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Determine all $f : \mathbb R^+ \rightarrow \mathbb R^+$ that satisfies

$f(x + y) = f(x^2 + y^2)$ $\forall x,y \in \mathbb R^+$

My Proof :

I define $g(x) + g(y) = f(x + y)$

Then , we get $g(x^2) - g(x) = c$ ; $c \in \mathbb R^+$

$f(x) = \frac{1 + \sqrt{4c + 1}}{2}$ , which is some constant $k$.

Hence , $f(x) = k$ $\forall k \in \mathbb R^+$.

Is my proof correct?

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This doesn't work; there's no reason why $f(x+y)$ can be written as $g(x) + g(y)$, and moreover, $g(x^2) - g(x) = c$ doesn't give you that solution for $f$, because it's not a quadratic in $g(x)$; you'd need $g(x)^2 - g(x) = c$ instead.

The correct solution is as follows: fix $x+y=c$, for some $c \in \mathbb R^+$. Then as we vary $x$ over $(0, c)$, $x^2+y^2$ takes all values between $\frac{c^2}{4}$ and $c^2$ (with an open interval at $c^2$).

So $f(c) = f(a)$ for all $a \in [c^2/4, c^2).$

We can then proceed by "induction" in some sense; $f(1)$ has the same value as $f(x)$ for $x \in [1/4, 1)$, so $f(x)$ is constant on $[1/4, 1]$. We then apply the same method to obtain that $f$ is constant on $[(1/4)^2/4, 1] = [1/64, 1]$, and so on. Similarly, we can extend the range which $f$ is constant to all positive reals greater than $1$, and conclude $f$ is constant. I'll leave this as an exercise.

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A bit of geometry:

Consider the first quadrant with $x,y>0$:

1) Let $y+x=c >0;$

$y=-x+c$ is a line with $x,y$ intercepts $c$.

Triangle: $(0,0); P(0,c); Q(c,0)$ has hypothenuse $\overline{PQ} = √2c;$ height from $(0,0)$ to hypothenuse has length $c/√2$.

$y=-x+c$ is tangent to the circle $x^2+y^2=c^2/2$;

The circle passing through $P,Q$ centre $(0,0)$:

$x^2+y^2=c^2$;

For the annulus defined by $c^2/2 \le x^2+y^2 <c^2$ we have:

$f(x^2+y^2)=f(c)$;

2) For $d=c/√2$ we find

$f(x^2+y^2)=f(d)=f(c/√2)$ for

$c^2/4 \le x^2+y^2 <c^2/2$.

3) Starting with an arbitrarily large $c>0$ we proceed to annuli with arbitrarily small radii.

Within these annuli the function $f$ is constant.

4) Note: The upper circle due to the $<$ is not included in the annulus.

5) We have disjoint annuli and within each region $f$ is some contant.

6) For $f$ continuos we can conlude $f =c$ (same constant) in the different regions

7) Bonus question: How to proceed without the above assumption?

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  • $\begingroup$ I don't think you can take $y+x=0$ since $f$ is defined on the positive reals. $\endgroup$ – Yes it's me Jun 8 '20 at 11:53

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