Let $u\in C[0,T]$ and $\beta\in(0,1)$. Suppose that $$ \int_0^x u(r)k(x,r)\,dr = 0,\quad \text{for all }x>0, $$ where $$ k(x,r):= (x-r)^{-\beta}-r^{-\beta},\quad x>r>0. $$ I want to prove that $u$ must be constant using the fact that $k(x,\cdot)$ is anti-symmetric around $x/2$, for each $x>0$, in the sense that
$$ k(x,r) = -k(x,x-r). $$
Comment: I can prove the statement by a maximum principle argument, after reducing the assumption to $$ \int_0^x (x-r)^{\beta-1}r^{-\beta}(u(r)-u(x))\,dr = 0,\quad x>0, $$ by applying the Riemann-Liouville integral $J^\beta_0f(x):=\int_0^x(x-r)^{\beta-1} f(z)$.