How to solve this functional equation: $2f(x) = f(x-1)+f(x+1)$?

Question

After some calculations, I came up with this functional equation: $f(x-1)+f(x+1)=2f(x)$. I found linear function is one possible answer, but don't know how to derive it.

I don't know much about the techniques to solve this kind of equation. Can anyone help?

Edited: $f$ is a probability function, that is $0 \leq f \leq 1$, and $f$ is continuous.

Answer 1

You have $f(x+1)-f(x)=f(x)-f(x-1)$ and so $f(x+2)-f(x+1)$ is equal to the same value, and similarly $$f(x+n) - f(x+n-1)=f(x)-f(x-1)$$ for all integer $n$. This gives $$f(x+n) = (n+1)f(x)-nf(x-1)=n(f(x)-f(x-1)) + f(x)$$ which is linear in integer $n$, though not necessarily linear in real $x$.

If you fix the values of $f(y)$ on $[-1,1)$ then you can give a general solution using rounding and $$f(y) = \lfloor y \rfloor\left(f(y-\lfloor y \rfloor)- f(y-\lfloor y \rfloor-1)\right) + f(y-\lfloor y \rfloor)$$

Answer 2

Hint: $f$ is completely determined by its values on $[0,2)$. Use recursion to solve for $f$ given its values on $[0,2)$.