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January 9

Alternate proof of the Weierstrass approximation theorem

Hello everyone,

I have been asked to provide ('complete') an alternate proof of the Weierstrass approximation theorem - for any continuous function on [a, b] we can uniformly approximate it by polynomials - which begins as follows (the 'original' was the common proof using Bernstein polynomials):

Let 0 < a < b < 1, and f:[a,b]->R the continuous function we wish to approximate by polynomials. Fix any continuous extension of f to all of R such that the function is identically zero outside of [0,1], and denote this again by f...

Problem is, I have absolutely no idea how to continue. How does extending f to all of R help? Surely that only makes it harder to approximate. Could anyone please help me? Thank you very much! 178.176.2.17 (talk) 00:07, 9 January 2011 (UTC)[reply]

Is that all you have to go by? No indications even of which kind of theory you're supposed to use?

The extension appears to indicate that you're supposed to do something with the function that requires f to be defined on the entire real line. My first hunch would be something like Fourier transform it, then Taylor approximate each frequency component and then sum them back together. But I don't actually know whether that would work, and even if it did, it would seem to be much easier to extend f periodically and use a plain Fourier series. So it's probably a false scent. –Henning Makholm (talk) 02:45, 9 January 2011 (UTC)[reply]

Extending the function to all of R may be a device to allow convolution. If you take the convolution product of f with a polynomial P, the resulting function will be a polynomial. If P is taken to be a good approximation of the Dirac delta function, then the convolution product will be uniformly close to f. Of course, to do this, you need to cut off part of the function p. But since you've extended f by zero, the part of p that you've cut off doesn't matter, and the convolution is still a polynomial. 82.120.58.206 (talk) 03:28, 9 January 2011 (UTC)[reply]

That is, it's a polynomial on the interval you're interested in. 82.120.58.206 (talk) 03:35, 9 January 2011 (UTC)[reply]

That makes much more sense than my wild guessing above. –Henning Makholm (talk) 05:15, 9 January 2011 (UTC)[reply]

What you wrote is the proof I first learned for Stone-Weierstrass, and still the one I like best, although I'm having trouble remembering sufficient conditions for uniform convergence of Fourier series at the moment. The function f would probably first need to be approached by a function satisfying those conditions. (Either that, or replace the Fourier series with its Cesàro sum and use Fejér's theorem.) 82.120.58.206 (talk) 06:54, 9 January 2011 (UTC)[reply]

That makes sense to me - the second approach sounds good certainly - but if our polynomials are going to tend to the delta function, don't we need the 'width' to get smaller and smaller? But at the same time, we presumably need the polynomial to be positive over [0,1], otherwise we can't guarantee effects at the edges will disappear in the convolution. So, does the polynomial in the convolution get sharper and sharper as it tends to the delta function, or does it get stay the same width, i.e. remain positive on [0,1]? Clarification would be much appreciated, thank you! 178.176.6.47 (talk) 04:36, 10 January 2011 (UTC)[reply]

Consider, for example

(1-x^{2})^{n}

scaled such that its integral over [-1,1] becomes 1. The peak becomes narrower with increasing n, but the useful domain of the function stays the same. –Henning Makholm (talk) 12:10, 10 January 2011 (UTC)[reply]

Yes, I agree with Henning's suggestion. Extend the polynomial by 0 outside [-1,1]. 82.120.58.206 (talk) 12:23, 10 January 2011 (UTC)[reply]

Got it, thank you very much! 178.176.10.8 (talk) 14:45, 10 January 2011 (UTC)[reply]