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G/ and gkp Ä kf k1 kgkp : kf Furthermore, if p D 1, then f g is also continuous. G/. 3 Representation Theory Let H1 and H2 be two Hilbert spaces (the corresponding norms and scalar product are simply denoted by k k and h ; i). Suppose that AW H1 ! H1 ; H2 /. Recall that A is an isometry if kAuk D kuk for every u 2 H1 . Since kAuk2 D hAu; Aui D hA Au; ui and kuk2 D hu; ui, the polarization identity implies that A is an isometry if and only if A A D idH1 . Hence, isometries are injective, but they are not necessarily surjective.

A; b/f ; gi ¤ 0 as a continuous function, because neither fO nor gO can identically vanish. R/ is irreducible. R/. R/ and that D C ˚ . R/. R/ that satisfies it is called a wavelet. 2 The Use of Representations 35 Let us now consider the full affine group. a /j2 à da jOg. /j2 d jaj This time, for any nonzero , as a ranges in R the numbers a cover R and the change of variable a 7! a/j2 da jaj à ÂZ R jOg. 30) which cannot be zero if both f and g are not zero. This proves that full is an irreducible unitary representation of Gfull .

61. G; ; H ; / is a reproducing system. G/, so that, Proof. G/. x/ i D hF; V . x/: The second statement is obvious, because K is of course in the range of V and hence coincides with its projection onto the range. t u We end this section with a classical result due to Duflo and Moore [15], later reviewed with a slightly different argument in [22]. The proof would require some results on unbounded operators that defy the scope of this chapter. Here we content ourselves with its statement2 and some comments.

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Scientific papers, vol.2 by Rayleigh.

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