What functions satisfy $f(x+1)+f(x-1)=x$?

Question

What functions satisfy $f(x+1)+f(x-1)=x$?

If we consider $f(x)=\frac{1}{2}x$, then we have $f(x+1)+f(x-1)=x$ . I tried to replace $x$ by $x+1, x-1$ in the equality,to get $f(x)$ but without success.

Any help is appreciated.

Answer 1

Since you have found a specific solution $f(x)=\frac12x$, let $f(x)=\frac12x+g(x)$, we get

$$g(x+1)+g(x-1)=0\Longrightarrow g(x)=-g(x+2)$$

therefore, any functions $$f(x)=\frac12x+g(x)$$

will satisfy your equation, where $g(x)=-g(x+2)$.

Answer 2

There are many such functions. Specify the value of $f : \mathbb{R} \mapsto \mathbb{R}$ on the interval $[0, 2)$. Then, define

$$f (x) = 2 \left\lfloor \frac{x}{4} \right\rfloor + h \left( x - 4 \left\lfloor \frac{x}{4} \right\rfloor \right)$$

where $h : [0, 4) \mapsto \mathbb{R}$ is defined by

$$h (y) = \begin{cases} f (y) & y \in [0, 2) \\ y - 1 - f (y - 2) & y \in [2, 4) \end{cases}$$

for all $x \notin [0, 2)$.