Orthonormierte Wavelets mit kompaktem Träger

Wir stehen vor der Aufgabe, Skalierungsfunktionen φ: ℝ →ℂ zu produzieren mit folgenden Eigenschaften: (a)φ ∈ L2, supp(φ) kompakt(b)% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqy1dyMaai % ikaiaadshacaGGPaGaeyyyIO7aaOaaaeaacaaIYaaaleqaaOWaaabu % aeaacaWGObWaaSbaaSqaaiaadUgaaeqaaOGaeqy1dygaleaacaWGRb % aabeqdcqGHris5aOGaaiikaiaaikdacaWG0bGaeyOeI0Iaam4Aaiaa % cMcacaaMf8UaamOyaiaadQhacaWG3bGaaiOlaiaaywW7daqiaaqaai % abew9aMbGaayPadaWaaeWaaeaacqaH+oaEaiaawIcacaGLPaaacqGH % 9aqpcaWGibWaaeWaaeaadaWcaaqaaiabe67a4bqaaiaaikdaaaaaca % GLOaGaayzkaaWaaecaaeaacqaHvpGzaiaawkWaamaabmaabaWaaSaa % aeaacqaH+oaEaeaacaaIYaaaaaGaayjkaiaawMcaaiaacYcaaaa!625E!$$\phi (t) \equiv \sqrt 2 \sum\limits_k {{h_k}\phi } (2t - k)\quad bzw.\quad \widehat \phi \left( \xi \right) = H\left( {\frac{\xi }{2}} \right)\widehat \phi \left( {\frac{\xi }{2}} \right),$$(c)% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacq % aHvpGzaSqabeqaniabgUIiYdGccaGGOaGaamiDaiaacMcacaWGKbGa % amiDaiabg2da9iaaigdacaaMf8UaamOyaiaadQhacaWG3bGaaiOlai % aaywW7daqiaaqaaiabew9aMbGaayPadaWaaeWaaeaacaaIWaaacaGL % OaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXaaabaWaaOaaaeaacaaIYa % GaeqiWdahaleqaaaaakiaacYcaaaa!5052!$$\int \phi (t)dt = 1\quad bzw.\quad \widehat \phi \left( 0 \right) = \frac{1}{{\sqrt {2\pi } }},$$(d)% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacq % aHvpGzaSqabeqaniabgUIiYdGccaGGOaGaamiDaiaacMcadaqdaaqa % aiabew9aMnaabmaabaGaamiDaiabgkHiTiaadUgaaiaawIcacaGLPa % aaaaGaamizaiaadshacqGH9aqpcqaH0oazdaWgaaWcbaGaaGimaiaa % dUgaaeqaaOGaaGzbVlaadkgacaWG6bGaam4Daiaac6cacaaMf8+aaa % buaeaadaabdaqaamaaHaaabaGaeqy1dygacaGLcmaadaqadaqaaiab % e67a4jabgUcaRiaaikdacqaHapaCcaWGSbaacaGLOaGaayzkaaaaca % GLhWUaayjcSdaaleaacaWGRbaabeqdcqGHris5aOWaaWbaaSqabeaa % caaIYaaaaOGaeyyyIO7aaSaaaeaacaaIXaaabaWaaOaaaeaacaaIYa % GaeqiWdahaleqaaaaakiaac6caaaa!66A0!$$\int \phi (t)\overline {\phi \left( {t - k} \right)} dt = {\delta _{0k}}\quad bzw.\quad {\sum\limits_k {\left| {\widehat \phi \left( {\xi + 2\pi l} \right)} \right|} ^2} \equiv \frac{1}{{\sqrt {2\pi } }}.$$