Limit everywhere function [Solution]

The problem was to find a function $f$ whose limit $g$ exists everywhere (i.e. $\lim_{t\to x} f(t) = g(x)$) and $f(x) = 2g(x) - \sin x$.

The key is to realize that $g$ must continuous!

(In fact one can show that $f$ is continuous almost everywhere, but that does not help this problem immediately).

If $g$ is continuous then, the functional equation implies that $f$ is too, and the result that $f(x) = \sin x$ follows easily.

To prove $g$ is continuous at $c$ we need to show that given an $\epsilon \gt 0$, there is a $\delta \gt 0$, such that $$|g(x) - g(c)| \lt \epsilon \quad \forall x, 0 \lt |x - c| \lt \delta$$  

Now we have that there is some $\delta$ such that

$$|f(x) - g(c)| \lt \frac{\epsilon}{2}, \quad \forall x, 0 \lt |x - c| \lt \delta$$

Also for any $x_1$ such that $0 \lt |x_1 - c| \lt \delta$ we have a $\delta_1$ such that $$|f(x) - g(x_1)| \lt \frac{\epsilon}{2} \quad \forall x, 0 \lt |x - x_1| \lt \delta_1$$

Thus we can find an $x$ such that for any $0 \lt |x_1 - c| \lt \delta$ we have the above two to be true, and hence adding them gives us that

$$|g(x_1) - g(c)| \lt \epsilon \quad \forall x_1, 0 \lt |x_1 - c| \lt \delta$$