# Power-like functional equation

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?

• How is $z$ related to $x$ and $y$ in your equation? – Kavi Rama Murthy Apr 25 at 23:27
• @KaviRamaMurthy: there is no relation. They can be any real number. – Arnaldo Apr 25 at 23:28
• Continuous functions satisfying $f(xy)=f(x)f(y)$ are power functions. – Kavi Rama Murthy Apr 25 at 23:29
• @mihaild OP is talking about functions of the type $x^{b}$ and you are talking about $b^{x}$. – Kavi Rama Murthy Apr 25 at 23:40
• @Arnaldo Yes, the only continuous solutions are power functions. – Kavi Rama Murthy Apr 25 at 23:40

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$$.

• It is right, by it is a particular case of Cauchy functional $f(xy)=f(x)f(y)$. Isn't it? – Arnaldo Apr 26 at 1:19

$$f(x+y)=f(x)+f(y)$$ 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 at 12:36
• @Arnaldo: which first ? – Yves Daoust Apr 27 at 7:02