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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2What is epsilon, what is p(x), what have you done so far, what's your effort and/or insights...?? – DonAntonio Dec 08 '13 at 06:51
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Oh, and write mathematics in this site with using LaTeX, otherwise it's easy to misunderstand what you write. – DonAntonio Dec 08 '13 at 06:51
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trigometric polynomial? – Haha Dec 08 '13 at 07:04
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Are you familiar with Weierstrass approximation theorem? – Mhenni Benghorbal Dec 08 '13 at 07:16
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the Weierstrass Approx. Thrm. says "Let I be a closed bounded interval and f:I-->R is continuous then for each positive epsilon there is a polynomial p:R-->R such that |f(x)−p(x)|<ϵ for all points x in I" – lola Dec 08 '13 at 07:19
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@lola: Do you know your interval? – Mhenni Benghorbal Dec 08 '13 at 07:26
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Is your function continuous? – Mhenni Benghorbal Dec 08 '13 at 07:29
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the function is continuous – lola Dec 08 '13 at 07:31
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@lola: Note that, in your problem, they use a different norm while in Weierstrass theorem they use the infinite norm $||\cdot||_{\infty}$. – Mhenni Benghorbal Dec 08 '13 at 07:39
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2@lola: Your function is not continuous at $0$. – Mhenni Benghorbal Dec 08 '13 at 07:45
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@lola: Your function is integrable and it is a well known fact that every integrable function can be approximated by a continuous function. – Mhenni Benghorbal Dec 08 '13 at 07:47
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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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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. – Pete L. Clark Dec 09 '13 at 00:30