# Small fractional parts of binary forms

Kiseok Yeon<EMAIL_ADDRESS>Department of Mathematics, Purdue University,
150 N. University Street, West Lafayette, IN 47907-2067, USA

###### Abstract.

We obtain bounds on fractional parts of binary forms of the shape

$\Psi(x,y)=\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}$

with $\alpha_{k},\alpha_{l},\ldots,\alpha_{0}\in{\mathbb{R}}$ and $l\leq k-2.$
By exploiting recent progress on Vinogradov’s mean value theorem and earlier
work on exponential sums over smooth numbers, we derive estimates superior to
those obtained hitherto for the best exponent $\sigma$, depending on $k$ and
$l,$ such that

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\Psi(x,y)\|\leq X^{-\sigma+\epsilon}.$

###### Key words and phrases:

Exponential sums, Diophantine approximation

###### 2020 Mathematics Subject Classification:

11J54,11L03

##  1\. Introduction

In 1914, Hardy and Littlewood [References, p 172] posed the question as to
whether, when $\alpha\in{\mathbb{R}}$, $k\in{\mathbb{N}}$ and $\epsilon>0$ is
any positive number, there exists $\sigma>0$ not depending on $\alpha$ such
that

$\min_{1\leq x\leq X}\|\alpha x^{k}\|\leq X^{-\sigma+\epsilon},$ (1.1)

provided that $X$ is sufficiently large in terms of $k$ and $\epsilon$. Here,
$\|\cdot\|$ denotes the distance to the nearest integer. Vinogradov
[References] gave a positive answer to this question by providing a specific
exponent $\sigma=k/(2^{k-1}k+1)$. This was quantitatively improved by
Heilbronn [References], who replaced the exponent $\sigma=2/5$ by $1/2$ in the
case $k=2$. Danicic [References] extended this conclusion to
$\sigma=1/2^{k-1}$ for all $k\in{\mathbb{N}}$. Subsequently, the exponent
$1/2$ in the case $k=2$ was improved to $\sigma=4/7$ by Zaharescu
[References]. Sharper results for $k\geq 6$ depend on recent progress
Vinogradov’s mean value theorem, and thus Baker $[\ref{ref3}]$ proved that the
exponent $\sigma$ can be replaced by $\sigma=1/(k(k-1))$. Furthermore, Wooley
[References] showed that the exponent $\sigma$ can be improved to
$\sigma=1/(k\log k+O(k\log\log k))$. However, it is conjectured in
[References] that the exponent $\sigma$ should be improvable to $1.$
Unfortunately, current techniques do not seem capable of achieving this
conjecture.

Turning our attention to problems in many variables, one is led to ask
analogous questions regarding the distribution mod $1$ of polynomials
$f(x_{1},\ldots,x_{s})\in{\mathbb{R}}[x_{1},\ldots,x_{s}]$. In [References,
Theorem 10.2], Baker proved that when
$\alpha_{1},\ldots,\alpha_{s}\in{\mathbb{R}}$ and $k\in{\mathbb{N}}$, there
exists $s_{0}$ such that whenever $s\geq s_{0}$, one has

$\min_{\begin{subarray}{c}0\leq\boldsymbol{x}\leq X\\\
\boldsymbol{x}\neq\boldsymbol{0}\end{subarray}}\|\alpha_{1}x_{1}^{k}+\cdots+\alpha_{s}x_{s}^{k}\|\leq
X^{-k+\epsilon}.$

As a quantitative generalization, Baker [References,References] provided an
explicit exponent $\sigma(s,k)$ such that

$\min_{\begin{subarray}{c}0\leq\boldsymbol{x}\leq X\\\
\boldsymbol{x}\neq\boldsymbol{0}\end{subarray}}\|\alpha_{1}x_{1}^{k}+\cdots+\alpha_{s}x_{s}^{k}\|\leq
X^{-\sigma(s,k)+\epsilon}.$

For polynomials more general than additive forms, Schmidt [References] made
progress on the analogous problem regarding forms of odd degree $k$.
Specifically, for given $E>0$, there exist $s_{0}=s_{0}(E,k)$ such that
whenever $s>s_{0}$, one has

$\min_{\begin{subarray}{c}0\leq|\boldsymbol{x}|\leq X\\\
\boldsymbol{x}\neq\boldsymbol{0}\end{subarray}}\|f(\boldsymbol{x})\|<X^{-E}.$
(1.2)

In the special case of cubic form $f(\underline{\mathbf{x}})$, Dietmann
[References] provided explicit formula $\sigma(s)$ such that

$\min_{\begin{subarray}{c}0\leq|\boldsymbol{x}|\leq X\\\
\boldsymbol{x}\neq\boldsymbol{0}\end{subarray}}\|f(\boldsymbol{x})\|<X^{-\sigma(s)}.$

