If $\int_0^1 e^{- \frac{nx}{1-x}} f(x) \; dx =0$ for $n \geq 0$, then $f=0$ on $[0,1]$ Suppose that $f$ is continuous on $[0,1]$. If
$$
\int_0^1 e^{- \frac{nx}{1-x}} f(x) \; dx =0
$$
for $n \geq 0$, show that $f(x)=0$ on $[0,1]$.
I am unsure what I should be hoping to do to show this. I know $e^{-nx/(1-x)}$ converges pointwise to 0 on $[0,1]$. But this does not help - so I think. In the case where $n=0$, we have $\int_0^1 f(x) \; dx=0$. But this shows $f(x)=0$ only where $f(x)$ is nonnegative or nonpositive. Stone-Weierstrass only seems to complicate the matter. 
What should I start thinking about to solve this problems?
 A: Note that the function
$$
g : [0,1]\to \Bbb {R}, x \mapsto \begin {cases}e^{-x/(1-x)}, &x\neq 1\\
0,&x=1\end {cases}
$$
is continuous.
Your assumption is $\int_0^1 g^n (x)f (x)dx=0$ for all $n \in \Bbb {N}_0$. By linearity, we get $\int h (x)f (x)dx =0$ for all $h\in A $, where $A $ is the linear span of all $g^n $ for $n\in\Bbb {N}_0$. Note that $A $ is an algebra which separates points on $[0,1] $ and which contains the constants. 
Furthermore,  the space of all $h $ with $\int_0^1 h (x)f (x)dx=0$ is closed under uniform convergence. By Stone-Weierstrass, we thus get $\int_0^1 h (x)f (x)dx=0$ for all continuous $h$, so that $\int_0^1 f^2dx =0$. Hence $f\equiv 0$ by continuity. 
EDIT: As noted above, $A$ separates points on $[0,1]$ and contains the constants (both not difficult to see, since $x \mapsto \frac{x}{1-x} = -\frac{1-x-1}{1-x} = -(1 - \frac{1}{1-x})$ is strictly increasing on $[0,1)$). Furthermore, since the set $\{g^n \mid n \in \Bbb{N}_0\}$ is closed under multiplication, $A$ is an algebra.
Thus, the Stone-Weierstrass theorem shows that for each continuous $h : [0,1]\to \Bbb{R}$, there is a sequence $(h_n)_n$ in $A$ with $h_n \to h$ uniformly. This implies $0 = \int h_n f\, dx \to \int h f \, dx$ and hence $\int h f \, dx = 0$ as claimed above.
A: You should first probably not include $n$ in your polynomial. You just assume that $|p(x)-f(x)|<\epsilon$. You then use the fact that $|e^{-nx/(1-x)}|\le 1$ so (regardless of $n$):
$$\left|\int_0^1e^{-nx/(x-1)}p(x)dx\right| < \epsilon$$
Then you set $q(x-1) = p(x)$ integrate by parts:
$$\int_0^1e^{-nx/(1-x)}q(x-1)dx = \left[e^{-nx/(1-x)}Q(x-1)\right]_0^1 + \int_0^1{ne^{-nx/(1-x)}Q(x-1)\over (x-1)^2}dx$$
If we know $q(0) = 0$ (which means it's $x^0$ term is zero) and $Q$ is any antiderivate of $q$ so we can choose it so that the constant term is zero we see that $Q(x-1)/(x-1)^2$ is again a polynomial. And we can continue to integrate by parts. This first step we get:
$$\int_0^1e^{-nx/(1-x)}q(x-1)dx = -Q(-1) + \int_0^1{ne^{-nx/(1-x)}Q(x-1)\over (x-1)^2}dx$$
What we then want is the last integral to converge to zero which would prove that $|Q(-1)| = |P(0)| < \epsilon$. Repeating this results in an estimate of the terms of $p(x)$.
Now one would of course need to compute the result of the repeated integration by parts so that this results in an uniform estimate of $p(x)$ only in terms of $\epsilon$. We know it's less than the absolute sums of it's coefficient, so we need to make sure that the estimate don't depend on the degree of $p$.
Now we used that $q(0)=0$, to handle the case where this isn't we have to separate out it from the polynomial. We would solve this by choosing a polynomial on the form $C+q(x)$ instead where $q(0)=0$ and we would get an extra term in the integration by parts:
$$C\int_0^1 e^{-nx/(1-x)}$$
and we would like this to converge to zero as well. The integrand converges uniformly on any compact subset of the interval $[0,1]$ and it's otherwise bounded. That is:
$$C\int_0^1 e^{-nx/(1-x)} = C\int_0^\eta e^{-nx/(1-x)}dx + C\int_\eta^1 e^{-nx/(1-x)}dx $$
where the second term aproaches zero and the absolute of the first term is no larger than $C\eta$.
A: Here's an easy way to  apply stone wierstrauss.
Based on the idea in the comments by @Thomas, let $g(x) = e^{-x/(1-x)}$. Then for any polynomial $p$, we have 
$$
0= \int_0^1 f(x) p(g(x))
$$
If $p=f\circ g^{-1}$, then we're done. But that may not be a polynomial. So let's approximate with polynomials and see what happens. 
First we need an inverse for $g$. It turns out 
$$
\frac{\log(x)}{\log(x) -1}
$$
does the trick. You can check that it's continuous, and is the inverse of $g$.
Next, note that $h = f\circ g^{-1}$ is continuous, so pick a sequence of polynomials, $p_n$ which converge uniformly to $h$ by stone-wierstrauss. 
Now we use the full power of stone-wierstrauss. Because the limit converges uniformly, we can commute limits. That is,
$$
\lim_n \int_0^1f(x) p_n(g(x))dx=  \int_0^1\lim_nf(x) p_n(g(x))dx
$$
So now all that's left to do it put it together. We know 
$$ \lim_n \int_0^1f(x) p_n(g(x))dx = 0$$ by the first paragraph. Also, 
$$
\int_0^1\lim_nf(x) p_n(g(x))dx = \int_0^1 f(x)f(x)dx = \int_0^1 f(x)^2
$$
So $\int_0^1 f(x)^2=0$. And the integral of a non-negative function equals $0$ implies that function was 0. So $f = 0$. 
