A First Course in Harmonic Analysis (Universitext)

The Hilbert-Schmidt norm ||T ||HS of T is defined by way of ||T ||2HS def T ej , T ej . = j This quantity is ≥ zero yet could be +∞. we need to express that it doesn't rely on the alternative of the orthonormal foundation. For this keep in mind 12. three. HILBERT-SCHMIDT OPERATORS one hundred sixty five that Theorem 2. three. 2 signifies that for each orthonormal foundation (φα )α and all v, w ∈ H we have now v, w = v, φα φα , w . α So permit φα be one other orthonormal foundation. now not figuring out the independence but, we denote the Hilbert-Schmidt norm hooked up to the orthonormal foundation (ej ) through ||T ||HS,(e) . Then ||T ||2HS,(e) = T ej , T ej j = T ej , φα φα , T ej j α α j = T ej , φα φα , T ej ej , T ∗ φα T ∗ φα , ej = α j T ∗ φα , T ∗ φα = = ||T ∗ ||HS,(φ) . α The interchange of order of summation is justified, because all summands are optimistic. For (φ) = (e) this specifically implies ||T ∗ ||HS,(φ) = ||T ||HS,(φ) , in order that ||T ||HS,(e) = ||T ∗ ||HS,(φ) = ||T ||HS,(φ) . this offers the specified independence, so ||T ||HS is well-defined. we are saying that the operator T is a Hilbert-Schmidt operator , if ||T ||HS < ∞. Lemma 12. three. five for each bounded operator T on H, ||T || ≤ ||T ||HS . for each unitary operator U we've got ||U T ||HS = ||T U ||HS = ||T ||HS . evidence: enable v ∈ H with ||v|| = 1. Then there's an orthonormal foundation (ej ) with e1 = v. We get ||T v||2 = ||T e1 ||2 ≤ ||T ej ||2 = ||T ||2HS . j 166 bankruptcy 12. THE HEISENBERG crew The invariance below multiplication by means of unitary operators is obvious, in view that (U ej ) is an orthonormal foundation while (ej ) is. the most instance we're attracted to is the subsequent. remember L2 (R), that's the Hilbert house finishing touch of L2bc (R) in addition to that of Cc (R). allow okay be a continual bounded functionality on R2 and feel R R |k(x, y)|2 dx dy < ∞. This double vital is to be understood as follows. We suppose that for each x ∈ R the necessary R |k(x, y)|2 dy exists and defines a continuing functionality on R that's integrable, and an analogous with x and y interchanged. below those conditions we name okay an L2 -kernel . Proposition 12. three. 6 consider k(x, y) is an L2 -kernel on R. For ϕ ∈ Cc (R) define Kϕ(x) def = k(x, y)ϕ(y) dy. R Then okϕ lies in L2bc (R), and ok extends to a Hilbert-Schmidt operator okay : L2 (R) → L2 (R) with ||K||2HS = R R |k(x, y)|2 dx dy. evidence: The functionality okϕ is obviously non-stop and bounded. We use the Cauchy-Schwartz inequality to estimate ||Kϕ||2 = R |Kϕ(x)|2 dx 2 = k(x, y)ϕ(y) dy R ≤ dx R R R R R = |k(x, y)|2 dx dy R |ϕ(y)|2 dy |k(x, y)|2 dx dy ||ϕ||2 . So okay extends to a bounded operator on L2 (R). allow (ej ) be an 12. four. THE PLANCHEREL THEOREM FOR H 167 orthonormal foundation of L2 (R). Then ||K||HS = Kej , Kej j = j R j R j R Kej (x)Kej (x) dx = = = R = R k(x, y)ej (y) dy R k(x, y)ej (y) dy dx k(x, . ), ej ej , k(x, . ) dx k(x, . ), ej ej , k(x, . ) dx j k(x, . ), k(x, . ) dx R = R 12. four R |k(x, y)|2 dx dy. The Plancherel Theorem for H permit G be a in the neighborhood compact staff and permit f ∈ Cc (G). repair a Haar degree on G. For a unitary illustration (π, Vπ ) of G we define, officially at first, π(f ) = f (x)π(x) dx G because the special linear operator on Vπ that satisfies π(f )v, w f (x) π(x)v, w dx = G for all v, w ∈ Vπ .