In this paper, we seek to make progress on making bounds of the shape
$(\ref{2'})$ explicit, concentrating for the present on the situation with
$s=2.$ Here, one may write
$f(x_{1},x_{2})=\alpha_{k}x_{1}^{k}+\alpha_{k-1}x_{1}^{k-1}x_{2}+\cdots+\alpha_{0}x_{2}^{k}$
with $\alpha_{k},\ldots,\alpha_{0}\in{\mathbb{R}}$. Hitherto, the only general
strategy available has been a trivial one that reduces this problem in two
variables, by specializing $x_{2}=0$ for example, to a problem in one
variable. In this direction, Wooley [References, Theorem 4] obtained a
nontrivial exponent $\sigma$ by restricting the polynomial to have the shape
$\alpha\Phi(x_{1},x_{2})$ where
$\Phi(x_{1},x_{2})\in{\mathbb{Z}}[x_{1},x_{2}]$ is a binary form of degree $k$
exceeding $1$. The main difficulties of the problem regarding general binary
forms in ${\mathbb{R}}[x_{1},x_{2}]$ arise from our relatively poor
understanding concerning how the monomials $\alpha_{l}x_{1}^{l}x_{2}^{k-l}$,
with $0\leq l\leq k$, combine to influence the fractional parts of
$f(x_{1},x_{2})$. Our goal in this paper is to explore approaches to
overcoming these difficulties and go beyond the trivial approach hitherto
applied, concentrating on the class of binary forms. Our hope is that the
arguments described here may be useful in studying fractional parts of
polynomials in many variables in general.

Our first theorem provides bounds for binary forms of degree $k$ useful for
small $k$.

######  Theorem 1.1.

Let $k$ and $l$ be natural numbers satisfying $1\leq l\leq k-2,$ and consider
real numbers $\alpha_{k}$, $\alpha_{j}$ ($1\leq j\leq l$). Then, for any
$\epsilon>0$, there exists a real number $X(k,\epsilon)$ such that whenever
$X\geq X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|<X^{-\sigma+\epsilon},$
(1.3)

where

$\sigma=\frac{l+2}{l+1}2^{1-k}.$

For comparison with earlier results, by incorporating the conclusion of
Danicic [References] with the trivial strategy mentioned above, one obtains

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\leq\min_{1\leq
x\leq X}\|\alpha_{k}x^{k}\|\leq X^{-2^{1-k}+\epsilon}$

The conclusion of Theorem 1.1 improves on the exponent $2^{1-k}$ here by a
factor which may be as large as $\frac{3}{2}$ in the case $l=1.$

######  Corollary 1.2.

Let $k$ be a natural number, and consider real numbers $\alpha_{k},\alpha_{j}\
(0\leq j\leq k-2)$. Let
$\varphi(x,y)=c_{k}x^{k}+c_{k-1}x^{k-1}y+\cdots+c_{1}xy^{k-1}+c_{0}y^{k}$ with
$c_{j}\in{\mathbb{Z}}.$ Then, for any $\epsilon>0$, there exist a real number
$X(\epsilon,c_{k},c_{k-1})$ such that whenever $X\geq
X(\epsilon,c_{k},c_{k-1})$ one has

$\min_{\begin{subarray}{c}0\leq|x|,|y|\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}\varphi(x,y)+\alpha_{k-2}x^{k-2}y^{2}+\cdots+\alpha_{0}y^{k}\|<X^{-\sigma+\epsilon},$

where $\sigma=(k/(k-1))2^{1-k}.$

In [References], Dietmann recorded that for sufficiently large $X,$ one has

$\min_{\begin{subarray}{c}0\leq|x|,|y|\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{3}+\beta x^{2}y+\gamma xy^{2}+\delta
y^{3}\|<X^{-1/4+\epsilon},$

which is obtained by specializing $y=0$ and applying the earlier work of
Danicic $[\ref{ref6}]$. In Corollary 1.2 with $k=3$, we obtain the superior
exponent $\sigma=3/8$ by imposing the condition that $\alpha$ and $\beta$ have
a rational ratio. By further restricting $\alpha,\beta,\gamma,\delta$ to have
rational ratio to each other, Wooley [References, Theorem 4] proved that this
can be improved to $\sigma=1/2.$

Our next two theorems provide bounds useful when $k$ is larger by exploiting
recent progress on Vinogradov’s mean value theorem and earlier work applying
smooth numbers. The statements of these results is facilitated by writing

$\rho(k,l)=2^{2}(k-1)+2^{3}(k-2)+\cdots+2^{l+1}(k-l),$

and

$\displaystyle l_{0}=\max\\{l\in{\mathbb{N}}|\ 7\rho(k,l)\leq k(k-1)\\}$ (1.4)
$\displaystyle l_{1}=\max\\{l\in{\mathbb{N}}|\ 7\rho(k,l)\leq k\log k\\},$
(1.5)

where the number 7 here may be replaced by smaller positive numbers.

