Let $\lambda\in\mathbb{R}$ be fixed and positive. Let $f$ be continuous on $[0,1]$ satisfy $\int_0^1 t^{n\lambda}f(t)dt=0$ for all but finitely many $n\in\mathbb{N}$. Prove $f=0$, for all $x\in[0,1]$.
*Since the domain is nonnegative, no concern about $f(t)t^{n\lambda}$ being odd.
Rewrite the integral. Let $x=t^\lambda$, and when $t=0$, $x=0$, and when $t=1$, $x=1$.
Then we have $\int_0^1 t^{n\lambda}f(t)dt=\int_0^1 x^nf(x^{\frac{1}{\lambda}})dx=0$ for all but finitely many $n\in\mathbb{N}$.
(I'm skipping steps) Since $f(x^{\frac{1}{\lambda}})=0$ for all but possibly not at $0$, and $f(x^{\frac{1}{\lambda}})$ is continuous, $f=0$ for all $x$.
Thus, $f(t)=0$ for $t\in[0,1]$.
Is the idea correct? Thank you.
Hints: Let $g(t)=t^\lambda f(t).$ It's enough to show $g\equiv 0.$ From our hypothesis there exists $N\in \mathbb N$ such that
$$\int_0^1 g(t)t^{n\lambda}\,dt = 0,\,\, n=N,N+1,\cdots.$$
It follows that the above integral is $0$ for $n\in \{N,2N,3N,\dots\}.$ By Stone-Weierstrass, polynomials in $t^{N\lambda}$ are dense in $C[0,1].$ So there are polynomials $p_k$ such that $p_k(t^{N\lambda})\to g(t)$ uniformly on $[0,1].$ Consider $p_k(t^{N\lambda})-p_k(0).$