Answer
The set of all negative functions: $f(x)<0$ is not a vector subspace of $C(-\infty, \infty)$.
Work Step by Step
The set of all negative functions: $f(x)<0$ is not a vector subspace of $C(-\infty, \infty)$ because it is not closed under scalar multiplication. For example, $f(x)=-x^2-1$ and $c=2z\in R$. One can see that $cf(x)=2x^2+2$ which is not negative function.