On the constant in the Pólya-Vinogradov inequality

Journal Pre-proof On the constant in the Pólya-Vinogradov inequality

Bryce Kerr

PII:

S0022-314X(19)30403-2

DOI:

https://doi.org/10.1016/j.jnt.2019.11.003

Reference:

YJNTH 6440

To appear in:

Journal of Number Theory

Received date:

8 April 2019

Revised date:

25 November 2019

Accepted date:

29 November 2019

Please cite this article as: B. Kerr, On the constant in the Pólya-Vinogradov inequality, J. Number Theory (2019), doi: https://doi.org/10.1016/j.jnt.2019.11.003.

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´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY BRYCE KERR Abstract. In this paper we obtain a new constant in the P´ olyaVinogradov inequality. Our argument follows previously established techniques which use the Fourier expansion of an interval to reduce to Gauss sums. Our improvement comes from approximating an interval by a function with slower decay on the edges and this allows for a better estimate of the 1 norm of the Fourier transform. This approximation induces an error for our original sums which we deal with by combining some ideas of Hildebrand with Garaev and Karatsuba concerning long character sums.

1. Introduction Given integers q, M and N and a primitive multiplicative character χ mod q we consider estimating the sums  χ(n), S(χ, M, N ) = M
and when M = 0 we write S(χ, 0, N ) = S(χ, N ). The first nontrivial result in this direction is due to P´olya and Vinogradov from the early 1900’s and states that (1)

S(χ, M, N )  cq 1/2 log q,

for some constant c independent of q. Up to improvements in the constant c this bound has remained sharpest known for the past 100 years and a fundamental question in the area of character sums is whether c can be taken arbitrarily small. Montgomery and Vaughan [19] have shown conditionally on the Generalized Riemann Hypothesis that S(χ, M, N )  q 1/2 log log q. This would be best possible since Payley [20] has shown that there exists an infinite sequence of integers q and characters χ mod q such Date: December 16, 2019. 1

2

B. KERR

that max S(χ, N )  q 1/2 log log q.

1N
Although making a o(1) improvement on the P´olya-Vinogradov inequality for all characters χ and intervals (M, M + N ] remains an open problem, there has been progress in determining general situations where such improvements can be made. Concerning short character sums, a classic result of Burgess [5, 6] states that for any primitive χ |S(χ, M, N )|  N 1−1/r q (r+1)/(4r

2 )+o(1)

,

provided r  3 and for any r  2 if q is cubefree. Hildebrand [15] has shown that one can improve on the constant in the P´olya-Vinogradov inequality given estimates for short character sums and Bober and Goldmakher [2] and Fromm and Goldmakher [10] have shown how improvements on the constant in the Polya-Vinogradov inequality may be used to obtain new estimates for short character sums. Concerning long character sums, Hildebrand [16] has shown that if χ(−1) = 1 then |S(χ, αq)| < εq 1/2 log q, for all α ∈ (0, 1) except for a set of measure q −c1 ε and that if α = o(1) and χ(−1) = 1 then S(χ, αq) = o(q 1/2 log q). Bober and Goldmakher [1] and Bober, Goldmakher, Granville and Koukoulopoulos [3] have obtained much more precise results concerning the distribution of long character sums and Granville and Soundararajan [13] have obtained results concerning the distribution of short character sums. Granville and Soundararajan [14] have also shown that S(χ, N )  q 1/2 (log q)1−δg /2+o(1) , if χ has odd order g, where π g sin , π g and the factor δg /2 occuring above has been improved by Goldmakher [12] and Lamzouri and Mangerel [18]. δg = 1 −

We consider the problem of estimating S(χ, M, N ) uniformly over χ, M and N in the P´olya-Vinogradov range. Since the work of P´olya and Vinogradov there have been a number of improvements to the constant c occuring in (1). The sharpest constant is due to Pomerance [21]

