I want to show that the set of all the solutions to this functional equation $f(x-t) = f(x) -h$ where $f$ is strictly monotone continuous function is the set of affine functions $ax +b$. Update: The constants $t$ can be arbitrarily chosen, and $h$ is only dependent on $t$ and $f$.
Is my statement correct? And what would be a good approach to this problem? I am not sure how to tackle this problem without adding other assumptions such as $f$ is differentiable everywhere.
Those are not the only solutions. For instance, for $f(x+1)=f(x)+1$, you have the solution $x+\frac{\sin(\pi x)}{2}$. Basically, given your affine function $ax+b$ that solves the equation, you can add any periodic, continuous, differentiable function with a period that divides $t$ and with derivative bounded above and below by $\pm a$. (You can relax the differentiable criterion somewhat. For instance, $\frac{|\sin(\pi x)|}2$ also works in the example above. But I don't know a more general, sufficient criterion.)