Functional calculus in weighted group algebras

Let now be two intervals [q, r] ⊂ [q − ε, r + ε]. It is then possible to construct a continuousfunction ϕ, supp ϕ ⊂ [q − ε, r + ε], ϕ ≡ 1 on [q, r] and ˆϕ(λ) ∈ L 1 (R, t). We just have tocheck that it is possible to find a, b, c, d, a < b, c < d, a + d < c + b, such that⎧a + d = q⎪⎨c + b = r⎪⎩ a + c = q − εb + d = r + ε.As a matter of fact, solving this system gives us c = k, b = r − k, d = ε + k, a = q − ε − kwith k ∈ R arbitrary. If |x r | ≤ K r t(x) for all x ∈ R, for all r ∈ N, for constants K r , thenϕ ∈ Cc∞ (R).Next we shall show that there are similar results for R/2πZ, resp. Z.4.4. Let w : Z → Z be such that(1) w ≥ 1, w increasing on N(2) w(−n) = w(n) ∀n ∈ Z(3) w(n + m) ≤ Cw k (n)w k (m) for some k ≥ 1, C ≥ 1, ∀n, m ∈ N(4)∑ ln w(n)1 + n 2 < +∞.n∈NAssume moreover that there exists l 1 ∈ N such that∑(5) w −2l1k (n) < +∞(6)n∈N∑w −(l 1−k 2) (n) < +∞.n∈NThis has the following consequences for the function w:(i) First of all one hasw(n + m) ≤ Cw k (n)w k (m) ∀n, m ∈ Z.For n, m ∈ Z − this follows from the symmetry of the function w. If n ≥ 0, m ≤ 0,n + m ≥ 0, thenw(n + m) ≤ w(n) ≤ Cw k (n) ≤ Cw k (n)w k (m)as C ≥ 1, w increasing on N and k ≥ 1. The other cases are treated similarly.(ii) As k ≤ k 2 , one also has∑w −(l1−k) (n) < +∞.n∈N(iii) If we put l = l 1 · k, then l 1 ≤ l and∑w −(l−k2) (n) < +∞.n∈N19