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I know that the function $e^{inx}$ can be uniformly approximated on $[-\pi,\pi]$ by polynomials in $x$. I want to use this to show that polynomials are dense in $L^2([-\pi,\pi])$.

Suppose that $f\in L^2([-\pi,\pi])$. I want to show that for any $\epsilon>0$, there is a polynomial $p$ such that $|\int_{-\pi}^\pi (f(x)-p(x))^2dx|<\epsilon$. I was thinking about writing $f$ in terms of its coefficients, i.e. $$f(x)=\sum_{n=-\infty}^\infty \hat{f}(n)e^{inx}$$ But I'm still not sure how this can lead to the polynomial $p$.

Mika H.
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3 Answers3

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Let $\epsilon >0$. Then, there exists some $N$ so that

$$\left\|f(x)- \sum_{n=-N}^N \hat{f}(n)e^{inx} \right\|_2 < \frac{\epsilon}{2}$$

Now for each $-N \le n \le N$ you can find some polynomial $P_n$ so that

$$\left\| \hat{f}(n) e^{inx} - \hat{f}(n)P_n \right\|_2 < \frac{\epsilon}{2(2N+1)}$$

Now prove that $$P= \sum_{n=-N}^N \hat{f}(n) P_n$$ works.

J.G.
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N. S.
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  • And here you're using that the partial sums of the Fourier series converge to the function itself in $L^2$, right? I remember this to be false in $L^1$, but for $L^2$ it should work well. – Mika H. Nov 02 '13 at 06:01
  • (there is no need to approximate each term in the truncated Fourier series by a polynomial: the truncated sum is continuous so you can approximate the whole thing at once. (and it is not easier to approximate an exponential than a finite sum of exponentials :-) )) – Mariano Suárez-Álvarez Nov 02 '13 at 06:03
  • @MikaH. Yes, the FS converges in $L^2$. – N. S. Nov 02 '13 at 06:04
  • I wonder how Mika H. Knows he can find those polynomials, as he said that he does not know the S-W theorem! – Mariano Suárez-Álvarez Nov 02 '13 at 06:04
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    @MarianoSuárez-Alvarez I think this one is easy, isn't it? Just by the expansion $e^{inx}=1+inx+(inx)^2/2!+\ldots$. Then the partial sum converges to $e^{inx}$ since the tail is clearly small. – Mika H. Nov 02 '13 at 06:13
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    Ah. That works nicely :-) (notice that the same argument gives you an approximation to the partial Fourier series, not only to each of its terms separately) – Mariano Suárez-Álvarez Nov 02 '13 at 06:20
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As noted here, the linear combinations of characteristic functions of intervals are dense in $L^2$, so it is enough to show that you can approximate these arbitrarily well in the norm of $L^2$ by polynomials.

Can you do that?

Community
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Mariano Suárez-Álvarez
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  • It seems too advanced for me. Would it be possible to prove using the fact I stated at the beginning? – Mika H. Nov 02 '13 at 05:40
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Here is another way.

The polynomials are dense in $C[\pi,\pi]$, and we can think of $C[\pi,\pi]$ as a subset of $L^2[\pi,\pi]$.

Furthermore, $L^2[\pi,\pi]$ is the completion of $C[\pi,\pi]$ with respect to the $L^2[\pi,\pi]$ norm. Hence the polynomials are dense in $L^2[\pi,\pi]$.

copper.hat
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  • It seems too advanced for me. Would it be possible to prove using the fact I stated at the beginning? – Mika H. Nov 02 '13 at 05:41
  • It is, @N.S. has provided such a proof. Note that the first part of the proof mirrors the fact that $L^2$ is the completion of $C[-\pi,\pi]$ with respect to the $L^2$ norm, and the second part utilizes the fact the the polynomials are dense in $C[-\pi,\pi]$. – copper.hat Nov 02 '13 at 05:54