Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 124 / 4 / a

Solution

B

m = n

contributes exactly

T \sum_{n \leq X} a_{n} \overline{b_{n}} / n^{2 σ}

. If

m \neq = n

\int_{T}^{2 T} e^{i t l o g (m / n)} d t = \frac{e ^{2 i T l o g (m / n)} - e ^{i T l o g (m / n)}}{i l o g ( m / n )},

whose absolute value is at most

2/∣ lo g (m / n) ∣

. Thus the mean-value formula is

\int_{T}^{2 T} A (σ + i t) \overline{B (σ + i t)} d t = T n \leq X \sum \frac{a _{n} b _{n}}{n ^{2 σ}} + O m, n \leq X m \neq = n \sum \frac{∣ a _{n} ∣∣ b _{m} ∣}{n ^{σ} m ^{σ} ∣ lo g ( m / n ) ∣} .

This also covers

T = 0

For

m > n

, integration of

1/ x

over

[n, m]

gives

lo g (m / n) \geq (m - n) / m

; the analogous inequality holds with

m, n

exchanged. Hence

∣ lo g (m / n) ∣^{- 1} \leq max (m, n) /∣ m - n ∣ \leq X

. This gives the requested first bound

O (X \sum_{m \neq = n} ∣ a_{n} ∣∣ b_{m} ∣/ (n^{σ} m^{σ}))

For the weighted bound, put

α_{n} = ∣ a_{n} ∣ n^{- σ}

and

β_{m} = ∣ b_{m} ∣ m^{- σ}

. The inequality

2 α_{n} β_{m} \leq \frac{n}{m} α_{n}^{2} + \frac{m}{n} β_{m}^{2}

reduces the estimate to a row sum and its symmetric counterpart. For a fixed

n

, the preceding logarithm bounds give

\sum_{m \leq X m \neq = n} \frac{n / m}{∣ l o g ( m / n ) ∣} \leq \sum_{m > n} \frac{n}{m - n} + \sum_{m < n} \frac{n ^{2}}{m ( n - m )} ≪ n lo g (2 X) .

Here

n^{2} / (m (n - m)) = n (1/ m + 1/ (n - m))

, and each resulting harmonic sum is

O (lo g (2 X))

. The symmetric row sum is bounded by

O (m lo g (2 X))

. Therefore the second error bound is

O (n \leq X \sum \frac{∣ a _{n} ∣ ^{2} n lo g ( 2 X )}{n ^{2 σ}} + m \leq X \sum \frac{∣ b _{m} ∣ ^{2} m lo g ( 2 X )}{m ^{2 σ}}) .

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