Proof of Theorem 1 Sample Clauses

Proof of Theorem 1. 2.2 1. This simplifies the corresponding steps in the proof. We fix the following notation for the Fourier transform ∫ fˆ(ξ) = Rn f (x)e—iξx dx, fˇ(x) = (2π)—n ∫ Rn f (ξ)eiξx dξ. (2.2.1)

Proof of Theorem 1. 5: By Proposition 4.16, λ is an algebraic integer, and, thus, all conjugates of λ are algebraic integers, since they satisfy the same monic, integral polynomial. Recall that 1/λ is a conjugate of λ. By Theorem 3.23, algebraic integers form a ring. So, r2 = λ · λ¯ and 1/r2 = 1/λ · 1/λ¯ are also algebraic integers. Since r and 1/r satisfy the polynomials x2 − r2 and x2 − 1/r2, r and 1/r are algebraic integers as well. Furthermore, algebraic integers are closed under multiplication. So, eiθ = 1/r · reiθ is also an algebraic integer. Let σ ∈ G(Nr,eiθ /Q). We will consider all σ that fix r. (The set of all such automorphisms is nonempty, since the identity map is an automorphism that fixes r.) Note that from Theorem 4.16 the conjugates of λ are known. If we assume that r is fixed under some embedding into C, then we can determine the image of eiθ under the embedding as well. Claim: If σ(r) = r, then σ(eiθ) = e±iθ. Suppose by way of contradiction, there exists σ such that σ(r) = r and σ(reiθ) = 1/λ, 1/λ¯, or eiφ. First, assume σ(λ) = 1/λ¯. (If σ(λ) = 1/λ, then compose σ with conjugation. Note that σ(r) will still equal r.) Then, σ(reiθ) = r · σ(eiθ), which implies σ(eiθ) = 1/r2 · eiθ and also σ(e2iθ) = 1/r4 · e2iθ. Thus, e2iθ has a conjugate with norm 1/r4, which is a contradiction by Lemma 6.1. Second, assume σ(reiθ) = eiφ. Then, σ(eiθ) = 1/r · eiφ and, furthermore, σ(e2iθ) = 1/r2 · e2iφ. By Lemma 6.1, the only possible conjugate of e2iθ with norm equal to 1/r2 is precisely 1/r2. But, this implies eiφ = ±1, which is a contradiction.

Proof of Theorem 1 mBIDH G1,G2 ,eˆ is negligible and qS, qH1 are finite.

Proof of Theorem 1. 4.1‌ This proof is analogous to the proof of [BO16, Theorem 2.1]. By well known properties of Bessel functions, such as the bounds in (9.8.4) of [AS72], for x ≥ 37.5 the modified Bessel function I1(x) is bounded by N ≤ x2 e−xI1(x) ≤ M where N = 0.394, M = 0.399. First, let 2 ≤ k ≤ 5, and let β := αk . Then by Theorem 5.1.2, for n ≥ 450 we have: 2π . k − 1 Σ1 . ▇ − 1 + 24n β . 2 −µΣ Σ eµ

Proof of Theorem 1. 1‌ The proof of Theorem 1.2.1 requires a well known result of ▇▇▇▇▇ [Stu80], which says that if a modular form with integer Fourier coefficients is nonvanishing modulo a prime A, then there is a bound on the index of the first coefficient which is nonzero modulo A. To state his theorem, for a rational prime A and a modular form f (z) = Σ∞n=0 a(n)qn ∈ Mk(Γ0(N ), χ) with coefficients in Z, we define ordA(f ) := minn{n : A ‡ a(n)}, and we say ordA(f ) := ∞ if A|a(n) for all n. Σ 3.2.1 (▇▇▇▇▇). For a modular form f (z) = ∞n=0 a(n)qn ∈ Mk(Γ0(N ), χ) ordA(f ) > 12 [Γ0(1) : Γ0(N )], then ordA(f ) = ∞.

Proof of Theorem 1. The results of the previous sections reduce the general case to the case of elements whose spectra are contained in sets of the form ε Σ =d=ef (εZ × εZ) ∩ ([—1; 1] × [—1; 1]). It turns out that if σ(T ) C Σε, then we can remove its off-diagonal elements with respect to P in such a way that the element remains normal. The idea is to map the spectrum onto a line segment, then remove off-diagonal elements from the resulting self-adjoint element (it will remain self-adjoint), and then map it back. The choice of particular maps is important since it is the only place where we get a loss in the power of δ. The following lemma describes these maps.

Proof of Theorem 1