If $f(x-y) + f(x+y) = 2f(x)f(y)$ for all $x, y \in \mathbb{R}$, then $f(x)$ is .....

Given the functional equation:

$f(x-y) + f(x+y) = 2f(x)f(y)$

Substituting $y = 0$, we get:

$f(x) + f(x) = 2f(x)f(0)$

$2f(x) = 2f(x)f(0)$

$f(x)(1 - f(0)) = 0$

This implies that either $f(x) = 0$ for all $x \in \mathbb{R}$, or $f(0) = 1$.

Case 1: If $f(x) = 0$ for all $x$, then $f(-x) = 0 = f(x)$. Thus, $f(x)$ is an even function.

Case 2: If $f(0) = 1$, substitute $x = 0$ in the given functional equation:

$f(0-y) + f(0+y) = 2f(0)f(y)$

$f(-y) + f(y) = 2(1)f(y)$

$f(-y) = f(y)$

This shows that $f(x)$ is an even function.

In all possible cases, $f(x)$ is an even function.