Fix a real number $\lambda>0$. Let $f\in C[0,1]$ satisfy $\int_0^1 t^{n\lambda}f(t)dt=0$ for all but finitely many $n\in\mathbb{N}$. What can be deduced about the function $f$?
Claim: $f$ will be $0$ for all $x\in [0,1]$.
*Let $A$ be the set of polynomials in $t^\lambda$. Note that $A$ is a subalgebra of $C[0,1]$ that separates points and vanishes at nowhere, so $A$ is dense in $C[0,1]$ by Stone-Weierstrass. Thus, we have a set of polynomials in $A$ such that converges uniformly to $f(t)$.
Since $\int_0^1 t^{n\lambda}f(t)dt=0$ for all but finitely many $n\in\mathbb{N}$, we do not know that if $\int_0^1p_n(t^\lambda)f(t)dt=0$ for all polynomial $p_n(t^\lambda)$.
Since by Stone-Weierstrass theorem there exists a set of polynomials in set $A$ that converges uniformly to $f$, $\forall\epsilon\geq 0$, $\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $|f-p_n|\leq\epsilon$.
Note $\exists N'\in\mathbb{N}$, such that $\int_0^1t^{n\lambda}f(t)dt=0$ for all $n\geq N'$. Let $M=max\{N,N'\}$, we have $\int_0^1p_{n\geq M}(t^\lambda)f(t)dt=0=\int_0^1f(t)^2dt=0$.
Since $f$ is continuous, so is $f^2$. Since $\int_0^1f(t)^2dt=0$ with $f^2\geq 0$, $f^2=0$ and $f=0$.
Is the above proof correct especially the highlighted ones (emboldened)?