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$$f(x_{2}-x_{1})= f(x_{2}-y)f(y-x_{1})$$ 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.

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

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  • $\begingroup$ OP is not assuming smoothness. $\endgroup$ – Kavi Rama Murthy Jan 22 '19 at 12:05
  • $\begingroup$ @KaviRamaMurthy Oh, yes, I forgot to add that explicitly. I'll edit, thanks! $\endgroup$ – Jose Brox Jan 22 '19 at 12:14
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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 $f(x)=e^{cx}$].

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