f(x + y) + f(x − y) = 2f(x)f(y)

$f(0)=1$. For $x=y=0$, we have $2f(0)=2f(0)^2$, so $f(0)\in\{0,1\}$. If $f(0)=0$, then $y=0$ gives $2f(x)=0$ for all $x$, excluded. So $f(0)=1$.

Evenness. $x=0$: $f(y)+f(-y)=2f(y)$, i.e. $f(-y)=f(y)$ and $f$ is even. Observe also that
f'(0) = \lim_{y \rightarrow 0} \frac{f(y) - f(-y)}{2y} =0$. **Differential equation.** Fix $x$ and differentiate twice with respect to $y$:f''(x+y)+f''(x-y)=2f(x)f''(y).Put $y=0$: $2f''(x)=2f(x)f''(0)$, i.e.f''(x)=kf(x),\qquad k=f''(0),$$
with $f(0)=1$, $f'(0)=0$. By uniqueness of solutions of linear ODEs:
- $k=0$: $f(x)=1$;
- $k=\omega^2>0$: $f(x)=\cosh(\omega x)$;
- $k=-\omega^2<0$: $f(x)=\cos(\omega x)$.

Verification. $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ and $\cosh(x+y)+\cosh(x-y)=2\cosh x\cosh y$; constant $1$ gives $2=2$ ✓.

Remark. This is d'Alembert's equation; the name "cosine functional equation" comes from the trigonometric identity. Continuity alone (without differentiability) is already enough, but needs a more delicate argument (integrate against a bump function to get smoothness).

Lesson. When a functional equation has a smoothness hypothesis, differentiating in one variable and plugging a special value often produces an ODE.