## Spring 2017, problem 43

Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x)f(y)=f(x-y)$ for all real numbers $x$ and $y$.

### Comments

Let $x=y$. Then $(f(x))^{2}=f(0)$.

On the other hand,

$f(x)f(0)=f(x)\iff f(x)(f(0)-1)=0.$

As $f(x)\not\equiv 0$ then $ f(0)=1$, and so $f(x) = \pm1$.

Claim: $ \not\exists x: f(x)= -1$.

Indeed, suppose $\exists x: f(x)=-1$. As $f(x)f(\frac{x}{2})=f(\frac{x}{2})$ we have $f(x)=1$, because $f(\frac{x}{2})\neq0$. A contradiction.

Hence, we conclude that $f(x)\equiv 1$.

Ex: $f(x)= (cos(x))^{2}+ (sin(x))^{2}$.

Assuming f(x) is differentiable, let's differentiate f(x)f(y) = f(x-y) with respect to x and y:

f'(x) f(y) = f'(x-y);

f(x) f'(y) = -f'(x-y)

which can be equated as: f'(x)/f(x) = -f'(y)/f(y) = A (where A is any real constant). Integrating this differential equation now produces:

ln|f(x)| = Ax + B => f(x) = exp{Ax + B}.

Taking x = y = 0, we can determine an initial condition for f(x) above: f(0)f(0) = f (0 - 0) => f(0)^2 = f(0) => f(0) = 0 or 1. Checking these into our solution for f(x) yields:

0 = e^B (not permissible), 1 = e^B => B = 0 (permissible)

and thus the required set of functions is f(x) = e^Ax. Checking this solution against the original functional equation yields:

e^Ax * e^Ay = e^A(x-y) => e^A(x+y) = e^A(x-y) => Ax + Ay = Ax - Ay => Ay = -Ay

which forces A = 0. Ultimately, f(x) = e^0x = 1 is the only nonzero function that solves f(x)f(y) = f(x-y) for all real x & y.

Call $P(x,y)$ the given property above, $a$ with $f_a\neq0$ and $\alpha=f_0$,

$P(a,o)\Rightarrow~\alpha f_a=f_a\Rightarrow~\alpha=1,$

$P(x,x)\Rightarrow~f^2_x=\alpha\neq 0,$

$P(x,\frac x2)\Rightarrow~f_x.f_{\frac x2}=f_{\frac x2}\neq 0\Rightarrow~f_x=1$ which works

$f\equiv1$.