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

3

and is based on ideas of Landau [17] and an unpublished observation of Bateman, see [15]. In particular, Pomerance [21, Theorem 1] shows that   1/2 2 log q, if χ(−1) = 1, 2 + o(1) q π  1/2 |S(χ, M, N )|   1 (2) + o(1) q log q, if χ(−1) = −1. 2π Pomerance gives the lower order terms explicitly and these have been improved by Frolenkov [8] and Frolenkov and Soundararajan [9]. In the case of intervals starting from the origin one may obtain better constants with the sharpest given by Granville and Soundararajan [14], improving on previous results of Hildebrand [15, 16]. We note that both [14] and [15] obtain sharper constants than our main result although these are restricted to sums of the form S(χ, 0, N ). The strength of our result lies in the estimation of S(χ, M, N ) for arbitrary M . In this paper we obtain a new constant in the P´olya-Vinogradov inequality for arbitrary intervals. Our argument follows previously established techniques which use the Fourier expansion of an interval to reduce to Gauss sums. Our improvement comes from approximating an interval by a function with slower decay on the edges which allows for a better estimate of the 1 norm of the Fourier transform. This induces an error for our original sums which we deal with by combining some ideas of Hildebrand [15] with Garaev and Karatsuba [11]. A new feature of our argument is that we use estimates for long character sums to improve on the constant in the P´olya-Vinogradov inequality. For example, if one could show that for any ε > 0 we have S(χ, M, N ) = o(q 1/2 log q), for arbitrary M whenever N < q 1−ε and sufficiently large q then it would follow from our argument that S(χ, M, N ) = o(q 1/2 log q), for arbitrary M and N . We end the introduction by mentioning some notation and conventions used throughout. Given a real number x we let e(x) = e2πix , and for integer q define eq (x) = e(x/q).

4

B. KERR

We will adopt the following convention to avoid ambiguity in order of operations when fractional indicies occur as exponents. Given a real number q and integers α1 , . . . , αj , β1 , . . . , βj we write α1 ...αk

q α1 ...αk /β1 ...βk = q β1 ...βk . In general, if a term of the form α1 . . . αk /β1 . . . βk , occurs in the exponent of some real number, we take the order of operations as all multiplications occur before the / symbol and all divisions occur after the / symbol. 2. Main result Our main result is as follows. Theorem 1. For integer q we define  1 if q is cubefree, c = 41 otherwise. 3 For any primitive character χ mod q and integers M and N we have      4c   χ(n)  (1 + o(1)) 2 q 1/2 log q.    π M
Comparing Theorem 1 with previous results, we improve on the estimate (2). We also note the bound of Granville Soundararajan [14]   1/2 69c log q, if χ(−1) = 1, 1/2 + o(1) q 70π3  1/2 |S(χ, N )|   c + o(1) q log q, if χ(−1) = −1, π which implies (3)

|S(χ, M, N )| 





69c 1/2 + o(1)  35π3  2c + o(1) q 1/2 π

q 1/2 log q, if χ(−1) = 1, log q, if χ(−1) = −1,

and hence Theorem 1 is sharper than (3) when χ(−1) = −1. 3. Preliminary estimates for character sums The aim of this section is to obtain estimates for long character sums which will be required for the proof of Theorem 1. The following Lemma is a consequence of the work of Burgess [4, 5, 6].

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

5

Lemma 2. Let q, V and r  2 be positive integers satisfying V  q 1/2r , and suppose χ is a primitive character mod q. Then we have   q   (λ + v ) . . . (λ + v )     1 r χ   q 1/2+o(1) V 2r ,   (λ + v r+1 ) . . . (λ + v2r )  1v ,...,v V λ=1 1

2r

for r  3 and any r  2 provided q is cubefree. For a proof of the following, see [7]. Lemma 3. Let q, M, N and U be integers satisfying 2N U < q. The number of solutions to the congruence n 1 u1 ≡ n 2 u2

mod q,

with variables satisfying M < n1 , n2  M + N,

1  u1 , u2  U,

is O(N U log q). The following is a variant of the Burgess bound for twists of characters to small modulus. Lemma 4. Let q, M, N, k and r be integers satisfying N  q 1/2+1/4r . Let χ be a primitive character mod q and ψ be any multiplicative character mod k. Then we have  2 ψ(n)χ(n)  kN 1−1/r q (r+1)/4r +o(1) , M
for r  3 and any r  2 provided q is cubefree. Proof. We fix an integer r  2 and a sufficiently small ε > 0 and proceed by induction on N . We formulate our induction hypothesis as follows. For any integers M and K we have      2   ψ(n)χ(n)  ckK 1−1/r q (r+1)/4r +ε ,    M
for some constant c to be determined later which may depend on ε. Since the result is trivial for K  q 1/4 this forms the basis of our induction. Define the integers

1/2r

q N , , V = U= 1/2r 16q k

6

B. KERR

and note that UV 

N . 16k

For any integer y < N we have   ψ(n)χ(n) =

ψ(n + y)χ(n + y)

M −y
M


=



ψ(n + y)χ(n + y) +

M


−

ψ(n + y)χ(n + y)

M −y
ψ(n + y)χ(n + y),

M +N −y
and hence by our induction hypothesis   θc 2 ψ(n)χ(n) = ψ(n + y)χ(n + y) + kN 1−1/r q (r+1)/4r +ε . 2 M
where W =



 

ψ(n + kuv)χ(n + kuv).

