Let $f:(0,1]\rightarrow \mathbb R_{\ge0}$ be a continuous function such that for every $t_0\in [0,1]$, $$\int_0^{t_0} [f(t)-f(t_0)]\,dt\le c_1 t_0^\gamma$$ for $0<\gamma<1$ and $c_1>0$. Does it follow that $$\int_0^{t_0} f(t)dt\le c_2t_0^\gamma$$ for some $c_2>0$?
I know that this fails to hold if for example $\gamma=1$ we can have $f(t)=-\log t$ but I have no idea how to prove this for $0<\gamma<1$.
Any help is appreciated.
Let $g(t)=\int_0^{t} f(\tau)d\tau$. Then $$g(t)+g'(t)t\le c_1 t^\gamma$$ i.e. $$\bigg(\frac{g(t)}{t}\bigg)'=\frac{g(t)+g'(t)t}{t^2}\le \frac{c_1}{t^{2-\gamma}}.$$ Integrating over $t\in [t_0,1]$, $$g(t_0)\le ct_0\bigg(\frac{1}{t_0^{1-\gamma}}+1\bigg)\le c_2t_0^\gamma$$ for all $t_0\in (0,1]$ for some $c, c_2>0$.