Does there exist a twice differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$f'(x) = f(x+1)-f(x)$$ for all $x$ and $f''(0) \ne 0$.
Some things that I have tried is that obviously any linear function satisfies the first condition but not the second. Any hints/motivation would be helpful. Thanks.
Since we are just looking for existence, this will do.
Assume $f(z)$ is a function on complex plane instead, and assume it have the form $f(z)=e^{cz}$ for some constant $c$. The condition $f''(z)\not=0$ simply means $c\not=0$, while the other condition give: $ce^{cz}=(e^{c}-1)e^{cz}$ which means we need to solve for $e^{c}-c-1=0$. Applying standard technique to reduce to: $-(c+1)e^{-(c+1)}=-\frac{1}{e}$ which allow us to use Lambert W-function to get $c=-1-W(-\frac{1}{e})$. Hence found the solution in complex number.
To reduce this back to real number just take the real part, so we get: $f(x)=e^{Re(c)x}\cos(Im(c)x)$ which satisfy all condition.
Since the question is about existence: you can think of the right side as a difference quotient. $$f^{'}(x) =\frac{f(x+1)-f(x)}{(x+1)-x}.$$ The right hand side is a slope of the secant line between the points with $x-$coordinates $x$ and $x+1$. So any affine function (with non-zero slope and non-zero constant term) will satisfy this relation. For sure they are twice differentiable.
By affine I mean: $f(x)=ax+b$
