If $\int_{a}^{b}f(x)g^n(x)dx=0, \quad \forall n \in \mathbb{N}, \; $ then $\; f \equiv 0$ Let g be continuous, not negative, and strictly increasing in [a, b]. Prove that if $f$ is continuous and
$$\int_{a}^{b}f(x)g^n(x)dx=0, \quad \forall n \in \mathbb{N},$$
then $f\equiv 0$.
With a change of variable I have arrived here but I could not continue: 
$$\int_{g(a)}^{g(b)}f(u)u^n \frac{du}{u'}=0, \quad \forall n \in \mathbb{N},$$
Two particular cases already resolved by the community are:
$g(x)=x$ and $x \in [0,1]$: $\quad \int_{0}^{1}f(x)x^ndx=0, \quad \forall n \in \mathbb{N}, \; $
then $f\equiv 0$.
$g(x)=e^x$: $\quad \int_{a}^{b}f(x)e^{nx} dx=0, \quad \forall n \in \mathbb{N}, \; $
then $f\equiv 0$.
 A: Per the problem statement, $g$ is a homeomorphism $[a,b]\to [c,d]$. The continuous function $f\circ g^{-1}\colon [c,d]\to\Bbb R$ can be approximated uniformly by polynomials. So for $\epsilon>0$, we find $p\in\Bbb R[X]$ with $|f(g^{-1}(t))-p(t)| <\epsilon$ for all $t\in[c,d]$, or equivalently $|f(x)-p(g(x))|<\epsilon$ for all $x\in[a,b]$. Then
$$\begin{align}\int_a^bf(x)^2\,\mathrm dx&=\int_a^b f(x)p(g(x))\,\mathrm dx+\int_a^bf(x)(f(x)-p(g(x))\,\mathrm dx \\
&\le\rlap{\qquad0}\hphantom{\int_a^b f(x)p(g(x))\,\mathrm dx+}\llap{+\;\epsilon}\int_a^b|f(x)|\,\mathrm dx.\end{align}$$
As $\epsilon$ was arbitrary $>0$, we conclude $\int_a^bf(x)^2\,\mathrm dx\le 0$ and therefore $f\equiv 0$.
A: For simplicity let assume $[a,b] = [0,1]$ and first suppose $g(x)=x$ taking into account the particular case, we have $\int_{0}^{1} f(x) p(x) = 0 $ for each $P(x) = polynomial$.
now since $f(x)$ is continuous on $[0,1]$ then there exists a sequence of polynomials, say $P_n$ such that $P_n \rightarrow f$ uniformly. This implies that
$$  0= \int_{0}^{1} f(x) p_n (x)  \rightarrow \int_{0}^{1} |f(x)|^2 $$ Hence, $\int_{0}^{1} |f(x)|^2 =0$, and because $f$ is continuous we arrive $f=0.$
Now let W.L.O.G assume $g:[0,1] \rightarrow [0,1]$ is onto (otherwise you can scale this g) this implies $0=g(0),~ 1=g(1)$ now by substitution $y=g(x)$ the condition of question becomes $$  \int_{0}^{1} F(y) y^n dy =0 $$ where $F(y) = f(g^{-1} (y)) (g^{-1} (y))'   $ hence according to my pervious argument $F=0$ which implies $f=0$
Some Clarifications :
Note that since $g$ is strictly increasing so $g^{-1}$ is. And it is a well-known fact that every increasing function has almost countable non-differentiable points, thus $g^{-1}$  is differentiable a.e.  Therefor it does not impact on the integrability and the value of integral. Thus this makes $F=0$ a.e and then $f=0$ a.e and since $f$ is continuous $f=0$ every where!.   
P.S for those who have hard time of scaling $g$, if $[g(0) , g(1)] \neq [0,1]$ you can scale this $g$ in this way $\frac{g(x)-g(0)}{g(1)-g(0)}.$   
A: Hagen von Eitzen above has indeed given an elegant solution valid when the given g is strictly increasing but no assumption that g be non-negative is needed or used. However it is not necessary assume the integral (1) I[a,b]f(x)(g(x))^n dx =0 for all non negative integers n= 0.1,2,3...   . It suffices to assume this for integers 
n >= M where M is any positive integer and by increasing M we  assume M is odd .We use the Stone Weierstrass theorem instead of the Weierstrass Theorem .Note that (1) implies that I[a,b]f(x) (g(x)^M (p(g(x))dx =0  for all polynomials If g(x) is never 0 in [a,b] then g(x)^M is strictly increasing hence the algebra of functions of the form (2)q(x)= (g(x)*M (p(g(x)) separates the points of [a,b] and don't all vanish at any one point and so  dense in the continuous functions for the mac norm .Using such a sequence converging to f(x) uniformly we see that I(f(x)^2 ) =0 so f vanishes identically as before
 What happens if g(c) = 0 for some c in [a,b] (exactly one ,g is strictly increasing . Then  the functions of the form C + q (q as in (2),C a real constant ) are a dense set . But if h(c)=0 H continuous then a sequence of functions c+q oonverges uniformly to h and all q(c)=0  hence the sequence of constants c tends to 0 so the the sequence of q 's =c+q -c tends to h uniformly . Take h = f T  where T(x) is between 0 and 1 ,T(c) =0 .and T=1 off and interval of length e containing c , T continuous  then see that I (f^2T) =0 Let e tend to 0 to get I(f*2)=0 and hence f=0 .  Stuart M.N.  
A: Okay, re-attempt. Assume $f$ is not identically zero. Then there exists at least one point $x_0 \in [a,b]$ such that $f(x_0) \neq 0$ and say w.l.o.g. that $f(x_0) > 0$ (otherwise just take the negation). Since $f$ is continuous, there is a small neighborhood around $x_0$ where $f>0$ in that interval. Since $g$ is non-negative and increasing, this implies that
$$
\int f(x)\,g^n(x)\,dx > 0
$$
Which is a contradiction to the original statement. Thus $f \equiv 0$.