######  Theorem 1.3.

Let $k$ and $l$ be natural numbers satisfying $1\leq l\leq l_{0}$ and $k\geq
3$, and consider real numbers $\alpha_{k}$ and $\alpha_{j}$ $(1\leq j\leq l)$.
Then, for any $\epsilon>0$, there exists a real number $X(k,\epsilon)$ such
that whenever $X\geq X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|<X^{-\sigma+\epsilon},$
(1.6)

where

$\sigma=\frac{2}{k(k-1)+\rho(k,l)}$

For comparison with earlier results, by combining the earlier work of Baker
[References, Theorem 3] and the trivial treatment mentioned above, one obtains

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|<\min_{\begin{subarray}{c}1\leq
x\leq X\end{subarray}}\|\alpha x^{k}\|\leq X^{-\sigma+\epsilon},$

where $\sigma=1/(k(k-1)).$ Notice that for all $l$ with $1\leq l\leq l_{0},$
the conclusion of Theorem 1.3 is superior to this earlier result stemming from
Baker [References, Theorem 3]. For instance, in the case $l=1$, the earlier
result, which is $\sigma=1/(k(k-1))$, can be improved to

$\sigma=\frac{2}{(k+2)(k-1)}.$

######  Theorem 1.4.

Let $k$ and $l$ be natural numbers satisfying $1\leq l\leq l_{1}$ and $k\geq
3$, and consider real numbers $\alpha_{k}$ and $\alpha_{j}$ $(1\leq j\leq l)$.
Then, for any $\epsilon>0$, there exist a real number $X(k,\epsilon)$ such
that whenever $X\geq X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|<X^{-\sigma+\epsilon},$
(1.7)

where

$\sigma=\frac{2}{k\log k+\rho(k,l)+Ck\log\log k}$

in which $C$ is a positive constant which may not depend on $k$.

For comparison with earlier results, by combining the earlier work of Wooley
$[\ref{ref18}$, Theorem 1.2] and the trivial method mentioned above, one
obtains

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|<\min_{\begin{subarray}{c}1\leq
x\leq X\end{subarray}}\|\alpha x^{k}\|\leq X^{-\sigma+\epsilon},$

where $\sigma=1/(k\log k+Ck\log\log k)$ in which $C$ is a positive constant
which may not depend on $k$. Notice that for all $l$ with $1\leq l\leq l_{1},$
the conclusion of Theorem 1.4 is superior to this earlier result stemming from
Wooley [References, Theorem 1.2].

The method underlying the proof of our main theorems is to exploit inductive
arguments based on a classical argument widely used in studying fractional
parts of polynomials. However, it is complicated by the need to obtain
quantitative inequalities useful for each of the inductive steps. Thus, these
and other technical complications may obscure the key ideas of our arguments,
and so we sketch this inductive argument in Section 2. Then, in Section 3, we
provide key lemmas about the quantitative inequalities required for the
success of our arguments. In Sections 4, 5 and 6, by exploiting these key
lemmas and the inductive arguments, we provide the proof of Theorem 1.1, 1.3
and 1.4, respectively. Throughout this paper, we use $\gg$ and $\ll$ to denote
Vinogradov’s well-known notation, and write $e(z)$ for $e^{2\pi iz}$. We adopt
the convention that whenever $\epsilon$ and $\delta$ appear in a statement,
then the statement holds for each $\epsilon,\delta>0$, with implicit constants
depending on $\epsilon$ and $\delta$, respectively.

## Acknowledgement

The author acknowledges support from NSF grant DMS-2001549 under the
supervision of Trevor Wooley. The author is grateful for support from Purdue
University. The author would like to thank Trevor Wooley for careful reading
and helpful comments.

##  2\. Outline of the proof

In this paper, we mainly follow a classical argument relating fractional parts
of polynomials to lower bounds of associated exponential sums [References,
Theorem 2.2]. We record here this theorem.

