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

  • $\begingroup$ I do not have Rudin's book right now, but I'd suggest an affine change of variable: $x \mapsto \alpha x + \beta$. $\endgroup$ – Siminore Aug 7 '12 at 8:14

(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 $t'$.

  • $\begingroup$ Actually, Rudin first defines $P_{n}(x) := \int_{-1}^{1}f(x+t)Q_{n}(t)dt$, for $0\leq x \leq 1$. He then says "Our assumptions about $f$ show, by a simple change of variable, that $P_{n}(x) = \int_{-x}^{1-x}f(x+t)Q_{n}(t)dt = \dots$", as your formula begins. It isn't clear to me how he arrives at that first integral in your formula; the second substitution is clear. $\endgroup$ – Kirk Boyer Dec 23 '12 at 17:59
  • $\begingroup$ @Kirk Boyer: See my edit. $\endgroup$ – Christian Blatter Dec 23 '12 at 18:44
  • $\begingroup$ Somehow I didn't see how that worked, before, but it is clear now. $\endgroup$ – Kirk Boyer Dec 23 '12 at 18:51

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