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I was studying real analysis with Abbott's book, and I came across section 6.7, which talks about Weirstrass approximation theorem. At the end of this section he explains that the sketch that he gave to the reader as exercises for the proof of the Weirstrass approximation theorem is due to Henri Lebesgue, and as I was not able to follow all the steps of this proof, nor I was able to find a solution online, I am here asking if anyone knows about the proof or could at least help me understand the steps. Here they are:

  1. Prove a weaker statement involving approximating continuous functions with polygonal functions (which I was able to do);
  2. Find the Taylor series for $\sqrt{1-x}$ centered at $0$ and show that the series converges to the function in the closed interval $[-1,1]$ (which I as able to do);
  3. Using number 2) and the fact that $|a|=\sqrt{a^2}$, prove that there exist a sequence of polynomials that uniformly converges to $|x|$ on $[-1,1]$, and then generalize this result for any interval $[a,b]$ (with this I tried to substitute $x$ with $1-x^2$ in the first series and I found out that the convergence should happen on $[-\sqrt 2,\sqrt 2]$, I don't know if it is a mistake and honestly don't know if it does matter. For the second part I was not able to provide a proof, the only thing I could think of was another kind of substitution based on the values a and b of the interval);

There are other steps to the proof, but I noticed that if I assumed the rightness of the previous ones, I was able to finish the proof. Could anybody help? I tried to search online other proofs of this and found the Bernstein's which I was able to comprehend, but still I really want to understand this one too.

Gary
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Evan
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  • Hi @Evan I just came across this question. I have been working through the same section although I am struggling a little bit with it. Is there any chance you can check my question out? https://math.stackexchange.com/questions/4417415/proving-the-weierstrass-approximation-theorem-using-polygonal-functions – Sergio Apr 01 '22 at 21:07

1 Answers1

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It should be enough to rescale the $x$-axis. If $f_n(x)$ converges to $f(x)$ uniformly on $[-1, 1]$, then $K f_n(x/K)$ converges uniformly to $Kf(x/K)$ on $[-K, K]$. But for the absolute value we have $K|x/K| = |x|$.

Adam Bartoš
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  • Thanks! And if I can ask something: my book specify that the original sequence of polynomials should converge on the interval [-1,1], and as I explained in my post I get a different one (even if it shouldn't matter as what I get contains it), did I do anything wrong with the calculation of the interval where the sequence converge? Or maybe the substitution that I made was wrong? Or is everything alright? – Evan Sep 07 '21 at 14:26
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    @Evan If you substitute $x = 1 - a^2$, you ask: for which $a$ I get $x \in [-1, 1]$, and $a \in [-\sqrt2, \sqrt2]$ looks correct to me. The fact that you need the corvergence only for $a \in [-1, 1]$ is no problem. – Adam Bartoš Sep 07 '21 at 14:37
  • Thanks! I was kinda of scared because I didn't get the exact numbers and I thought I fucked up somewhere – Evan Sep 07 '21 at 14:39
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    @Evan Here you in fact don't need to solve for covergence interval, it is enough if you try the given interval: if $a \in [-1, 1]$, then $a^2 \in [0, 1]$, and so $x = 1 - a^2 \in [0, 1] ⊆ [-1, 1]$. This is an alternative way to get what you need and to convince yourself that the previous calculation was also right. – Adam Bartoš Sep 07 '21 at 14:43