I am reading Rudin's proof (3rd edition) and am wondering what substitution is made to make it true that $P_n(x)=$ the integral from $-x$ to $1-x$ is equal to the same function integrated from -1 to 1. He says there's a substitution but I haven't found the right one. Thanks a lot
Stone-Weierstrass theorem proof (Rudin)
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real-analysis
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0I do not have Rudin's book right now, but I'd suggest an affine change of variable: $x \mapsto \alpha x + \beta$. – 2012-08-07
1 Answers
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(Edit after a comment by Kirk Boyer) The formula in question is $P_n(x):=\int_{-1}^1 f(x+t)Q_n(t)\ dt=\int_{-x}^{1-x}f(x+t)Q_n(t)\ dt=\int_0^1f(t)Q_n(t-x)\ dt\ .$ One obtains the second integral on account of the assumption that $f(t)$ is $\equiv0$ outside $[0,1]$, the third integral substituting $t:=t'-x$ $\,(0\leq t'\leq1)$, and finally writing again $t$ in place of $t'$.
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0Somehow I didn't see how that worked, before, but it is clear now. – 2012-12-23