Stone-Weierstrass theorem applied to Fourier series - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:22:23Z http://mathoverflow.net/feeds/question/70339 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70339/stone-weierstrass-theorem-applied-to-fourier-series Stone-Weierstrass theorem applied to Fourier series nareto 2011-07-14T15:51:15Z 2011-08-13T09:06:17Z <p>This is a question on Fourier series convergence. The problem is, in the applications of the Stone Weierstrass approximation theorem on <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Applications_2" rel="nofollow">wikipedia</a>, there's stated that as a consequence of the theorem the space of trigonometrical polynomials is dense (with the sup norm) in the space of continous functions in [0,1] - i.e. for every continous function its fourier series converges. This boggles me: isn't continuity not enough for the convergence (let alone uniform) of a Fourier series? What about du-Bois Reymond [and many others] example of continous function with non convergent Fourier series in a point?</p> http://mathoverflow.net/questions/70339/stone-weierstrass-theorem-applied-to-fourier-series/72814#72814 Answer by JR for Stone-Weierstrass theorem applied to Fourier series JR 2011-08-13T06:01:52Z 2011-08-13T06:01:52Z <p>The density of trigonometric polynomials in $C[0,1]$ with respect to the sup norm does not imply that the Fourier Series of some $f\in{}C[0,1]$ must converge pointwise. </p> <p>Let $e_k:=e^{2\pi{}ikx}$ for $k\in{}\mathbb{Z}$. Then it can be shown that </p> <ol> <li>$\{e_k|k\in\mathbb{Z}\}$ is an orthonormal basis for $L^2[0,1]$ <em>with respect to the $L^2$ norm</em>. That is, $=\delta_{jk}$ and the span of the $e_k$ is dense in $L^2[0,1]$.</li> <li>If $f\in{}L^2[0,1]$ and $V_n:=span\{e_k|k=-n,...,n\}$, then the nth partial sum of the Fourier Series of $f$, $P_{n}f:=\sum_{k=-n}^{n}$$e_k is the L^2 projection of f onto V_n, i.e. for any g\in{}V_n we have ||P_{n}f-f||$$<em>{2}$ $\leq$ $||g-f||$$</em>{2}$</li> </ol> <p>So the partial sums of a Fourier Series are a good approximation of a general $L^2$ function, and hence of $C[0,1]$ function, but only in the $L^2$ sense. To get pointwise convergence, one needs a stronger condition than continuity (e.g. differentiability), as you pointed out.</p> <p>What goes wrong in an attempted proof? One would like to argue that </p> <ol> <li>if for some trigonometric polynomial $p\in{}V_n$ and $f\in{}C[0,1]$ we have $||p-f||&lt;\varepsilon$ (sup norm), then for the nth partial sum $P_{n}f$, $||P_{n}f-f||&lt;\varepsilon$.</li> <li>$||P_{n+1}f-f||\leq{}||P_{n}f-f||$ for all $n$</li> </ol> <p>The fact (2) above facilitates these arguments in the case of the $L^2$ norm, but not for the sup norm.</p> <p>I hope this was helpful.</p> http://mathoverflow.net/questions/70339/stone-weierstrass-theorem-applied-to-fourier-series/72820#72820 Answer by Julien Puydt for Stone-Weierstrass theorem applied to Fourier series Julien Puydt 2011-08-13T08:51:43Z 2011-08-13T09:06:17Z <p>The key point is that you're confusing uniform convergence and $L^2$ convergence ; indeed as $\mathcal{C}([0;1])$ is both a subspace of $\mathcal{B}([0;1])$ with $|.|_\infty$ and of $L^2([0;1])$ with $|.|_2$, you get two norms on the same vector space.</p> <p>But as it isn't a finite-dimensional space, it can have non-equivalent norms - and indeed, those two norms definitely aren't equivalent, which in particular means that a sequence which has a good behaviour for the $L^2$ norm (the partial sums of the Fourier series) doesn't necessarily have a good $|.|_\infty$ behaviour.</p> <p>EDIT: I should have said a little more ; there's an obvious inequality between the two norms (the mean inequality) so they are not that unrelated. But there is no reverse inequality, as can be shown by considering a sequence of piecewise linear functions : for $n\in\mathbb N$, consider $f_n$ as $t\mapsto n^\alpha-n^{\alpha+\beta}t$ on $[0;n^{-\beta}]$ and zero elsewhere ; if you choose $\alpha,\beta>0$ carefully, then you'll get a sequence which converges to zero for the $L^2$ norm, and won't converge uniformly.</p>