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Is there a function $g:[0,1]to mathbb R$ such that $$int_0^1 x^n g(x) , mathrm d x$$ is equal to $1$ if $n=0$ and equal to $0$ for $n=1,2,3, ldots$ ?

If there is, what would be an example of such a function? What if we require that $g$ be continuous?

I know I am expected to state what I have tried but I am honestly stuck. I wanted to integrate by parts but given that $g$ is not differentiable, this is rather useless, I think. Hints would be appreciated too.

Let $g$ be a Lebesgue-integrable function on $[0, 1]$. Assume that there exists $N geq 0$ such that

$$ forall n geq N : int_{0}^{1} x^n g(x) , dx = 0. $$

Then we prove the following claim.

Claim. $g$ vanishes almost everywhere. Consequently, $int_{0}^{1} x^n g(x) , dx = 0$ for all $n geq 0$.

Proof. For each $varphi$ in the set $C([0, 1])$ of all continuous functions on $[0, 1]$, Stone-Weierstrass theorem allows to find a sequence $p_n$ of polynomials such that $p_n(x) to varphi(x)$ uniformly on $[0, 1]$. This implies

$$ forall varphi in C([0, 1]) : int_{0}^{1} x^N varphi(x) g(x) , dx = 0. $$

Now for any $0 < a < b < 1$, we may choose $0 leq varphi_n(x) uparrow x^{-N} mathbf{1}_{[a, b]}(x)$, thus by the dominated convergence theorem, we have

$$ forall 0 < a < b < 1 : int_{a}^{b} g(x) , dx = 0. $$

This is enough to show that $g equiv 0$ a.e. on $[0, 1]$ as required.

There is no continuous function with this property: $int p(x) (xg(x)), dx =0$ for all polynomials $p$ so Weierstrass theorem tells you that $int (xg(x))^{2}, dx =0$ from which you get $g equiv 0$ . So the integral is $0$ for $n=0$ also. Actually, continuity is not required. Using the fact that $xg(x)$ can be approximated in $L^{1}$ norm by continuous functions (hence by polynomials) you can show, by a similar argument, that $xg(x)=0$ almost everywhere which forces the integral to be $0$ for $n=0$. Hence there is no such function as long as the integrals in he question exist for all $n geq 1$.

Assuming $gin L^2(0,1)$ we are allowed to write
$$ g(x) stackrel{L^2}{=} sum_{ngeq 0} c_n P_n(2x-1),qquad c_n=(2n+1)int_{0}^{1}g(x)P_n(2x-1),dx.$$
Our constraints give $c_0=1$ and
$$ c_n = (2n+1)int_{0}^{1}g(x)left[(-1)^n+x q_n(x)right],dx = (-1)^n (2n+1) $$
so, formally,
$$ g(x) stackrel{L^2}{=}sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1) $$
but the RHS of the last line is not a square-integrable function over $(0,1)$:
$$ int_{0}^{1}left[sum_{ngeq 0}(-1)^n (2n+1) P_n(2x-1)right]^2,dx = sum_{ngeq 0}frac{1}{2n+1}=+infty $$
so there are no solutions in $L^2(0,1)$. A fortiori, no continuous solutions.