M
Since ψ has modulus k, we have         −1 |W |  χ(nu + kv)     M
λ=1

where I(λ) counts the number of solutions to the congruence nu−1 ≡ λ

mod q,

M < n  M + N, u ∈ U .

By H¨older’s inequality 2r ⎞ ⎛ q   q 2r−2  q        |W |2r  I(λ) I(λ)2 ⎝ χ(λ + kv) ⎠ .    λ=1

λ=1

λ=1 1vV

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

7

We have q 

I(λ) = N |U|  N U,

λ=1

and by Lemma 3 q 

I(λ)2  N U log q,

λ=1

since N U  q. By Lemma 2  2r q        χ(λ + kv)    

1v1 ,...,v2r V

λ=1 1vV





  q   (λ + kv ) . . . (λ + kv )    1 r χ    (λ + kvr+1 ) . . . (λ + kv2r )  λ=1   q   (λ + v ) . . . (λ + v )    1 r χ    (λ + vr+1 ) . . . (λ + v2r ) 

1v1 ,...,v2r kV

q

1/2+o(1) 2r

λ=1

2r

k V ,

since V  q 1/2r /k, provided r  3 or r  2 and q cubefree. Combining the above estimate, we arrive at |W |2r  k 2r (N U )2r−1 q 1/2+o(1) V 2r , which after recalling the choice of U and V implies U 1−1/2r 1/4r+o(1) |W |  kN 1−1/2r q , |U|V |U| and since |U|  U q o(1) , we get |W | 2  kN 1−1/r q (r+1)/4r +o(1) , |U|V and hence |W | 2  c0 kN 1−1/r q (r+1)/4r +ε/2 , |U|V for some c0 which may depend on ε, provided q is sufficiently large. Combining the above with (4) gives      c 2 2   ψ(n)χ(n)  c0 kN 1−1/r q (r+1)/4r +ε/2 + kN 1−1/r q (r+1)/4r +ε    2 M
 ckN 1−1/r q (r+1)/4r

2 +ε

,

on taking c = 2c0 and assuming q is sufficiently large.



8

B. KERR

We may remove the upper bound N  q 1/2+1/4r occuring in Lemma 4 by partitioning summation into smaller intervals. Corollary 5. For integer q we define  1 if q is cubefree, c = 41 otherwise. 3 Let χ be a primitive character mod q and ψ be any multiplicative character mod k. For any ε > 0 there exists some δ > 0 such that if N  q c+ε , then we have



ψ(n)χ(n)  kN q −δ .

M
Proof. We note that Lemma 4 implies  (5) ψ(n)χ(n)  kN q −δ , M
if q c+ε  N  q 1/2 , which follows by taking r sufficiently large in terms of ε if r is cubefree and r = 3 otherwise. If N  q 1/2 then partition the interval (M, M +N ] into disjoint intervals I1 ∪ . . . IJ = (M, M + N ], satisfying |Ij | = q 1/2 , By (5)



1  j  J − 1,

|IJ |  q 1/2 ,

ψ(n)χ(n)  kq 1/2−δ ,

J

N . q 1/2

1  j  J − 1,

n∈Ij

and



ψ(n)χ(n)  kq 1/2−δ + q c+ε ,

n∈IJ

where the term q c+ε accounts for the possibility the interval IJ satisfies |IJ |  q c+ε .

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

9

The above estimates imply 

ψ(n)χ(n) =

M
and hence 

J  

ψ(n)χ(n)  kN q −δ + q c+ε ,

j=1 n∈IJ

ψ(n)χ(n)  kN q −δ + N q −ε

provided N  q c+2ε .

M
The result follows after replacing ε by 2ε and redefining δ.



The following is due to Montgomery and Vaughan [19]. Lemma 6. Let N be a positive integer and α a real number satisfying     α − a   1 ,  q  q2 for integers a and q satisfying (a, q) = 1. Suppose that N , R for some positive number R. Then for any multiplicative function f satisfying |f |  1 we have      N (log R)3/2 N   + f (n)e(αn)  .    log N R1/2 1nN 2Rq

The proof of the following estimate follows the proof of Hildebrand [15, Lemma 3] and is based on Lemma 4 and Lemma 6. Lemma 7. For integer q we define  1 if q is cubefree, c = 41 otherwise. 3 For any primitive character χ mod q, any ε > 0, any real number α and any integer N satisfying (6) we have

q c+ε  N  q,      N   χ(n)e(αn)  ,    log q 1nN

provided q is sufficiently large.

