$.


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