Let $f:[0, \infty)\to [0, \infty)$ be non-decreasing and satisfy for all $t>t\_{0}$, $$f(t)+C\int\_{t\_{0}}^{t}f^{\gamma}(s)ds\leq \frac{1}{t-t\_{0}}\int\_{t\_{0}}^{t}f(s)ds,$$ where $0<\gamma<1$ and $C>0$.


Of course, if $f$ is differentiable and the somewhat similar differential inequality $$f'+Cf^{\gamma}\leq 0$$ holds, we have that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$. which follows from comparing $f$ with the function $g(t)=a(T-t)^{1/(1-\gamma)}$ where $a=((1-\gamma)C)^{1/(1-\gamma)}$.


**My question**: Is it possible to show that that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$ which does not use differentiation of $f$, that is, only the inequality in the first display math holds?


The differential inequality is, for example, used in the paper **Bénilan, Philippe, and Michael G. Crandall, "The Continuous Dependence on $p$ of Solutions of $u\_{t}— \Delta \varphi(u) = 0$" Indiana University Mathematics Journal 30.2 (1981): 161-177** to show finite extinction time of a solution $u(x, t)$ of a non-linear parabolic PDE. There, $f$ is an integral over $\mathbb{R}^{n}$ of some power of $u$ and the differentiation is with respect to $t$.


Thanks in advance!



Property so that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?
As $f$ is non-decreasing, the integral inequality implies that $f$ is $0$.


**Proof.** The integral inequality implies, in particular, that the same inequality is satisfied for $C=0$, so
$$
 \int\_{t\_0}^t f(t) \, ds = (t-t\_0)f(t) \le \int\_{t\_0}^t f(s) \, ds
$$
for all $t \ge t\_0$. Now fix $t > t\_0$. Then it follows that the non-increasing function $g: [t\_0,t] \ni s \mapsto f(t) - f(s) \in [0,\infty)$ has integral $\le 0$ over the interval $[t\_0,t]$, and thus this integral is actually $0$. So $g(s) = 0$ for all $s \in (t\_0,t]$ and hence, $f(t) = f(s)$ for all $s \in (t\_0,t]$. Since $t > t\_0$ was arbitrary, it follows that $f$ is constant on $(t\_0,\infty)$.


Using the integral inequality from the question again, but now with $C > 0$, one can readily check that the constant value of $f$ on $(t\_0,\infty)$ is $0$. As $f$ is non-decreasing, we conclude that $f$ is constantly $0$ on $[0,\infty)$. $\square$


