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Let $g: \mathbb R \to \mathbb R$ be continuous and $g(0)=0$, $\forall x \gt 0$, $g(x)g(-x) \gt 0$. Find all solutions $f: \mathbb R \to \mathbb R$ to the functional equation

$$g(f(x+y))=g(f(x)) +g(f(y)) \quad x,y \in \mathbb R$$

I concluded that $f(x)=0$ $\forall x$ is the only solution. My argument goes as the following:

$g(x)=0 \implies x=0$.
Substituting $y=0$ we get that $g(f(0))=0$. Next, substitute $y=-x$ to get $g(f(0))=g(f(x))+g(f(-x))$, and so $g(f(x))=-g(f(-x))$ $\forall x \in \mathbb R$.

Then, suppose $\exists x \in \mathbb R$ such that $f(x) \neq 0$. Take $g(x)=x^2$ (it is continuous and satisfies the conditions) we get that $f(x)^2=-f(-x)^2$ which is a contradiction, since the left hand side is strictly positive, and the right hand side is non-positive.

Is this correct? Thank you very much!

Sean Thrasher
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    so $g o f$ must be linear in its argument. Thus $g o f(x) = ax$. For any $a \neq 0$ this will contradict $g(x)g(-x)>0$ thus $g o f(x) = 0$, so $f(x) = 0$ – fGDu94 Nov 23 '19 at 19:17
  • This is Cauchy's functional equation with $f$ substituted for $g\circ f$. – Jam Nov 23 '19 at 20:27
  • Sean, I don't think your contradiction is valid. Just because $f$ satisfies the equation when paired with a particular $g$, it doesn't mean that it would do so for all $g$. So even if $f$ can't work for $g=x^2$, it doesn't exclude there being another $g$ that works for it. – Jam Nov 23 '19 at 20:39
  • @GeorgeDewhirst $g\circ f$ may not necessarily be linear, there are nonlinear solutions to Cauchy's functional equation. But I still don't follow your logic. How would $g\circ f=ax$ contradict $g(x)g(-x)>0$? – Jam Nov 23 '19 at 21:48
  • it looks like it would be linear as $gof(x+y) = gf(x)+gf(y)$. Just letting $h = gof$ for a moment and studying it on its own, the function must be a linear one.

    Supposing $g o f = ax, a \neq 0$, then $0 = g o f(x+-x) = g o f(x) - g o f(-x)$. Hence $g(ax)= -g(-ax)$ but they are both positive (or both negative) thus must both be zero.

     – fGDu94 Nov 23 '19 at 21:51
  • @GeorgeDewhirst Thank you for the explanation. This is a nice argument, though I think you mean $g\circ f (x) = ax=-ax$ for the last equation since we don't necessarily have $f$ linear. The case where $g\circ f$ is nonlinear is covered in (Question 131197) but we must have $g\circ f$ dense in $\mathbb{R}^2$ and it's dependent on whether the axiom of choice is used. – Jam Nov 23 '19 at 22:01
  • yeah i just had a look at these more pathological examples. If g o f is nonlinear can you get a standard form for it? It may still be quite doable – fGDu94 Nov 23 '19 at 22:02

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