10

B. KERR

Proof. Let R = (log q)3 , and apply Dirichlet’s theorem to obtain integers r and k satisfying 1  k  N/R, (r, k) = 1 and  R r   (7) . α −   k kN If k  R then by Lemma 6      N (log R)3/2 N N   + , χ(n)e(αn)    1/2   log N R log q 1nN and hence we may suppose k  R. By (7) and partial summation                r rn      χ(n)e(αn)  1 + N α −  max  χ(n)e     k M N 1nM k  1nN      rn     (log q)3  χ(n)e (8) ,  k  1nM

for some M  N . If M  q c+ε/2 then we bound summation over m trivially to get      N   , χ(n)e(αn)  q c+ε/2     log q 1nN

by (6). Suppose next that M  q c+ε/2 .

(9) We have (10)



χ(n)e

 rn 

1nM

k

where

=

k  ar   e S(a), k a=1



S(a) =

χ(n).

1nM n≡a mod k

Fix some 1  a  k and consider S(a). Let d = (a, k) and write a a = , d

k =

k , d

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

so that S(a) =



χ(n) =

1nM n≡a mod k n≡0 mod d

 1 ψ(a ) φ(k  ) ψ mod k



11

ψ(n)χ(n).

1nM n≡0 mod d

If (d, q) = 1 then S(a) = 0. If (d, q) = 1 then by (9) and Corollary 5 we have         1    kN q −δ , |S(a)|  ψ(n)χ(n)  φ(k  ) ψ mod k   1nM/d for some δ > 0 depending on ε. By (8) and (10) this gives      N   , χ(n)e(αn)  k 2 (log q)3 N q −δ  (log q)9 N q −δ     log q 1nN 

which completes the proof.

The proof of the following is based on some ideas of Garaev and Karatsuba [11]. Lemma 8. For integer q we define  1 if q is cubefree, c = 41 otherwise. 3 For any primitive character χ mod q, any ε > 0 and integers M and N with N < q 1−c−ε we have  χ(n)  q 1/2 . M
Proof. Expanding into Gauss sums, we have            1     χ(n) = 1/2  χ(m)eq (mn)   q    M




0
χ(m)eq (m(M + n)),

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B. KERR

and



S2 =



χ(m)eq (m(M + n)),

0
so that

  1  χ(n)  1/2 (|S1 | + |S2 |) .  q M
    



Bounding S1 trivially gives |S1 |  q, and hence

    

(11)

  |S2 |  χ(n)  q 1/2 + 1/2 .  q M
Considering S2 , we have  S2 =

χ(m)eq (M m)

q/N <|m|(q−1)/2

(12)

eq ((N + 1)m) − eq (m) eq (m) − 1

= S2,1 − S2,2 ,

where



S2,1 =

ρ(m)χ(m)eq ((M + N + 1)m),

q/N <|m|(q−1)/2

and



S2,2 =

ρ(m)χ(m)eq ((M + 1)m),

q/N <|m|(q−1)/2

and ρ(m) is given by 1 . eq (m) − 1 Considering S2,1 , by partial summation  S2,1 = (ρ(t) − ρ(t + 1))T (t) + ρ((q − 1)/2)T ((q − 1)/2), ρ(m) =

q/N
where T (x) =



χ(m)eq ((M + N + 1)m).

q/N <|m|t

Since (13)

T ((q − 1)/2)  q 1/2 log q,

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

13

and (ρ(t) − ρ(t + 1)) 

q , t2

we have 

S2,1  q 1/2 log q + q

q/N
T (t) . t2

We note that the inequality (13) is a variant of the standard P´olyaVinogradov inequality and has a similar proof by expanding into Gauss sums. Let  χ(m)eq ((M + N + 1)m), T0 (t) = 0<|m|t

so that T0 (t) = T (t) + O

q , N

and hence S2,1  q

 q/N
q

 q/N
T0 (t) q2 1/2 + q log q + t2 N

 q/N
1 t2

T0 (t) + q. t2

Since N  q 1−c−ε we have q/N > q c+ε and hence by Lemma 7  1 q + q  q. S2,1  log q t q/N
A similar argument shows that S2,2  q, and hence by (11) and (12)      

M
   χ(n)  q 1/2 , 

which completes the proof.



