Let $f:[0, \infty)\to [0, \infty)$ be **non-increasing** 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 $C>0$.


**My question**: Is it true that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?


**Remark**: 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 $T>0$. This follows from comparing $f$ with the function $g(t)=a(T-t)^{1/(1-\gamma)}$ where $a=((1-\gamma)C)^{1/(1-\gamma)}$.


See also [Property so that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?](https://mathoverflow.net/questions/440661/property-so-that-ft-equiv-0-for-all-t-geq-t-for-some-finite-t0) for the case when $f$ is non-decreasing.


Thanks in advance!



Integral inequality implies $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?
Since $f$ is nonnegative and nonincreasing, we have $f(t)\to L$ as $t\to\infty$ for some $L\in[0,\infty)$. Then from your first display we get
$$L+C\int\_{t\_{0}}^\infty f^{\gamma}(s)\,ds\le L.$$
So, $f=0$ on $(t\_0,\infty)$, as desired.


