$A=\left\{\sum_{k=1}^n a_ke^{kx}:a_k\in\mathbb R, n\in\mathbb N\right\}$ is dense in $C[a,b]$

I need to prove that $$A=\displaystyle\left\{\sum_{k=1}^n a_ke^{kx}:a_k\in\mathbb R, n\in\mathbb N\right\}$$ is dense in $C[a,b]$. I had in mind to use the Stone–Weierstrass' theorem, I know that $A$ is an algebra that separates points, but we also need $A$ to contain non-null constants which is extremely complicated for me or at least I can't see the way. I have also tried to prove the density directly, but fall to prove that every polynomial can be uniformly approximated by elements of $A$. Any ideas?

Solution 1:

As you probably noticed, we have that $$ B = \left\{ \sum_{k=0}^n a_k e^{kx} : a_k \in \mathbb{R}, n \in \mathbb{N}_0 \right\} $$ is dense in $C[a, b]$, thanks to the Stone-Weierstrass theorem. This easily implies the density of $A$, as follows.

Let $f(x) \in C[a, b]$, and let $p_i(x) \in B$ be elements such that $p_i(x) \to f(x)/e^x$ uniformly, which exists as $B$ is dense. Then $e^xp_i(x) \to f(x)$ uniformly, as $e^x$ is bounded in $C[a, b]$. Now, if $p_i(x)=\sum_{k=0}^n a_k e^{kx}$ is in $B$, then $e^xp_i(x)=\sum_{k=1}^{n+1}a_{k-1}e^{kx}$ is in $A$. This proves that $A$ is dense, as required.