Let $f:[0,1]\rightarrow \mathbb{R}$ be continuous. Assume $\int_{0}^{1}f(t)e^{-nt}dt=0$ whenever $n$ is a non-negative integer. Does it follow $f$ is identically $0$? Does the answer change if $-nt$ is replaced by $+nt$? $2nt?$
My work: Consider the sub-algebra of $C[0,1]$ generated by $e^{-nt}$. Let $P_n(x)=a_nx^n+a_{n-1}x^{n-1}+⋯+a_1x+a_0$. Consider $A=\{P_n(e^{-t})∣n=0,1,2,3…\}$. It's not difficult to see that $A$ is a sub-algebra of $C[0,1]$ since $A$ is closed under multiplication & addition.
Moreover, $A$ vanishes at no point in $[0,1]$ since $e^{-t} \neq 0 $ for all $t$ and the same choice of element in $A$ proves that $A$ separates point in $[0,1].$
Now, the uniform closure of $A$ consists of all continuous functions on $[0,1]$. Since $f\in{C[0,1]}$, there exists $P_n(e^{-t})$ s.t $P_n(e^{-t})\rightarrow f(t)$ uniformly on $[0,1]$. So, $P_n(e^{-t})f(t)\rightarrow f(t)^2$ uniformly. Therefore, $$\int_{0}^{1}P_n(e^{-t})f(t)\,\mathrm dt \rightarrow \int_{0}^{1}f(t)^2\,\mathrm dt\text{ as }n\rightarrow \infty$$ But, $\int_{0}^{1}P_n(e^{-t})f(t)\,\mathrm dt=0$ for all $n$. Hence, $\int_{0}^{1}f(t)^2\,\mathrm dt=0$. By continuity, we obtain $f≡0$.
The result holds true even if we replace $-nt$ by $+nt$. But, I wonder if this is true for all even exponents & similarly for odd exponents?
Is my approach to this problem correct?