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For some function $f \in C(R)$ the following equality holds: $f(x) + f(y) = f(\sqrt{x^2 + y^2})$ for any $x, \ y \in R$.

Prove that $f(x) = ax^2 \ \forall x\in \mathbb{R}$, where $a = f(1)$.

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The crux of this proof is to show that $f$ agrees on $\mathbb{Q}$ with $ax^2$.

Inductively applying the functional equation, we have ($n \in \mathbb{N}$) $$ \underbrace{f(1) + \dots + f(1)}_{n^2 \text{ times}} = n^2 f(1) = f(\sqrt{n^2}) = f(n) $$

Hence $f(n) = an^2$ at every positive integer $n$. Observe that $f$ is symmetric about the origin since $$ f(-x) + f(x) = f(\sqrt{x^2 + x^2}) = f(x) + f(x) \implies f(-x) = f(x) $$ Hence $f(n) = an^2$ on $\mathbb{Z}$. If $p/q$ is a rational number (without loss of generality let $p,q > 0$), we have $$ p^2 f(1/q) = \sum_{r=0}^{p^2-1} f(1/q) = f(\sqrt{p^2/q^2}) = f(p/q) $$ set $p = q$, then $q^2 f(1/q) = f(1)$. Hence $f(x) = ax^2$ on $\mathbb{Q}$. Since continuous functions agreeing on a dense subset of $\mathbb{R}$ must be the same, $f(x) = ax^2$ everywhere.

A bit rough on the details but I think you can work it out yourself.

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The functional equation $$ f(x) + f(y) = f \left( \sqrt{x^2 + y^2} \right) $$ implies $$ f(0) + f(0) = f \left( \sqrt{0^2 + 0^2} \right) = f(0) $$ so $f(0) = 0$. Further $$ f(-\lvert x \vert) = f(-\lvert x \vert) + f(0) = f \left( \sqrt{(-\lvert x \rvert)^2 + 0^2} \right) = f(\lvert x \rvert) $$ which means $f$ is symmetric regarding the $y$-axis: $$ f(x) = f(-x) $$ So we can write \begin{align} f(x) + f(y) &= f(\lvert x \rvert) + f(\lvert y \rvert) \\ &= f(\sqrt{X}) + f(\sqrt{Y}) \\ &= f(\sqrt{X + Y}) \\ \end{align} so $g = f \circ \sqrt{.}$ is subject to Cauchy's functional equation $$ g(x+y) = g(x) + g(y) \quad (x,y \in \mathbb{R}, x,y \ge 0) $$ which over the reals has many interesting solutions, among them the family $$ g(x) = a x $$ with $a \in \mathbb{R}$. If a continuous solution is requested, which is the case here, it is the solution. This means $$ f(\sqrt{X}) = a X \iff \\ f(x) = a x^2 $$ for non-negative $x$. It is symmetric and of course $f(1) = a$.

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This assumes that $f$ is differentiable.

$f(x) + f(y) = f(\sqrt{x^2 + y^2}) $.

If $y=0$, then $f(x)+f(0) = f(x)$, so $f(0) = 0$.

Since $f(x)+f(-y) =f(\sqrt{x^2 + y^2}) =f(x)+f(y) $, $f(y) = f(-y)$.

Differentiating wrt $x$,

$\begin{array}\\ f'(x) &=(\sqrt{x^2+y^2})'f'(\sqrt{x^2+y^2})\\ &=\dfrac{x}{\sqrt{x^2+y^2}} f'(\sqrt{x^2+y^2})\\ \text{so}\\ \dfrac{f'(x)}{x} &=\dfrac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\\ \end{array} $

Therefore $\dfrac{f'(x)}{x} $ is constant. Setting $\dfrac{f'(x)}{x} =c$, $f'(x) = cx$ so $f(x) = cx^2/2+d$.

Since $f(0) = 0$, $f(x) = cx^2/2$.

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    $\begingroup$ How do you know that $f$ is differentiable? $\endgroup$ – copper.hat Dec 27 '16 at 21:11
  • $\begingroup$ I added the statement that this is assumed. $\endgroup$ – marty cohen Dec 27 '16 at 21:13

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