Almost-all results on the p^\lambda problem

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� 1 − e � �� −τ hx ≪ hx −τ−1 , ∂ ∂x and taking into consideration that the O-constant in (1) depends only on the O-constant in (3). Lemma 2: We have (2) if for a suitable η > 0. sup N
We sh<strong>all</strong> prove the following lemma by adapting a method of Heath-Brown (see [11, page 257]). Lemma 4: Let Qh be any positive integer. Then |Lh| 2 � ≪ QhX � hX λ Y λ Q −1 �1/2 � h Q −1 h Y 2 + Y � Then Proof: For q ∈ � we define the interval Iq by Iq := � Y λ (q − 1)/Qh, Y λ � q/Qh . |Lh| ≪ � q≤2Qh � m∼X | � n ∼ Y, n λ ∈ Iq, mn ∼ N � 2−λ + QhX h −1 Y 2−λ + Y X λ�� (log N). bne � hm λ n λ� |. From that, using the Cauchy-Schwarz inequality, we obtain |Lh| 2 ≪ QhX � � | � bne � hm λ n λ� | 2 q≤2Qh ≪ QhX � q≤2Qh ≪ QhX � m∼X � n ∼ Y, n λ ∈ Iq n, r ∼ Y, |t| ≤ 2Y λ /Qh n ∼ Y, nλ ∈ Iq, mn ∼ N � r ∼ Y, r λ ∈ Iq | � m ∼ X, mn ∼ N, mr ∼ N | � m ∼ X, mn ∼ N, mr ∼ N e � h(n λ − r λ )m λ� | e � htm λ� |, (5) where we put t := n λ − r λ . Applying [11, Lemma 1] to the inner sum in the last line of (5), we get � m ∼ X, mn ∼ N, mr ∼ N e � htm λ� � ≪ min X, (ht) −1 X 1−λ + � htX λ�1/2 � ≤ � htX λ� 1/2 + X min � 1, (ht) −1 X −λ� . (6) Combining (5) and (6), and using partial summation, we obtain where |Lh| 2 ≪ QhX � hX λ Y λ Q −1 �1/2 h S � 2Y λ Q −1 � h + QhX 2 S � h −1 X −λ� + h −1 QhX 2−λ (log N) max ∆≥h−1X −λ S(∆)∆−1 , (7) 5
Page 1 and 2: Almost-all
Page 3: 2 Reduction to exponential sums We
Page 7 and 8: � 2 Kh ≪ (log N) N 1−λ h −
Page 9 and 10: Then the conditions in (9), (10), (
Page 11: Using this inequality, (22) and Tay

Almost-all results on the p^\lambda problem

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