Stone-Weierstrass Theorem

Question

In the proof of Stone-Weierstrass theorem provided in Rudin's Principles of Mathematical Analysis, why do we only need to show that there exists a sequence of polynomials $P_n$ that converges uniformly to the continuous complex function on $[0,1]$?

Answer 1

I think you are just asking why, "We may assume, without loss of generality, that $[a, b] = [0,1]$." The reason is that if $f$ is continuous on $[a,b]$, then we can uniformly approximate $f(a+(b-a)x)$ on $[0,1]$ with a sequence of polynomials $(p_n)_{n=1}^\infty$, and then $(p_n((x-a)/(b-a)))_{n=1}^\infty$ will converge uniformly to $f$.

Answer 2

Define $\phi:[a,b]\rightarrow [0,1]$ by \begin{align*} t = \phi(x) = \frac{x-a}{b-a},\quad\text{and}\quad x = \phi^{-1}(t) = a + (b-a)t \end{align*} It is continuously increasing function so that $\phi$ is bijective and $\phi^{-1}$ is also continuous. Then let $f(x) = f(\phi^{-1}(t))=g(t)$ so that $g:[0,1]\rightarrow R$ is also continuous by composition theorem. Suppose the theorem holds for $g$. Let $\varepsilon > 0$, there exists $P_n$ such that $\|P_n - g\|_{\infty} < \varepsilon$. Let $t\in[0,1]$, then \begin{align*} P_n(t) =& a_0 + a_1t + a_2 t^2 +\cdots + a_n t^n\\ =& a_0 + a_1 \phi(x) + a_2 \phi(x)^2 +\cdots + a_n \phi(x)^n\\ =& a_0 + a_1 \left(\frac{x-a}{b-a} \right) + a_2 \left(\frac{x-a}{b-a} \right)^2 +\cdots + a_n \left(\frac{x-a}{b-a} \right)^n\\ =&P_n\circ \phi(x) \end{align*} is also a polynomial. Thus, $\|P_n\circ \phi - f\| = \|P_n - f\circ\phi^{-1}\|=\|P_n - g\| < \varepsilon$.