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I would like to know what are the especifications of a functional equation that give us a power function as a solution.

For example, if $f:\Bbb R \to \Bbb R$ is continuous and monotonic, such that $$f(x)+f(y)=f(z)$$ iif $$f(\lambda x)+f(\lambda y)=f(\lambda z)$$ for all $\lambda>0 $, then $f(x)=ax^b$.

Does anyone know another functional equation that gives a power function as a solution?

Arnaldo
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For all $\ x\ $ and $\ y\ne 0,\ $ the only continuous solutions of the equation $\ f(x)^2 = f(xy)f(x/y)\ $ is $\ f(x) = ax^b.$

An alternative equation similar to Cauchy's functional equation is $\ f(1)f(x\ y) = f(x)f(y)\ $ for all $x,y$.

Somos
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$$f(x+y)=f(x)+f(y)$$ gives a linear function and $$f(x+y)+f(0)=f(x)+f(y)$$ an affine one.

Then using logarithmic transformations,

$$f(xy)f(1)=f(x)f(y)$$ is your answer.

By the same reasoning, the exponential $ab^x$ is the solution of

$$f(x+y)f(0)=f(x)f(y),$$ and the logarithm $a+\log_bx$ that of

$$f(xy)+f(0)=f(x)+f(y).$$

  • I don't get how you go from $f(x+y)+f(0)=f(x)+f(y)$ into $f(xy)f(1)=f(x)f(y)$. The first functional equation shouldn't be $f(x+y)=f(x)f(y)$? – Arnaldo Apr 26 '19 at 12:36
  • @Arnaldo: which first ? –  Apr 27 '19 at 07:02