Finding a constant to bound a function

Question

I'm currently working on a problem about differential equations and I came across the following problem.

Let $f$ be a continuous function defined on $[0,\infty)$ such that $f(x)\ge 0$, $f(0)\ne 0$. Let $g$ be a integrable function defined on $[0,\infty)$ such that $\int_0^\infty g(x)\mathrm{d}x<\infty$. Let $0<p<1$, suppose $$ f(x)\le f(0)+\int_0^x (x-y)^{p-1}f(y)g(y)\mathrm{d}y, $$ does it follow $f(x)\le Kf(0)$ for some constant $K$?

I was unable to prove this or find a counterexample. However I personally believe that there may exist a counterexample.

Answer 1

This inequality does not imply that $f$ is bounded. Obviously any $f$ satisfying $f(0)=1$, $$ 1\le f(x)\le 1 + \int_0^x (x-y)^{p-1}g(y)\, dy \equiv 1 +I(x) $$ will also satisfy your inequality. Here we can easily make $I(x)$ unbounded.

More specifically, let's try $g(y)=h(y)=\chi_{(a-1,a+1)}(y)|y-a|^{-p}$. Then $I(a)=\infty$, and $I(x)$ stays large near $x=a$ since (for example) $$ I(a+h)\ge \int_{a+h-1}^{a+h}(a+h-y)^{p-1}|y-a|^{-p}\, dy =\int_0^1 t^{p-1} |t-h|^{-p}\, dt , $$ and now Fatou's lemma shows that $\liminf_{h\to 0+} I(a+h)\ge\infty$.

We can then take $g(y)=\sum w_n h(y-a_n)$, with $w_n>0$, $a_n\to\infty$. As just discussed, this will allow us to make $f$ unbounded. On the other hand, we can still satisfy conditions such as $g\in L^1$ or $\int e^{2x}g^2<\infty$ if the $w_n$ go to zero sufficiently rapidly (in the latter case, we also need $p<1/2$).