Is the span of $e^{-nx}$ dense in $C([0,1])$? Prove or disprove: finite linear combinations of the form $\{e^{-nx}\}_{n \in \mathbb{N}}$ are dense in $C([0,1])$. So by finite linear combinations, does it mean say for some vector in $\mathbb{R}^d$, say $\overline{a}=(a_1,a_2,...,a_d)$, then is $\sum_{k=0}^d a_ke^{-kx}$ dense in $C([0,1])$? And by dense does it mean the closure is the entire space?
 A: Proof: Take any function $f\in C([0,1])$.
Create a new function $g(t)=f(-\ln(t))$ for $t\in[e^{-1}, 1]$. This means that $f(x)=g(e^{-x})$ for $x\in[0,1]$.
That new function $g$ can now be uniformly approximated by polynomials, as per Weierstrass' Approximation theorem. i.e. for every $\epsilon>0$ there is a polynomial $p(t)=a_0+a_1t+\ldots+a_nt^n$ such that $|g(t)-p(t)|<\epsilon$ for all $t\in[e^{-1}, 1]$
Now, going back to $t=e^{-x}, x\in[0,1]$: we have $|g(e^{-x})-p(e^{-x})|<\epsilon$ too, for all $x\in[0,1]$. However, notice that $g(e^{-x})=f(x)$ and $p(e^{-x})=a_0+a_1e^{-x}+a_2e^{-2x}+\ldots+a_ne^{-nx}$. Thus, we have found a finite linear combination of $1, e^{-x}, e^{-2x},\ldots$ such that $|f(x)-(a_0+a_1e^{-x}+a_2e^{-2x}+\ldots+a_ne^{-nx})|<\epsilon$.
As this can be done for every $\epsilon>0$, this means that the set of finite linear combinations of $1, e^{-x}, e^{-2x},\ldots$  is dense in $C([0,1])$. $\quad\blacksquare$

Note: You can also drop $1$ (and even drop finitely many other functions of the form $e^{-nx}$) from the set, and the statement will still be valid.
Proof: This is because the interval $[e^{-1}, 1]$ does not contain zero, and it is easy to prove an extension of Weierstrass' theorem that, on an interval $[a,b]$ not containing zero, any continuous function can be uniformly approximated by a finite linear combination of $x^d, x^{d+1}, x^{d+2},\ldots$, where $d\ge 1$ is an integer.
Namely, for $\epsilon>0$, take a function $f\in C([a,b])$,  make a continuous function $g(x)=\frac{f(x)}{x^d}$, take $M=\max_{t\in[a,b]}\frac{1}{|x|^d}$, approximate $g(x)$ on $[a,b]$ by a polynomial $p=a_0+a_1x+\ldots+a_nx^n$ to precision $\frac{\epsilon}{M}$, i.e. $\left|\frac{f(x)}{x^d}-(a_0+a_1x+\ldots a_nx^n)\right|<\frac{\epsilon}{M}$, and multiply this inequality by $|x|^d$ to get $|f(x)-(a_0x^d+a_1x^{d+1}+\ldots+a_nx^{d+n})|<\epsilon$. $\quad\blacksquare$
