Given the vector space, $ C(-\infty,\infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{a^x\mid a \ge 1 \}$, a subspace of the given vector space?
As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).
My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-\infty,\infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).
Best Answer
As originally written, no, since $0\notin U.$ In addition, your $U$ seems to be the set of monic monomials in $x,$ not exponential functions, which would instead have the form $a^x$ for some positive constant $a.$ Even then, this would not be a subspace, for exactly the same reason.
Edit: Another simple approach would be to show that $U$ (in either version) is not closed under scalar multiplication.
Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original $U.$