How do we know this is the unique solution to this functional equation?

Question

I have the functional equation $$F(x)+F(y)=F(\sqrt {x^2+y^2})$$

I know that one solution to this equation is $F(x)=a\cdot x^2$. However, I am told that this is the unique solution, but I don't know how to show this?

How do we know that this is the unique solution?

Angina Seng · Accepted Answer · 2017-07-28T17:25:23.200

4

Replacing $x$ by $-x$ we see that $F(x)=F(-x)$: $F$ is an even function. Define $G(u)=F(\sqrt{u})$ for $u\ge0$. Then $$G(u)+G(v)=F(\sqrt u)+F(\sqrt v)=F(\sqrt{u+v})=G(u+v).$$ This functional equation has the only continuous solutions $G(u)=au$. But it also has non-continuous solutions. As $\Bbb R$ is an infinite-dimensional $\newcommand{\Q}{\Bbb Q}\Q$-vector space, there is a nonzero $\Q$-linear map $\phi:\Bbb R\to\Q$. Define $F(x)=\phi(x^2)$. Then $F$ is non-continuous and satisfies the functional equation.

The trouble is, that functions like $\phi$ can only be "constructed" by some form of the Axiom of Choice.

edited Jul 28 '17 at 17:25

answered Jul 24 '17 at 06:27

Angina Seng

158,341

I wondered why I was struggling so hard to prove that $F$ sends $x$ to $x^2$ for anything other than a set of measure $0$. – Patrick Stevens Jul 24 '17 at 06:28
How do we know that it only has thw continuous solution $G(u)=au$? – user56834 Jul 28 '17 at 17:23
From the additivity, one easily gets $G(u)=uG(1)$ for rational $u$, so if $G$ is continuous this holds for real $u$ too. @Programmer2134 – Angina Seng Jul 28 '17 at 17:26

How do we know this is the unique solution to this functional equation?

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