Let $f:[0,1]\rightarrow\mathbb{R}$ be continuous such that $\int_0^1 x^{2k}f(x)dx=0$, $\forall k\geq 1$, integers.

Show that $f(x)=0$, $\forall x\in[0,1]$.

*Let $y=x^2$, then this function can be written as $\int_0^1 yy^nf(\sqrt y)dx$, where $n\geq 0$.

Since $yf(\sqrt y)$ is continuous on a closed interval, by Weierstrass Approximation theorem, we can find a polynomial $p(x)$ converging to it uniformly.

Thus, we have $\int_0^1yf(\sqrt y)p(x)dx=0$, for any polynomial.Then we can get $f(\sqrt y)=0$ for all $y\in [0,1]$.

Therefore, $f(x)=0$, for all $y\in [0,1]$, i.e. for all $x\in[0,1]$.

Is this correct?

  • $\begingroup$ Seeing your recent questions, I would like to ask if you have heard of the Muntz-Szasz theorem, which gives very strong conditions of powers of a variable being dense in the space of continuous functions over a closed interval, endowed with the integral inner product. $\endgroup$ – астон вілла олоф мэллбэрг Aug 3 '17 at 0:49
  • $\begingroup$ No I have not... but I have an exam coming up, so I'm justing doing problems with only things that I have learned. $\endgroup$ – 2ndaccount Aug 3 '17 at 0:51
  • $\begingroup$ Do read it up once your exam is over, and best of luck for that. I think that a few steps are skipped, for example I want to see how you have got the first line, the expression seems incorrect. The idea after that is fine. Speaking of which, I have a proof of this that cleverly uses Bernstein polynomials. $\endgroup$ – астон вілла олоф мэллбэрг Aug 3 '17 at 0:56
  • $\begingroup$ By letting $y=x^2$, $x^{2k}f(x)dx$ is now $y^kf(\sqrt y)dy$. Since $x\in[0,1]$, and $y=x^2$, $y\in[0,1]$ also. $\endgroup$ – 2ndaccount Aug 3 '17 at 1:02
  • $\begingroup$ Since $k$ is an integer starting form $1$, in order to make this similar to $f(x)x^n$ problem with $n\geq 0$, I'm rewriting $y^kf(\sqrt y)dy$ as $yy^nf(\sqrt y)dy$, with $n\geq 0$. $\endgroup$ – 2ndaccount Aug 3 '17 at 1:05

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