2013
DOI: 10.4064/aa159-1-2
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Exotic approximate identities and Maass forms

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Cited by 2 publications
(1 citation statement)
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“…The key argument is a simple one and it is summarized in the following result. We merely sketch the details of the proof because it is essentially contained in that of [, Lemma 2.3] (see also [, § 5] and [, Lemma 2.2]). Lemma For 0<v<V consider the function F:double-struckH2R given by Ffalse(z,wfalse)=14πvdouble-struckHχVufalse(z,ζfalse)χvufalse(ζ,wfalse)dμfalse(ζfalse).Then there exists f:[0,)R such that F(z,w)=f(u(z,w)) and χVfχV+whereV±=()V(1+v)±v(1+V)2.Moreover, the S‐HC transform of f is hVhv.…”
Section: Proof Of the Main Resultsmentioning
confidence: 99%
“…The key argument is a simple one and it is summarized in the following result. We merely sketch the details of the proof because it is essentially contained in that of [, Lemma 2.3] (see also [, § 5] and [, Lemma 2.2]). Lemma For 0<v<V consider the function F:double-struckH2R given by Ffalse(z,wfalse)=14πvdouble-struckHχVufalse(z,ζfalse)χvufalse(ζ,wfalse)dμfalse(ζfalse).Then there exists f:[0,)R such that F(z,w)=f(u(z,w)) and χVfχV+whereV±=()V(1+v)±v(1+V)2.Moreover, the S‐HC transform of f is hVhv.…”
Section: Proof Of the Main Resultsmentioning
confidence: 99%