In Rudin's Principles of Mathematical Analysis, a proof of the Stone-Weierstrass theorem in its original statement is included (3ed, p159):

My question is about the step after (51), $P_n(x)=\int_{-1}^1f(x+t)Q_n(t)\operatorname{d}t$. How does one proceed from this, by a change of variable, to the next step, namely $P_n(x)=\int_{-x}^{1-x}f(x+t)Q_n(t)\operatorname{d}t$?

And another question is why $P_n(x)=\int_0^1f(t)Q_n(t-x)\operatorname{d}t$ is a polynomial.


Well the first equality, namely $\int_{-1}^{1}f(x+t)Q_n(t)dt = \int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.

Now $\int_{-x}^{1-x}f(x+t)Q_n(t)dt = \int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.

The fact that $\int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = \sum_{k=0}^{n}a_i(t+x)^k=\sum_{k=0}^{n}b_i(t)x^k$ and now $\int_{0}^{1}f(t)Q_n(t-x)dt = \sum_{k=0}^{n}(\int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.

  • $\begingroup$ Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$\sum_{i=0}^{2n}c_nk(i)\left(\int_0^1f(t)\cdot t^{2n-i}\operatorname{d}t\right)x^i$$ where $k(i)$ are the merged binomial coefficients. $\endgroup$ – Sayako Hoshimiya Dec 16 '18 at 11:46
  • $\begingroup$ But I am still wondering about the $\int_{-1}^{1}f(x+t)Q_n(t)dt = \int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks. $\endgroup$ – Sayako Hoshimiya Dec 16 '18 at 11:50
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    $\begingroup$ Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear? $\endgroup$ – Sorin Tirc Dec 16 '18 at 16:04
  • $\begingroup$ Yes, now I get it. Thank you. $\endgroup$ – Sayako Hoshimiya Dec 17 '18 at 2:18

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