I tried substituting some values of $x$. I also know that $f(x)=x$ is a solution to the function but I am just not able to systematically prove it. Solve the functional equation $f(x+1)+f(x-1)=2f(x)$ for $f(x)$
I tried substituting for $x= x-1; x-2; x+1;x+2$
But I can't seem to get to a systematic method to solve this question.
I do know that $ f(x)=x $ works but can't prove it.
 A: If you don't have more condition on f (continuity, derivability, convexity...), you will have a lot of possibilities...
With no more assomption, using usual sequences theories with linear recurrent condition (I only have the french reference), we can say that there exist two fonction g,h (with no condition on it) such that, if we note E(x) the greatest natural number smaller than x and {x}=$x-E(x) \in [0,1[$ the fractional part of x, we have : $$f(x)=g(\{x\})+h(\{x\})E(x)$$
To see that, consider the sequences $a_n=f(\{x\}+n)$, and note that the hypothesis become $a_{n+1}-2a_n+a_{n-1}=0$, and then use the theory (here 1 is a double zero of the caracteristic polynome)
Another way to state this result (via some manipulation) is to say that  there is two 1-periodic fonction k,l such that $$f(x)=k(x)+l(x)*x$$
We can then take such a fonction, and check that the equation is valid.
I don't think we can do better than that without more hypothesis.
A: Hint: If $f$ is $C^2$ then Taylor's Theorem shows that $$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x+h)}{h}=cf''(x),$$where $c$ is $1/2$ or $1$ or $2$ or something, I never recall the constant without working it out.
A: I can't prove whether or not there are extra solutions, so here's a partial answer:
Let $a$ be any real number and $f(x)=ax$
Now $f(x+1) + f(x-1) = a(x+1) + a(x-1) = ax + a +ax - a = 2ax = 2f(x)$
Therefore, $f(x)=ax$ is always a solution to the equation
A: Assuming continuity, we have, $$\frac {f(x+1)+f(x-1)}{2}=f\left(\frac {(x+1)+(x-1)}{2}\right)$$
Hence, from the equality condition of Jensen's inequality, it is clear that $f''(x)=0$ which means $f(x)$ is linear. Substitute $f(x)=ax+b$, you'll get $b=0$, and $a\in R$.