4. Estimate for the 1 norm of an exponential sum In this section we estimate the 1 norm of the Fourier transform of an approximation to an interval. The following is [21, Lemma 3]

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B. KERR

Lemma 9. For any real number x and positive integer n we have n  2 | sin jx|  log n + O(1). j π j=1 Lemma 10. For integers M, N and K satisfying K  q 1−c ,

N + 2K < q,

for some 0 < c < 1 we define the function f by

M +1 x +1− K K

if

M + 1 − K  x  M + 1,

x M +N −1 +1+ K K

if

M + N − 1  x  M + N − 1 + K,

f (x) = f (x) = −

M + 1  x  M + N − 1,

if

f (x) = 1

otherwise,

f (x) = 0

and let f(a) denote the Fourier transform of f q   f (x)eq (ax). f (a) = x=1

We have

q 

|f(a)| 

a=1

4q log (q/K) + o(q log q). π2

Proof. For a ≡ 0 mod q we have f(a) = S1 + S2 + S3 , where



S1 =

eq (ax),

M +1xM +N −1

S2 = and S3 =



x−M −1+K eq (ax), K M +1−KxM +1 

M +N −1xM +N −1+K

M +N −1+K −x eq (ax). K

We have S1 = eq ((M + 1)a)

1 − eq (a(N − 1)) , 1 − eq (a)

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

S2 = eq ((M + 1)a) =−

15

1  (K − x)eq (−ax) K 0xK

eq ((M + 2)a) eq (M a)(1 − eq (−Ka)) − , 1 − eq (a) K(1 − eq (−a))2

and S3 = eq ((M + N − 1)a)

1  (K − x)eq (ax) K 0xK

eq ((M + N − 1)a) eq ((M + N )a)(1 − eq (Ka)) − . 1 − eq (a) K(1 − eq (a))2 This implies that (1 − eq ((N + K − 2)a))(1 − eq (Ka)) f(a) = eq ((M − K + 2)a) + O(1), K(1 − eq (a))2 and hence | sin (π(N + K − 2)a/q)|| sin (πKa/q)| + O(1). |f(a)|  K| sin(πa/q)|2 Summing over a = 1, . . . , q gives (14) q   | sin (π(N + K − 2)a/q)|| sin (πKa/q)| |f(a)|  2 + O(q) K| sin(πa/q)|2 y=1 =

1a(q−1)/2

(15)

= 2T1 + 2T2 + O(q),

where



T1 =

1aq/(K log1/4 q)

and

| sin (π(N + K − 2)a/q)|| sin (πKa/q)| , K| sin(πa/q)|2



T2 =

q/(K log1/4 q)a(q−1)/2

| sin (π(N + K − 2)a/q)|| sin (πKa/q)| . K| sin(πa/q)|2

We have q T1 = π



 1aq/(K log1/4 q)

| sin (πKa/q)| πKa/q



πa/q | sin(πa/q)|

| sin (π(N + K − 2)a/q)| . a Since K  q 1−c , for any a satisfying ×

1  a  q/(K log1/4 q),

2

16

B. KERR

we have



 2 | sin (πKa/q)| πa/q = 1 + o(1), πKa/q | sin(πa/q)| which follows from the approximation sin x = 1 + O(x), as x → 0. x This implies  q | sin (π(N + K − 2)a/q)| , T1  (1 + o(1)) π a 1/4 1aq/(K log

q)

hence by Lemma 9 T1  (1 + o(1)) Considering T2 , we have  q2 T2  K 1/4 q/(K log

q)a(q−1)/2

2q log (q/K). π2

1  q log1/4 q = o(q log q). 2 a

Combining the above with (14) gives q  4q |f(a)|  (1 + o(1)) 2 log (q/K) + o(q log q) π y=1 4q log (q/K) + o(q log q), π2 and completes the proof. 

5. Proof of Theorem 1 Considering the sum (16)



S=

χ(n),

M
since



χ(n) = 0,

M
by modifying M if necessary we may assume that q (17) N< . 2 Define c by  1 if q is cubefree, c = 41 otherwise, 3



´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

17

and for a sufficiently small ε we let K = q 1−c−ε .

