An odd function satisfying $g(1-t)+g(1+t)=-t$

Question

I am looking for a continuously diffferentiable odd function $g$ such that $$g(1+t)+g(1-t)=-t$$ for all $t\in\mathbb{R}$.

Is this possible?

Answer 1

$g(1+t)+g(1-t) = g(1+t)-g(1) + g(1-t)-g(1) + 2 g(1) = -t$. Dividing across by $t$ gives $\frac{g(1+t)-g(1)}{t} - \frac{g(1-t)-g(1)}{-t} + 2 \frac{g(1)}{t} = -1$. Taking limits as $t \to 0$ gives $2 \lim_{t \to 0} \frac{g(1)}{t} = -1$, which is impossible (either $g(1) =0$, which cannot be or $g(1) \neq 0$, which cannot be either). Hence no such function exists.

Answer 2

If $t=1$, the equation reduces to $g(0)+g(2)=-1$. If $t=-1$, it reduces to $g(2)+g(0)=1$. It follows that we cannot have a $g$ as desired, regardless of whether it is odd, differentiable, or continuous.

Answer 3

Try differentiating and solving for $g$ from the resulting two equations. Now check for $g(1)$.