By Leopoldo Nachbin
ISBN-10: 0442059108
ISBN-13: 9780442059101
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Extra info for The Haar Integral
Example text
6 . 0,,,)(x)| 2/2+1 2 According to (17) it is sufficient to prove the relations p n ( l ) = l = | P n ( _ l ) | and \Pn(x)\^Pn{\). The first of these relations follows immediately from Rodrigues' formula. To prove the second one, let us put n(n+\)f(x) = n(n + \)Pfi(x) + (\-x2)[P;i(x)Y. It then appears that / ( l ) = / ^ ( i ) = / ( - - i ) and /(g) = />*(§) at every point ξ, where Ρ,'(ξ) = 0, consequently in every point, where Pn(x) has an extremum. Thus we have ( 18) max Pi (x) ^ max /(*). Differentiating the equation defining f(x) we find that n(n+\)f(x) = 2P;i(x)[n(n + \)Pn(x)-xPn(x) + ( 1 - JC2) P'n'(x)] = 2Fn(x) [(\-χΐ)Ρ'η>(χ)-2χΡ'η(χ) + + + n(n+ \)Rt(x)] + 2x[P'n(x)Y.
First of all we have \ç(t)p«u(t)tpn(t)dt=— + I Q{f)pn+\(f)qn(t)dt, \ç(t)pt^dt where qn(t) is a polynomial of degree ±gn. For this reason the last integral on the right-hand side vanishes. Applying Schwarz's inequality (6) it follows at once that b o(t)\Pn+i(i)\\tPn(t)\dt; b (16) 1 b ^ max ( | 4 |Ô[){J*^(0^+1 (0^^ JC>(0^1 (0^^! ^ = max (|a|, \b\)==A. We then obtain from (13), (14), (15) and (16) that ξ\δ \sn(ï)-m\^2AC^ Î-Ô f(t)-m t-ξ dt + 4s AC. FUNDAMENTAL IDEAS 29 Furthermore, by (12), we have i^m^h ζ-δ dt ξ-δ -ï\\\og\t-ï\ 2Κ (α—1)| log ô\a~x Since a—1 > 0 and d>0 may be chosen to be arbitrarily small, the quotient on the right-hand side can be made arbitrarily small, e.
N\2nPn(x), 1 . 4 . 6 . 0,,,)(x)| 2/2+1 2 According to (17) it is sufficient to prove the relations p n ( l ) = l = | P n ( _ l ) | and \Pn(x)\^Pn{\). The first of these relations follows immediately from Rodrigues' formula. To prove the second one, let us put n(n+\)f(x) = n(n + \)Pfi(x) + (\-x2)[P;i(x)Y. It then appears that / ( l ) = / ^ ( i ) = / ( - - i ) and /(g) = />*(§) at every point ξ, where Ρ,'(ξ) = 0, consequently in every point, where Pn(x) has an extremum. Thus we have ( 18) max Pi (x) ^ max /(*).