# Question about the proof of Stone-Weierstrass theorem (Weierstrass approximation theorem) in Rudin

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 $$(t+x)^k$$.

• 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. – Sayako Hoshimiya Dec 16 '18 at 11:46
• 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. – Sayako Hoshimiya Dec 16 '18 at 11:50
• 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? – Sorin Tirc Dec 16 '18 at 16:04
• Yes, now I get it. Thank you. – Sayako Hoshimiya Dec 17 '18 at 2:18