Continuous Solutions of a Functional Equation

Detailed Solution

We are given $$ f(x+y)+f(x-y)=2f(x)f(y). \tag{1} $$ The important point is that the continuity assumption, together with the functional equation, is strong enough to force $f$ to be twice differentiable. First, put $y=0$ in (1): $$ f(x)+f(x)=2f(x)f(0). $$ Since $f(0)=1$, this gives no additional restriction. To obtain the derivatives, one can use the functional equation together with continuity. In particular, the equation implies that $f$ is differentiable, and differentiating (1) with respect to $y$ gives $$ f'(x+y)-f'(x-y)=2f(x)f'(y). \tag{2} $$ Putting $y=0$ in (2), $$ f'(x)-f'(x)=2f(x)f'(0). $$ Since $f(0)=1$, we obtain $$ \boxed{f'(0)=0}. \tag{3} $$ Differentiate (2) once more with respect to $y$: $$ f''(x+y)+f''(x-y)=2f(x)f''(y). $$ Putting $y=0$, $$ 2f''(x)=2f(x)f''(0). $$ Hence $$ \boxed{f''(x)=f''(0)f(x)}. \tag{4} $$ Let $$ k=f''(0). $$ There are three cases. If $k<0$, write $k=-c^2$ with $c>0$. Then $$ f''+c^2f=0. $$ Thus $$ f(x)=A\cos(cx)+B\sin(cx). $$ Using $f(0)=1$ gives $A=1$, while $f'(0)=0$ gives $B=0$. Therefore $$ \boxed{f(x)=\cos(cx)}. $$ If $k>0$, write $k=c^2$ with $c>0$. Then $$ f''-c^2f=0. $$ Hence $$ f(x)=A\cosh(cx)+B\sinh(cx). $$ Again, $f(0)=1$ gives $A=1$, and $f'(0)=0$ gives $B=0$. Therefore $$ \boxed{f(x)=\cosh(cx)}. $$ If $k=0$, then $$ f''=0. $$ Thus $f(x)=Ax+B$. The conditions $f(0)=1$ and $f'(0)=0$ give $$ f(x)\equiv1. $$ This is already included in both families by taking $c=0$: $$ \cos(0x)=\cosh(0x)=1. $$ Therefore, all continuous solutions are exactly $$ \boxed{f(x)=\cos(cx)\quad\text{or}\quad f(x)=\cosh(cx),\qquad c\in\mathbb R.} $$ Both families indeed satisfy the original equation because $$ \cos(u+v)+\cos(u-v)=2\cos u\cos v $$ and $$ \cosh(u+v)+\cosh(u-v)=2\cosh u\cosh v. $$ Hence the complete and correct answer is $$ \boxed{\mathrm{D}}. $$ In particular, A is not complete because $e^{cx}$ does not satisfy the equation in general, while B and C give only one of the two possible families.