# An introduction to abstract algebra, - download pdf or read online By F. M. Hall

ISBN-10: 0521070554

ISBN-13: 9780521070553

ISBN-10: 0521298628

ISBN-13: 9780521298629

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Example text

The functions h and f p are in L1 , so the function w → h z−1 w 1/p | f (w)| is in L p by the (1, 1) case of the proposition treated first. Furthermore, 1/q w → h z−1 w is in Lq . 7 we obtain h z−1 w | f (w)| dw = h z−1 w 1/p | f (w)| h z−1 w 1/q dw 1/q 1/p ≦ h z−1 w | f (w)| p dw h z−1 w dw . Raising both sides to the p power and using Haar measure invariance w → zw, we get p/q h z−1 w | f (w)| p dw · ||h||1 . |(h # f ) (z)| p ≦ So integrating dz and using Fubini gives p/q ||h # f || pp ≦ ||h||1 || f || pp ||h||1 = ||h||1p || f ||qp because 1 + p/q = p.

We have 2 (i) (i) (i) (i) (i) E˜α E˜−α + E˜−α E˜α ≡ −2 E˜α (i) ˜ + H˜ α mod E˜α k. Proof. It will be convenient to use the operator θ (Cartan involution) defined on the Lie algebra by ¯ θZ = −t Z. Then k consists of those Z ∈ g such that θZ = Z (fixed point set of θ). We write (i) (i) (i) (i) (i) (i) (i) (i) E˜−α E˜α = E˜−α , E˜α + E˜α E˜−α = H˜ α + E˜α E˜−α . Furthermore, (i) (i) (i) (i) (i) E˜−α = θ E˜α = θE˜α + E˜α − E˜α , so (i) (i) (i) E˜α E˜−α ≡ − E˜α 2 (i) ˜ mod E˜α k. This proves the lemma.

By a change of variables v → −v, we can omit the term with −e−irv provided we multiply the integral by 2, which combined with (ev − e−v )−1 gives 1/sinh v. Then the integral can be evaluated, say by plugging into the calculus identity (1), showing that t Φp, HC Sp ϕc = ϕc . As to the second statement, we first make explicit the L2 -scalar product on G, for the polar measure. For two basic Gaussians ϕc , ϕc′ , we get, integrating over A ∼ = (−∞, ∞); ∞ e−v < ϕc , ϕc′ >p = 2 /2c −∞ 2 ′ 2v 2v e−v /2c v v −v e −e e − e−v ev − e−v 2 2 dv ∞ ′ ′ 2 e−v (c+c )/2cc v2 dv = −∞ ∞ ′ 3/2 = cc −∞ 2 ′ e−v (c+c )/2 v2 dv [by v → v (cc′ )1/2 ].