I have been attempting to understand the logic of the Weierstrass Theorem in the Principle of Mathematical Analysis by Rudini. enter image description here enter image description here

I know the proof is long but I am having problems understanding the following:

1-Why is $\int_{-1}^{1} Q_n(x) dx=1$?

2-Why use integrals to define the sequence of polynomials $P_n$, is it because we want to define the polynomial in the smallest partition of the interval $[0,1]$?

3- Can $sup(f(x+t)-f(x))=M$ because of the last inequality on the 51 step?

Thanks for reading!

  • $\begingroup$ you can see a proof of this theorem in the book Analysis I of Amann and Escher without the need to use integrals, but it is longer and more complex. $\endgroup$ – Masacroso Mar 11 '17 at 19:10
  • $\begingroup$ Rudini. I kind of like that. $\endgroup$ – zhw. Mar 12 '17 at 0:20

Here are answers to your question. I personally prefer this proof (french version here) of Weierstrass' theorem from Sergi Bernstein (1912), although both proofs rely on the same underlying principles.

  1. The polynomial $Q_n$ is explicitly normalized by a constant $c_n$ so that $\int Q_n = 1$. This implies $$ f(x) = \int_{-1}^1 f(x) Q_n(t) dt, $$ an equality used to write $|P_n(x) - f(x)|$ as an integral.

  2. The question is unclear.

  3. No. The function $f$ is continuous on a closed interval and is thus bounded by some constant $M$. Therefore by the triangle inequality $$ \sup_x | f(x+t) - f(x)|\le \sup_x (|f(x+t)| + |f(x)|) \le \sup_x|f(x+t)| + \sup_x|f(x)| \le 2M . $$

  • $\begingroup$ Thanks for the answer but you are not answering my questions? Why is $\int Q_n=1$ despite the fact there is constant $c_n$ that makes it equal to 1, why choose 1? Could you expand on point number 2? I am not getting what you wrote. Point number 3 I think the inequality holds if you remove the 2, I guess the 2 comes from the interval. However I am not certain, could you further explain it? $\endgroup$ – Pedro Gomes Mar 11 '17 at 19:00
  • $\begingroup$ Thank you for precising your question. See my edit. $\endgroup$ – Olivier Mar 11 '17 at 19:07
  • $\begingroup$ Thank you too, this was really clear. Only one problem arose me now which I think you could clarify me please. Why write $\int_{-1}^{1} f(x+t)Q_n(t) dt$ for $P_n$ which is equal to $\int_{0}^{1} f(t)Q_n(t-x) dt$ can $f(t)$ in this case be treated as a constant? $\endgroup$ – Pedro Gomes Mar 11 '17 at 19:23
  • $\begingroup$ The integral is with respect to $t$, so you cannot treat $f(t)$ as a constant here. To better understand $P_n$, I suggest you draw the graph of $Q_n(t-x)$ as a function of x. Note that it is bell shaped around $t$. If you think about it, you will be able to see that $P_n(t)$ is an average of the values of $f$ giving a lot of importance to the value $f(x)$ when $x$ is close to $t$, and very litlle importance to the value $f(x)$ when $x$ is far from $t$. $\endgroup$ – Olivier Mar 11 '17 at 19:31
  • $\begingroup$ Thank you! Your assistance has been incredible! $\endgroup$ – Pedro Gomes Mar 11 '17 at 19:34

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