# Use Stone-Weierstrass theorem to show that $f$ will be $0$ for all $x\in [0,1]$.

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)?

• I think this is not quite right: there are some equalities that should involve epsilons somehow. Specifically you should have something like $\left | \int_0^1 f(t) p_\epsilon(t) dt - \int_0^1 f(t)^2 dt \right |<\epsilon$, but then the first term is given to be zero and the result follows by taking $\epsilon$ to be arbitrary. – Ian Aug 7 '17 at 17:47
• That's not Weierstrass, it's Stone-Weierstrass. – zhw. Aug 7 '17 at 17:54
• @zhw Ok, Thank you. Will be corrected. But does the proof look ok? Is there any problem with the statements up there? Thank you! – 2ndaccount Aug 7 '17 at 17:55
• Your fourth paragraph: "Not true that" should be changed to "we do not know that" – zhw. Aug 7 '17 at 17:58
• @zhw. ok, thank you so much for correcting! Does everything else look ok? needs to be corrected? – 2ndaccount Aug 7 '17 at 18:01