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Prove or disprove: for every real-valued continuous function $f$ on $[0,1]$ such that $f(0)=0$ and every $\epsilon $, there is a polynomial $p$ having only odd powers of $x$, i.e., $p$ is of the form

$p(x)=a_1x+a_3x^3+a_5x^5+...+a_{2n+1}x^{2n+1},$

such that sup$_{x\in[0,1]}|f(x)-p(x)|<\epsilon$.

It is hard for me to prove, please give me some hints

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  • $\begingroup$ You left out the condition on $f(0)$. $\endgroup$ – Robert Israel Jun 25 '14 at 14:54
  • $\begingroup$ sorry, i edited $\endgroup$ – Toeplitz Jun 25 '14 at 14:55
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Hint: extend $f$ to an odd function on $[-1,1]$.

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  • $\begingroup$ that means, this odd function can be approximated by odd powers polynomials on$[-1,1]$ and later we come back to $[0,1]$. Is it true?, Please elaborate your hint little bit more.. $\endgroup$ – Toeplitz Jun 25 '14 at 15:04

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