Unique solution to an Algebraic equation

$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ I need to find a unique solution to this algebraic equation. Any hints as to how to proceed. The Exponential function works in this case, but I explicitly need to see the details. Thanks in advance.

• There is no unique solution, as $f(x)=b^x$ works for any $b\in(0,1)\cup(1,\infty)$. – Ben W Jan 22 '19 at 11:33
• @BenW Why do you exclude the case $b=1$? – Jose Brox Jan 22 '19 at 11:39
• Are $x_1,x_2$ supposed to be fixed real numbers, or is the equation supposed to hold for all $x_1,x_2\in\mathbb{R}$? – Jose Brox Jan 22 '19 at 11:44
• @JoseBrox just habit – Ben W Jan 22 '19 at 16:39

I'm going to suppose that $$x_1,x_2\in\mathbb{R}$$ are variables and not fixed real numbers, and that $$f$$ is differentiable.

First, if $$y=x_2$$ then $$f(x_2-x_1)=f(0)f(x_2-x_1)$$, hence $$f(0)=1$$ unless $$f(x_2-x_1)=0$$ for every $$x_1,x_2$$, i.e., unless $$f(x)=0$$ in all of $$\mathbb{R}$$.

Now suppose $$f\neq0$$ and let us differentiate the equation with respect to $$y$$:

$$0=-f'(x_2-y)f(y-x_1)+f(x_2-y)f'(x_1-y)$$,

so $$f'(x_2-y)f(y-x_1)=f(x_2-y)f'(x_1-y)$$ and substituting $$y=x_1$$ we get

$$f'(x_2-x_1)=f'(0)f(x_2-x_1)$$

and therefore

$$f'(x)=\alpha_f f(x) \,\,\,\, (1)$$

for some constant $$\alpha_f:=f'(0)\in\mathbb{R}$$.

Since the derivative of $$f$$ is proportional to $$f$$, the unique family of solutions for (1) when $$f\neq0$$ is $$f(x)=a^x$$ for $$a\in(0,\infty)$$.

Finally we check that $$f(x)=0$$ and $$f(x)=a^x$$, $$a>1$$ are solutions to the original equation, so they are the only ones.

• OP is not assuming smoothness. – Kavi Rama Murthy Jan 22 '19 at 12:05
• @KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks! – Jose Brox Jan 22 '19 at 12:14

The equation is equivalent to $$f(a+b)=f(a)f(b)$$ which is closely related to $$g(a+b)=g(a)+g(b)$$. It is well know that there are pathological examples of functions satisfying the equation which are not even measurable. However the only continuous solutions are the exponential functions mentioned in the comments. [Note that $$f(a)=0$$ for some $$a$$ implies $$f \equiv 0$$. If $$f$$ is continuous and never vanishes then it is always positive and we can take logarithms to reduce the given equation to $$g(a+b)=g(a)+g(b)$$ where $$g=\log \, f$$. In this case $$g(x)=cx$$ for some $$c$$ and $$f(x)=e^{cx}$$].