If $a>0,A>0$, $f(x)\geq 0$ on $\Bbb R$, $f(x)+\int_{x-1}^xf(t)dt=A,\ \forall\ x\in\Bbb R$. Can we show $e^{ax}f(x)$ is increasing.

Question

If $a,A$ are positive real numbers, $f(x)$ being nonnegative on $\Bbb R$, $f(x)+\int_{x-1}^xf(t)dt=A,\ \forall\ x\in\Bbb R$. Can we show $e^{ax}f(x)$ is increasing.

Roughly, if $f$ is differentiable, then $f'(x)+f(x)-f(x-1)=0$, thus $f'(x)+a f(x)\geq 0$, if $a\geq 1$. the result follows.

However, if $0<a<1$ or $f$ is not differentiable, can we still deduce that result?