Can D'Alembert's functional be derived from Cauchy functionas?

Question

Is it possible to derive D'Alembert's functional equation from Cauchy's functional equations? If so, can somebody kindly point me to a reference?

Edit: Can $f(x + y) + f(x − y) = 2f(x)f(y)$

be derived from either of

$f(x+y)=f(x)+f(y)$,

$f(xy)=f(x)f(y)$,

$f(x+y)=f(x)f(y)$,

$f(xy)=f(x)+f(y)$.

PS: I am not a mathematician.

Answer 1

Not in a meaningful way.

You cannot derive d'Alembert's functional equation from the single Cauchy functional equation $f(x+y)=f(x)+f(y)$, because this latter equation has the solution $f(x)=x$, and if you make algebraic manipulations to the equation $f(x+y)=f(x)+f(y)$ you can only end up with equations which have the solution $f(x)=x$, and possibly more solutions which the original equation did not have, but $f(x)=x$ is not a solution of d'Alembert's equation (which is easy to check), so no matter which manipulations you make, you can not end up with an equation which only has the solutions which d'Alembert's equation has. Likewise none of the other Cauchy functional equations can individually imply d'Alembert's functional equation, since they have solutions which the latter does not.

It can be done in a silly way however. The only continuous solutions to $f(x+y)=f(x)+f(y)$ are those of the form $ax$ for different $a$, and the only continuous solutions to $f(x+y)=f(x)f(y)$ are those of the form $b^x$ for different $b$. (See http://www.jstor.org/stable/2972575 .) If we assume both of these equations to be true, then the only common continuous solution is $f(x)=0$, and from the functional equation $f(x)=0$ it of course follows that $f(x+y)+f(x-y)=2f(x)f(y)$ since $0+0=2\cdot0\cdot0$.

Answer 2

It is quite an interesting observation that if $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfies $f(x+y)=f(x)f(y)$ and $f$ is even (about zero), then $f(x)$ also satisfies D'Alemberts functional equation. Since $f(x-y) = f(x)f(-y) = f(x)f(y)$ also.

So there are relationships between the equations!