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 $[0,1]$?
Stone-Weierstrass Theorem
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analysis
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0Could you please elaborate a little? Maybe give the precise statement of the conclusion and say why you don't see how it follows? – 2010-11-16
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0Because you can stretch it and so on? – 2010-11-16
1 Answers
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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 $f$.