By Ionin Y. J., Shrikhande M. S.

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**Sample text**

Write Step(G) for the sub-algebra of locally constant functions, which is easily seen to be everywhere dense. We now explain how to integrate any continuous Cp -valued function on G against an element λ of Λ(G). We begin with locally constant functions. Suppose that f in Step(G) is locally constant modulo the subgroup H of G. 1) x∈G/H where the cH(x) lie in Zp . We then deﬁne f dλ = G cH(x)f (x). x∈G/H One sees easily that this is independent of the choice of H. Since the cH(x) lie in Zp , we have ≤ f .

1. We have (1 − ϕ)R = T R. Proof. The inclusion of (1 − ϕ)R in T R is plain. Conversely, if h is any element of T R, let us show that it lies in (1 − ϕ)R. For each n ≥ 0, n we deﬁne ωn (T ) = (1 + T )p − 1. 4 The Logarithmic Derivative 21 where hn is a polynomial in Zp [T ] of degree less than pn , and rn is an element of R. Deﬁne n−1 ϕi (hn−i ). ln = i=0 Clearly, we have ln+1 − ϕ(ln ) = hn+1 . Since hn+1 converges to h in R, it suﬃces to show that ln converges to some l in R, because then we would have h = (1 − ϕ)l.

7, there exists a unique u = (un ) in U∞ such that f = fu and hence we have fu ((1 + T )p − 1) = fu (T )p . This implies that upn = un−1 for all n ≥ 1 and that fu (0) is in µp−1 . But then fu (0) = 1 since fu ≡ 1 mod p and so (un ) ∈ Tp (µ). Thus there exists a in Zp such that u = (ζn )a , whence f (T ) = (1 + T )a . This proves that ker(L) = A. It is clear that α ◦ L = 0, and the surjectivity of α follows from noting that ψ(1 + T ) = 0 and that α(1 + T ) = 1. Hence it only remains to prove that ker(α) ⊂ Im(L), which is the delicate part of the proof of the theorem.

### (2s1) Designs withs intersection numbers by Ionin Y. J., Shrikhande M. S.

by Jason

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