2
$\begingroup$

I'm taking a course of mathematical analysis and my professor just told us that Stone-Weierstrass theorem it's really important, but he didn't say why.

Can someone explain the idea of the proof and how to visualize the theorem.

And why is so important?

THEOREM:

Let A $\subset C(K)$ such that

1) A is a subalgebra with unity 1

2) For each $\ x_1, x_2 \in K $ with $\ x_1 \neq x_2 $, exist $f \in A$ such that f($\ x_1$) $ \neq $ f($\ x_2$).

Then $ \overline A = C(K)$, where C(K) is the space of continuous functions over a compact space

$\endgroup$
  • 1
    $\begingroup$ I dont know if it can be "visualized" (I think that no) but the point of the theorem is that with just a phew conditions we are able to know if a subset is dense, what means that we can approach points of the space using points of this dense subset. Of course this is used to know that we can approach (uniformly) some space of functions using functions of a dense subset of this space. Here the concept of "uniformly" is redundant because approaching points, under the norm of some space, is indeed more than just uniformly. $\endgroup$ – Masacroso Apr 18 '17 at 3:25
  • 1
    $\begingroup$ You should be able to visualise the conditions: a unital subalgebra just means that the subalgebra contains all the constant functions. The separating points condition is easy enough too: for example, the subalgebra consisting only of constant functions is not enough. $\endgroup$ – Joppy Apr 18 '17 at 4:06
1
$\begingroup$

The Stone-Weiestrass Theorem allows one to approximate continuous functions, and thus is very useful. One example of usage is for Fourier series, to prove the span of $\{\frac{1}{\sqrt{2\pi}}e^{inx}\}$ is a dense subset of $L^2([0,2\pi])$ I'm not sure how to effectively visualize this though.

The proof is a bit involving, and I recommend you to refer to Section $20.12$ of this book.

$\endgroup$
  • $\begingroup$ the book looks good, thanks. Also I think it's a little bit advanced for me, i'm an undergraduate student. $\endgroup$ – Aaron Martinez Apr 18 '17 at 4:22
  • $\begingroup$ @AaronMartinez Yea, could be a bit advanced for undergraduate.. Rudin's principles of mathematical analysis has a proof that might be a bit less advanced, but it only proves for compact metric space, instead of compact Hausdorff space. (still is a long proof though) $\endgroup$ – Yujie Zha Apr 18 '17 at 5:00
  • $\begingroup$ But if it works for metric spaces, it works for hausdorff spaces.(Every metric space is a Hausdorff space ) .But I feel Rudin's book advanced for me too. :D $\endgroup$ – Aaron Martinez Apr 18 '17 at 19:14
  • $\begingroup$ @AaronMartinez, You are right that every metric space is Hausdorff, but then the logic of your first sentence needs to be reversed, i.e. "if it works for Hausdorff spaces, it works for metric spaces" - not the other way around, because there are Hausdorff space that is not a metric space :) Could you give me an example of what knowledge seems advanced to you? $\endgroup$ – Yujie Zha Apr 18 '17 at 19:55
  • $\begingroup$ Oh I didn't knew I had to do the reverse too.Answering your question, Well almost every proof of Rudin's book. And maybe those proofs aren't that difficult but Rudin makes it hard to understand because he doesn't write every step of the proof (he makes it the shorter as possible, and in general that's how the books's proofs are written :( which is sad for people like me (beginners in math) ), so the student must 'complete' the proof to fully understand it. $\endgroup$ – Aaron Martinez Apr 19 '17 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.