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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.

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    $\begingroup$ I have no idea what's going on in between and after your step skipping, but there is something deeper going on here; you need some kind of Weierstrass approximation theorem type argument to make this go through. $\endgroup$ – Ian Aug 3 '17 at 17:56
  • $\begingroup$ I am not very sure after $x = t^\lambda$ that what you have is correct $\endgroup$ – Youem Aug 3 '17 at 17:57
  • $\begingroup$ @Ian I skipped the process of using Weierstrass to proximate this continuous function $x^Mf(x^{\frac{1}{\lambda}})$ for some finite $M$. $\endgroup$ – P. Rubin Aug 3 '17 at 18:04
  • $\begingroup$ That...doesn't work... $\endgroup$ – Ian Aug 3 '17 at 18:10
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    $\begingroup$ See "On Muntz' Theorem and Completely Monotone Functions", William Feller The American Mathematical Monthly Vol. 75, No. 4 (Apr., 1968), pp. 342-350 . $\endgroup$ – kimchi lover Aug 3 '17 at 18:30
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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 $p_k(t^{N\lambda})-p_k(0).$

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  • $\begingroup$ I like the new version better:) (+1) $\endgroup$ – clark Aug 7 '17 at 19:49
  • $\begingroup$ @clark Thanks, yes looking at $tf(t)$ makes life easier. $\endgroup$ – zhw. Aug 7 '17 at 19:54
  • $\begingroup$ @zhw. Given $\int_0^1 f(t) t^{n\lambda}dt=0$, for all $n\geq some N$, how did you get $\int_0^1 tf(t)t^{n\lambda}dt=0$ under the same condition? $\endgroup$ – P. Rubin Aug 8 '17 at 15:51
  • $\begingroup$ @P.Rubin Good point. Instead of $tf(t)$ it should have been $t^\lambda f(t).$ Will edit, thanks. $\endgroup$ – zhw. Aug 8 '17 at 16:06
  • $\begingroup$ I changed the hint. What don't you see? Where are you stuck? $\endgroup$ – zhw. Aug 8 '17 at 16:12

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