# Gronwall-like inequality with double integral

Assume that a non-negative function $f(t)$ on $[0,\infty)$ satisfies $f(0)=0$ and for certain $0<\alpha<1$, $$f(t)\leq \int_0^t\int_0^se^{-(s-u)}|f(u)|^\alpha \, du \, ds.$$ Can we get $f(t)=0$ for every $t>0$?? Or is there exists a counter example? Thanks for your help!

• Note that the right hand side may be written as $\int_0^t(1-e^{u-t}) |f(u)|^\alpha \, du$. – Hans Engler Dec 4 '17 at 20:42
• Yes! So this means that $f(t)$ may not equal to 0, right? – Wenguang Zhao Dec 4 '17 at 20:53

The answer is no. It is possible that $f(t) > 0$ for $t > 0$. As a counterexample, consider $f(t) = c_0t^{2/(1-\alpha)}$ for suitable $c_0$, to be chosen later. Then $$\frac{1}{2}\int_0^t(t-u)f(t)^\alpha \, du = c_0^\alpha C_1 t^{2/(1-\alpha)} = c_0^{\alpha - 1}C_1 f(t)$$ for a suitable $C_1 > 0$. Choose $c_0$ such that $c_0^{\alpha-1}C_1 = 1$. Therefore for all $t$ $$f(t) = \frac{1}{2}\int_0^t(t-u)f(t)^\alpha \, du \,.$$ Now for $0 \le t-u \le \frac{3}{2}$ $1 - e^{u-t} \ge 1/2 (t-u)$ and therefore $$f(t) = \frac{1}{2}\int_0^t(t-u)f(t)^\alpha \, du \le \int_0^t(1-e^{u-t}) f^\alpha(u) \, du \quad \text{on} \quad [0,3/2] \, .$$