Let $f:[0, \infty)\to [0, \infty)$ be non-increasing (and not necessarily differentiable nor continuous) and satisfy $$f(t)\leq f(0)-C\int\_{0}^{t}f(s)^{1/2}ds,$$ where $C>0$. How can one show that then $$f(t)\leq g(t)\quad \text{for all}~t\leq t\_{\*},$$ where $t\_{\*}>0$ and $g$ is a differentiable function on the interval $[0,t\_\*)$ such that $$g'(t)=-Cg(t)^{1/2}\quad \text{for all}~t\leq t\_{\*}, \quad g(0)=f(0)?$$


Obviously if we would have equality in the above integral inequality and differentiability of $f$, we would have $f=g$, but is it still true that $f\leq g$ under this weaker assumptions?


Thanks in advance!



Integral inequality implies majorization by solution of ODE
The problem with this question, compared to [this one of yours](https://math.stackexchange.com/questions/4658708/integral-inequality-implies-majorization-by-solution-of-ode), is that the vector field on the right hand side of the ODE is not a non-decreasing function of $g$. If you try to make the example of @fedja rigorous, you can see that you can manage to build even smooth counterexamples. To make everything work, you need an ODE of the form $g’(t)=v(g(t))$, where $v$ is non-decreasing.




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Another example, this time with a positive vector field:
$$ g’(t)=g(t)(2-g(t)), $$
with initial data $f(0)=g(0)=1$. Then
$$ g(t)=\frac{2e^{2t}}{e^{2t}+1}. $$
The corresponding integral inequality
$$ f(t)\leq 1+\int\_0^t f(s)(2-f(s))ds $$
admits as a solution, for instance,
$$ f(t)=\frac{1+2e^{-(x-10)^2}}{1+2e^{-(10)^2}}. $$
Clearly, $f(10)=3>2>g(10)$. The problem is that the vector field $v(x)=x(2-x)$ fails to be non-decreasing in its argument.




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Heuristic explanation. The second term of your integral equation is like a ‘bag’ that saves you a quantity of energy $v(f(s))$ per second. The inequality you want to prove is disproved as soon as you make $f$ gain more total energy than $g$.


If $v$ is a decreasing function, then as soon as $f(t)<g(t)$, you have $v(f(t))>v(g(t))$. That is, $f$ collects more energy than $g$ per second. So to make a large $f$, it is convenient to keep $f$ small for a suitable amount of time (so that it satisfies the inequality) and at the same time make it stay in a favorable range where you can collect more energy (in my previous example, $v$ is maximized at $x=1$), and use that energy later to grow past $g$ as soon as you have collected enough energy (while $g$ is forced to grow by the ODE and as soon as it gets large it will start collecting energy with a smaller rate with respect to $f$).


