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!

  • $\begingroup$ What do you mean by this discretization. Define $f_k$. Any way $f_k \leq g_k $ does not imply $f\leq g$ if $C>0$. $\endgroup$
    – Medo
    Mar 16 at 12:10
  • $\begingroup$ What could be an appropriate discretization of this problem? $\endgroup$
    – Shaq155
    Mar 16 at 12:17
  • 2
    $\begingroup$ 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 $[\delta,1]$. $\endgroup$
    – fedja
    Mar 16 at 14:43
  • $\begingroup$ @fedja: that should be an answer? $\endgroup$ Mar 16 at 17:34
  • $\begingroup$ @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 :-) $\endgroup$
    – fedja
    Mar 16 at 17:40

1 Answer 1


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 $f(t)<g(t)$, 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 $f$).


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