# $f(x + y) = f(x^2 + y^2)$

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?

## 2 Answers

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.

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 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 .

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?

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