SLIDE 1
Problem
Fn : fn(t) = a0+
n
- k=1
(ak cos kt + bk sin kt), ak, bk ∈ C the conjugate to fn:
- fn(t) =
n
- k=1
(ak sin kt − bk cos kt) We study the operator Λθ,Afn(t) = Aa0 +
n
- k=1
(ak cos(kt + θ) + bk sin(kt + θ)) = Aa0 + cos θ(fn(t) − a0) − sin θ fn(t), θ ∈ R, A ∈ C
- Λθ,Afn
- ∞ Cn(θ, A)fn∞,