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Let $$f(x)= \begin{cases} \sin(1/x) & x \ne 0 \\ 0 & x = 0 \end{cases} $$ Prove that for any $\epsilon>0 $ there exists a polynomial $p(x)$ such that $$\int_0^1 |f(x) - p(x)| \, dx \lt \epsilon.$$

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    $\begingroup$ What is epsilon, what is p(x), what have you done so far, what's your effort and/or insights...?? $\endgroup$
    – DonAntonio
    Dec 8, 2013 at 6:51
  • $\begingroup$ Oh, and write mathematics in this site with using LaTeX, otherwise it's easy to misunderstand what you write. $\endgroup$
    – DonAntonio
    Dec 8, 2013 at 6:51
  • $\begingroup$ trigometric polynomial? $\endgroup$
    – Haha
    Dec 8, 2013 at 7:04
  • $\begingroup$ Are you familiar with Weierstrass approximation theorem? $\endgroup$ Dec 8, 2013 at 7:16
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    $\begingroup$ @lola: Your function is not continuous at $0$. $\endgroup$ Dec 8, 2013 at 7:45

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Hint: Everything happens on $[0,1]$ and the function $f$ is continuous except at $0$ hence one can consider the functions $f_n:[0,1]\to\mathbb R$ defined by $f_n(x)=f(x)$ if $x\geqslant1/n$ and $f_n(x)=f(1/n)$ if $x\leqslant1/n$. These are continuous hence Weierstrass gives some polynomials $p_n$ such that $\|f_n-p_n\|_\infty\leqslant1/n$.

Can you estimate $\int\limits_0^1|f-p_n|$? You might want to decompose the integral into $\int\limits_0^{1/n}+\int\limits_{1/n}^1$ and to bound each part separately...

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  • $\begingroup$ This is a very nice answer, as usual. I especially like the fact that it seems not to use any unnecessary tools, so it should be useful to the broadest possible audience. $\endgroup$ Dec 9, 2013 at 0:30

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