4. <strong>The</strong> decomposition <strong>of</strong> ρ into a sum <strong>of</strong> irreducible representation is H = ⊕ n∈I Y d n forsome set I ⊂ N 0 . Moreover,∀f, g ∈ H k , 〈f, g〉 Hk = ∑ n∈IWhere {a n } n∈I are positive numbers.5. It holds that ∑ n∈IN d,n|S d−1 | a−2 n = 1.a 2 n〈P d,n f, P d,n g〉 L 2 (S d−1 )Pro<strong>of</strong> Let f ∈ H k , A ∈ O(d). To prove part 1, assume first thatn∑∀x ∈ S d−1 , f(x) = α i k(x, y i ) (9)For some y 1 , . . . , y n ∈ S d−1 and α 1 , . . . , α n ∈ C. We have, since k is symmetric, thatn∑f ◦ A(x) = α i k(Ax, y i )==i=1i=1n∑α i k(A −1 Ax, A −1 y i )i=1n∑α i k(x, A −1 y i )Thus, by <strong>The</strong>orem 5.1, f ◦ A ∈ H k . Moreover, it holds that||f ◦ A|| 2 H k= ∑α i ᾱ j k(A −1 y j , A −1 y i )i=11≤i,j≤n= ∑1≤i,j≤nα i ᾱ j k(y j , y i ) = ||f|| 2 H kThus, part 1 holds for function f ∈ H k <strong>of</strong> the form (9). For general f ∈ H k , by <strong>The</strong>orem 5.1,there is a sequence f n ∈ H k <strong>of</strong> functions <strong>of</strong> the from (9) that converges to f in H k . From whatwe have shown for functions <strong>of</strong> the form (9) if follows that ||f n −f m || Hk = ||f n ◦A−f m ◦A|| Hk ,thus f n ◦ A is a Cauchy sequence, hence, has a limit g ∈ H k . By <strong>The</strong>orem 5.1, convergencein H k entails point wise convergence, thus, g = f ◦ A. Finally,||f|| Hk = limn→∞||f n || Hk = limn→∞||f n ◦ A|| Hk = ||f ◦ A|| HkWe proceed to part 2. It is not hard to check that ρ is group homomorphism, so itonly remains to validate that for every f ∈ H the mapping A ↦→ ρ(A)f is continuous. Letɛ > 0 and let A ∈ O(d). We must show that there exists a neighbourhood U <strong>of</strong> A such that∀B ∈ U, ||f ◦ A −1 − f ◦ B −1 || Hk < ɛ. Choose g(·) = ∑ ni=1 α ik(·, y i ) such that ||g − f|| Hk < ɛ . 3By part 1, it holds that||f ◦ A −1 − f ◦ B −1 || Hk ≤ ||f ◦ A −1 − g ◦ A −1 || Hk + ||g ◦ A −1 − g ◦ B −1 || Hk + ||g ◦ B −1 − f ◦ B −1 || Hk= ||f − g|| Hk + ||g ◦ A −1 − g ◦ B −1 || Hk + ||g − f|| Hk< ɛ 3 + ||g ◦ A−1 − g ◦ B −1 || Hk + ɛ 325
Thus, it is enough to find a neighbourhood U <strong>of</strong> A such that ∀B ∈ U, ||g◦A −1 −g◦B −1 || Hk
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