On sums of hyper-Kloosterman sums

A.1 Simple global bounds on the direct functions

In this section, we prove Lemma 6.1.

For x > 0 and t ∈ ℝ , [19, 7.13.2 (17)] gives

where

satisfies a ~ j ⁢ ( 1 + x 2 / t 2 ) ≪ j 1 . In particular,

and for | t | ≪ x 1 / 2 + ϵ we have

so

The J-Bessel function at real order is related to the Hankel functions via J σ ⁢ ( x ) = Re ⁡ ( H σ ( 1 ) ⁢ ( x ) ) , so for x > σ > 0 , [19, 7.13.2 (11)] gives

where

In particular, for 0 < d ≪ x 1 / 2 + ϵ ,

For the first Hankel function, the asymptotics [19, 7.13.2 (13), (15), (22)] imply

From (A.5) and the power series [22, 8.402], we see that

holds on the boundary of the sector

uniformly on x > 0 . For a fixed x, the bound

holds inside the sector (by the power series expansion), so by Phragmén–Lindelöf for a sector, we have (6.1). Also, using [22, 8.471.1], for - 1 < Re ⁡ ( ν ) < 0 , we have

and this implies (6.2).

We may express the Y-Bessel function in terms of the Hankel functions as i ⁢ Y i ⁢ t ⁢ ( x ) = H i ⁢ t ( 1 ) ⁢ ( x ) - J i ⁢ t ⁢ ( x ) , so (A.5), the bounds on J i ⁢ t ⁢ ( x ) , and the above arguments give (6.3). Note that on 0 < x ≤ 1 , the Y-Bessel function displays an additional logarithmic dependence on x, so we simply avoid that here.

A.2 Bounds on the order derivative of the J-Bessel function

In this section, we prove Lemma 6.2. First, let us recall some well-known asymptotics for the Y-Bessel functions: From [30, 10.8.1, 10.8.2, 10.17.4], some simple, global bounds for Y 0 and Y 1 are

and more precisely, we have [30, 10.17.4]

As in the previous section, we prove the bound for ν ≥ 0 ; then for ν purely imaginary and Phragmén–Lindelöf implies the lemma. More precisely, we first show the bound for x < | t | 1 / 3 and then for | t | ≤ 2 to avoid some technical complications in the difficult case of ν purely imaginary.

When x < | t | 1 / 3 , we can simply use the power series expansion for (the ν-derivative of) J ν ⁢ ( x ) ; one can easily derive the bound

in this case. When | t | ≤ 2 , we apply a formula of Dunster [18, (2.1), (2.8)]:

for x ≠ 0 , | arg ⁡ ( x ) | < π , provided the unbounded parts of the contours lie on the positive real axis and the contours avoid ( - ∞ , 0 ] . This formula is effective in the case | Im ⁡ ( ν ) | ≪ 1 , but the opposite case requires the recovery of exponential factors resulting from the products of three Bessel functions. Again, a simple application of the J- and Y-Bessel bounds gives the lemma in this case, with the integrals adding at most a logarithmic factor in case x is near zero.

Now we handle the case where ν is purely imaginary. In [1, (42)], we have

This integral representation does not apply at Re ⁡ ( ν ) = 0 , but if we take 0 < η < x and N ≥ 0 , we can write this as

Then integration by parts gives

and this has an analytic continuation to Re ⁡ ( ν ) = 0 .

Now suppose x ≥ 4 ⁢ t 1 / 3 , t ≥ 3 and ν = i ⁢ t , and in place of the sharp cut-off we apply a smooth, dyadic partition of unity as y or x - y becomes small, stopping when y < t 1 / 3 or x - y < x / t (more precisely, we apply the partition, which depends on t, then analytically continue to ν = i ⁢ t ), so that

where each f i is smooth, f i ( n ) ⁢ ( y ) ≪ n 1 and, say, f 0 is supported on [ - 1 , 1 ] (in constructing the partition, f 0 will be nonzero at 0) while f 1 is supported on [ 1 2 , 3 2 ] ; the C sums are dyadic.

For the first term, we use (A.6) so that

For the fourth term, note that for 0 < z < 2 ⁢ t 1 / 3 we have

so we take N = 2 , giving

For the fifth term, we apply (A.6) and (A.7) so that

Repeated integration by parts implies

and we have

For the second and third terms, we apply the asymptotic expansion of the J-Bessel function (A.1):

using

So we may take M a = M b = 2 , but actually the trivial bound on j ≥ 1 is sufficient for

Assume for the moment that C ≥ t ϵ (this is only new for the T 1 ⁢ b integral). Then applying the asymptotic expansion (A.8) of Y 0 ⁢ ( y ) yields

We have

So by Lemma 5.1 with T = 1 , U = C , Q = C + t , and Y = t 2 ⁢ ( C + t ) C 2 , the integrals are O ⁢ ( t - 1 ⁢ ( x 2 + t 2 ) - 1 / 4 ) (we have an x - 1 / 2 in both cases, just need to recover the t 3 / 2 , hence just t ϵ ) unless

for some y ∈ [ C / 2 , 3 ⁢ C / 2 ] . This is impossible unless t ≪ C , in which case we must have

so that x ≫ C ≫ t 2 - ϵ . The trivial bound in that case becomes

For T 1 ⁢ b , in case x / t ≤ C < t ϵ , we perform a single integration by parts on the integral

Trivially bounding the result gives

in this case as well.