real valued functions with composition If $G$ is the set of all functions $f : \mathbb{R} \to \mathbb{R}$ such that $f(x) \ge 0$ for all $x \in \mathbb{R}$ with $f ∗ g = f \circ g$ (here $\circ$ denotes the operation of composition), for $f, g \in G$. my question: Does $G$ form a group?
 A: We need to show that $f \circ g$ is a function from $\mathbb{R}\to\mathbb{R}$ and that all its values are non-negative.
Certainly, $f \circ g=f(g(x))$ is well defined, since $g(x)$ is well defined and real valued for all real numbers. Also since $f(x) \ge 0$ for all $x \in \mathbb{R}$ we have $f(g(x)) \ge 0$ for all $x\in \mathbb{R}$ since $g(x) \in \mathbb{R}$. Hence $g$ is closed under composition.
The problem is whether you can find inverses or an identity. Let $f(x) \equiv 1$. $f$ is not a $1-1$ function, and it doesn't have an inverse with respect to composition, but the problem is deeper than that.
As @AmiteshDatta pointed out, there is no identity element. Assume there is a function, $e \in G$, for which $e \circ f = f$ and $f \circ e = f$. Since $g(x)=x^2 \in G$ we have $g\circ e = g$, and $(e(x))^2 = x^2$. Thus $e(x) = \pm |x|$ however since $e(x) \ge 0$, we conclude that $e(x)=|x|$.
Returning to the identity, $f \circ e = f$, this tells us that $f(|x|) = f(x)$ for all $x$ for all $f \in G$. This tells us that all functions in $G$ are even. This is a contradiction since $e^{x} \in G$.