(18) Define the function f by

f (x) = 1 if M + 1  x  M + N − 1,

f (x) =

f (x) = −

M +1 x +1− K K

M +N −1 x +1+ K K

if M + 1 − K  x  M + 1,

if M + N − 1  x  M + N − 1 + K,

f (x) = 0 otherwise. Considering (16), we have S=



f (n)χ(n) −

n



 M +1−KnM +1





−

M +N −1nM +N −1+K

M +1 n +1− K K



n M +N −1 − +1+ K K

χ(n)  χ(n).

By partial summation and Lemma 8    n M +1 χ(n)  q 1/2 , +1− K K M +1−KnM +1 and



 M +N −1nM +N −1+K

so that S=

n M +N −1 − +1+ K K



 χ(n)  q 1/2 ,

f (n)χ(n) + O(q 1/2 ).

n

Hence it is sufficient to show    q    4c   (19) f (n)χ(n)  + o(1) q 1/2 log q.    n=1 π2

18

B. KERR

Expanding f into a Fourier series and using Lemma 10, we get  q   q  q   1       f (n)χ(n)  |f(a)|  χ(n)eq (−an)    q   n=1

a=1

n=1

4q  (1 + o(1)) 2 log (q/K)q 1/2 π 4c = (1 + o(1)) 2 q 1/2 log q, π and completes the proof. References [1] J. W. Bober and L. Goldmakher, The distribution of the maximum of character sums, Mathematika, 59 (2013), 427–442. [2] J. W. Bober and L. Goldmakher, P´ olya-Vinogradov and the least quadratic nonresidue, Math. Ann. 366 (2016), 853–863. [3] J. W. Bober, L. Goldmakher, A. Granville and D. Koukoulopoulos, The frequency and the structure of large character sums, J. Eur. Math. Soc. (to appear). [4] D. A. Burgess, On character sums and L-series, Proc. London Math. Soc. 12 (3) (1962), 193–206. [5] D. A. Burgess, On character sums and L-series II, Proc. London Math. Soc. 13 (3) (1963), 524–536. [6] D. A. Burgess, The character sum estimate with r=3, J. London Math. Soc. 33 (2) (1986), 524–536. [7] J. Friedlander and H. Iwaniec, Estimates for character sums, Proc. Amer. Math. Soc. 119 (2) (1993), 265–372. [8] D. A. Frolenkov, A numerically explicit version of the P´ olya-Vinogradov inequality, Mosc. J. Comb. Number Theory 1 (3) (2011), 25–41. [9] D. A. Frolenkov and K. Soundararajan, A generalization of the P´ olyaVinogradov inequality, Ramanujan J. 31 (3) (2013), 271–279. [10] E. Fromm and L. Goldmakher, Improving the Burgess bound via P´ olyaVinogradov, arXiv:1706.03002. [11] M. Z. Garaev and A. A. Karatsuba, On character sums and the exceptional set of a congruence problem, J. Number Theory 114 (2005), 182–192. [12] L. Goldmakher, Multiplicative mimicry and improvements of the P´ olyaVinogradov inequality, Algebra Number Theory 6 (1) (2012), 123–163. [13] A. Granville and K. Soundararajan, Large Character Sums, J. Amer. Math. Soc. 14 (2001), 365–397. [14] A. Granville and K. Soundararajan, Large character sums: pretentious characters and the P´ olya-Vinogradov theorem, J. Amer. Math. Soc. 20 (2) (2007), 357–384. [15] A. Hildebrand, On the constant in the P´ olya-Vinogradov inequality, Canad. Math. Bull. 31 (1988), 347–352. [16] A. Hildebrand, Large values of character sums, J. Number Theory 29 (1988), 271–296.

´ ON THE CONSTANT IN THE POLYA-VINOGRADOV INEQUALITY

19

[17] E. Landau, Absch¨ atzungen von Charaktersummen, Einheiten und Klassenzahlen, Nachrichten K¨onigl. Ges. Wiss G¨ ottingen (1918), 79–97. [18] Y. Lamzouri and A. P. Mangerel, Large odd order character sums and improvements of the P´ olya-Vinogradov inequality, arXiv:1701.01042. [19] H. L. Montgomery and R. C. Vaughan, Exponential sums with multiplicative coefficients, Invent. Math. 43 (1977), 69–82. [20] R. E. A. C. Paley, A theorem on characters, J. London Math. Soc. 7 (1932), 28–32. [21] C. Pomerance, Remarks on the P´ olya-Vinogradov inequality, Integers, 11A, (2011). School of Physical, Environmental and Mathematical Sciences, The University of New South Wales Canberra, Australia Email address: [email protected]