Find all solutions $f:\mathbb R \to \mathbb R$ to the functional equation $g(f(x+y)) = g(f(x)) + g(f(y))$ Let $g:\mathbb R \to \mathbb R$ be continuous and such that $g(0) = 0$ and $g(x)g(−x)&gt;0$ for any $x &gt; 0$. 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 have deduced that (if $f$ is continuous) we must have $g(f(x)) = \lambda x$, because $g \circ f$ is a continuous linear function and all such functions must be of that form. I am then tempted to say that $f(x) = g^{-1}(\lambda x)$ but I cannot do this because I do not know that $g$ is bijective. Unfortunately I do not understand the significance of the conditions $g(0) =0$ and $g(x)g(-x)&gt;0$ for $x &gt; 0$. Any help would be appreciated.
First, notice that $g$ does not vanish except at $x=0$. So, by continuity, $g$ keeps a constant sign on $(0, +\infty)$, and a constant sign on $(-\infty, 0)$. Adding the condition $g(x)g(-x) &gt; 0$, you get that $g$ keeps a constant sign on $\mathbb{R}$.
Now, let $x=y=0$ in the equation. You get $g(f(0))=2g(f(0))$, therefore $g(f(0))=
0$. For all $x \in \mathbb{R}$ then, you have
$$0 = g(f(x-x))=g(f(x))+g(f(-x))$$
But $g$ keeps the same sign everywhere as stated at the beginning, so $g(f(x))=g(f(-x))=0$. Because the only point where $g$ vanishes is $0$, you deduce that
$$f(x)=0$$
So the only solution is the function $f$ constant equal to $0$.
Assume that $f$ is continuous and $y_1 , y_2 \in f(\mathbb{R})$ and $$g(y_1 ) =g(y_2 )$$ then $$y_1 =f(x_1) , y_2 =f(x_2)$$ and  hence $$\lambda x_1 =g(f(x_1)) =g(y_1 ) =g(y_2) =g(f(x_2)) =\lambda x_2 $$ thus $$x_1 =x_2 $$ and consequently $$y_1 =y_2.$$
Therefore $g|_{f(\mathbb{R} )}$ is an bijection and you can use $g^{-1}.$