 Unsmoothing a logarithmically weighted sum

ID: unsmoothing-a-logarithmically-weighted-sum

Suppose

∣ a_{n} ∣ \leq (lo g (2 n))^{k}

and

A (x) = \sum_{n \leq x} a_{n} lo g (x / n) = O (x (lo g x)^{β})

, with

0 \leq β \leq k

. For

0 < h \leq 1

, direct subtraction gives

A (x e^{h}) - A (x) = h \sum_{n \leq x} a_{n} + \sum_{x < n \leq x e^{h}} a_{n} lo g (x e^{h} / n) .

The last term has absolute value

O ((x h^{2} + h) (lo g x)^{k})

. Choose

h = (lo g x)^{(β - k) /2}

to obtain the displayed bound, with an additional

O ((lo g x)^{k})

term harmless as

x \to \infty

for fixed

k

. This elementary finite-difference argument converts a logarithmically smoothed Perron formula bound to cancellation in its original partial sum.

 New to topics? Read the docs here!