Integral inequality implies majorization by solution of ODE

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?

• What do you mean by this discretization. Define $f_k$. Any way $f_k \leq g_k$ does not imply $f\leq g$ if $C>0$.
– Medo
Mar 16 at 12:10
• What could be an appropriate discretization of this problem? Mar 16 at 12:17
• You cannot. Take $g(t)=(1-t)^2$ (with $C=2$) and $f(t)$ that drops extremely quickly from $f(0)=1$ to $f(\delta)=1/16$ with very small $\delta>0$ and stays there afterwards on the whole interval $[\delta,1]$. Mar 16 at 14:43
• @fedja: that should be an answer? Mar 16 at 17:34
• @WillieWong Well, the question was "how to show...", not "is it true that...". Besides, this post hardly passes the criteria for "being appropriate for MO", so a full answer would not be a right thing to post. I just did what I usually do in such situations: left a helpful comment and let other people decide if they want to close the thread :-) Mar 16 at 17:40

The problem with this question, compared to this one of yours, 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.
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.
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 $$f(t), you have $$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$$).