######  Lemma 2.1.

Let $x_{1},\ldots,x_{N}$ be real numbers. Suppose that $\|x_{n}\|\geq H^{-1}$
for every $n$ with $1\leq n\leq X$. Then,

$\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{n=1}^{N}e(hx_{n})\biggl{|}\geq N/6$

###### Proof.

See [References, Theorem 2.2]. ∎

Suppose that for all $x,y$ with $x,y\in[1,X]\cap{\mathbb{Z}}$ one has

$\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|>1/H.$
(2.1)

Then, by Lemma 2.1, one obtains

$\displaystyle\sum_{1\leq h\leq H}|\displaystyle\sum_{1\leq x,y\leq
X}e(h(\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}))|\gg
X^{2}.$ (2.2)

Our first goal is to obtain information about Diophantine approximations to
$\alpha_{l}$ from the lower bound (2.2). In Section 3, we shall obtain upper
bounds for the exponential sum on the left hand side of (2.2) in terms of
$\alpha_{k}$ and $\alpha_{l}$. Combining these upper bound with the lower
bound (2.2), one obtains the information concerning $\alpha_{l}.$

The next goal is to reduce the problem of bounding the fractional parts on the
left hand side of (2.1) to the corresponding diagonal problem through
inductive arguments. Note that by the triangle inequality, one has

$\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\\\
\leq\|\alpha_{k}x^{k}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|+\|\alpha_{l}x^{l}y^{k-l}\|.$
(2.3)

By putting $y=qy^{\prime}$ where $q$ is the denominator associated with a
rational approximation to $\alpha_{l},$ and again applying the triangle
inequality, we see that the right hand side in (2.3) is bounded above by

$\|\alpha_{k}x^{k}+\alpha_{l-1}x^{l-1}(qy^{\prime})^{k-l+1}+\cdots+\alpha_{0}(qy^{\prime})^{k}\|+x^{l}(qy^{\prime})^{k-l-1}y^{\prime}\|q\alpha_{l}\|.$
(2.4)

From the information about $\alpha_{l}$ obtained in the first step, one finds
that $\|q\alpha_{l}\|$ is small enough so that the second term in $(\ref{9})$
is $\leq 1/(2H)$. Thus, the inequality (2.1) implies that for all
$x,y^{\prime}$ one has

$\|\alpha_{k}x^{k}+\alpha_{l-1}x^{l-1}(qy^{\prime})^{k-l+1}+\cdots+\alpha_{0}(qy^{\prime})^{k}\|\gg
1/H.$

By applying this argument inductively, we ultimately infer from (2.1) that for
all $x$ and $y_{0}$ one has

$\|\alpha_{k}x^{k}+\alpha_{0}^{\prime}y_{0}^{k}\|\gg 1/H$

for all $x$ and $y_{0}.$ Experts will recognize that the diagonal polynomial
here is accessible to classical methods.

##  3\. Exponential sums in two variables

Let $A$ and $B$ be subsets of $[1,X]\cap{\mathbb{Z}}.$ Suppose that

$\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\geq
H^{-1}$

for all $x\in A,\ y\in B.$ Then, by Lemma 2.1, one finds that

$\displaystyle\sum_{h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}x\in A\\\ y\in
B\end{subarray}}e(h(\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}))\biggr{|}\gg|A||B|,$
(3.1)

where $|A|$ and $|B|$ are cardinalities of the sets $A$ and $B$, respectively.

Our goal in this section is to obtain upper bounds for the exponential sum
above in terms of $\alpha_{k}$ and $\alpha_{l}$. To obtain these upper bounds,
we introduce a variant of conventional Weyl differencing applicable to
exponential sums in many variables, and exploit an argument used in earlier
works on exponential sums in many variables (See, for example, [References,
Lemma 4.3]). To facilitate concision throughout this section, we define
$g(x,y)=g(x,y;\boldsymbol{\alpha})\in{\mathbb{R}}[x,y]$ by

$g(x,y)=\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}$

with $l\leq k-2.$ Furthermore, we define differencing operators
$\Delta_{1}^{x}$ and $\Delta_{1}^{y}$ by

$\Delta_{1}^{x}(g(x,y);h)=g(x+h,y)-g(x,y)$
$\Delta_{1}^{y}(g(x,y);h)=g(x,y+h)-g(x,y),$

and so we define $\Delta_{j}^{x}$ and $\Delta_{j}^{y}$ for $j\geq 2$
recursively by means of the relations

$\Delta_{j}^{x}(g(x,y);\mathbf{h})=\Delta_{j}^{x}(g(x,y);h_{1},...,h_{j})=\Delta_{1}^{x}(\Delta_{j-1}^{x}(g(x,y);h_{1},...,h_{j-1});h_{j})$
$\Delta_{j}^{y}(g(x,y);\mathbf{h})=\Delta_{j}^{y}(g(x,y);h_{1},...,h_{j})=\Delta_{1}^{y}(\Delta_{j-1}^{y}(g(x,y);h_{1},...,h_{j-1});h_{j}).$

The following lemma provides an upper bound for the exponential sum in (3.1)
with $A=[1,X]\cap{\mathbb{Z}}$ and $B=[1,Y]\cap{\mathbb{Z}}$. This upper bound
is useful for small $k.$ In advance of the statement of this lemma, we define
the exponential sum $T(\boldsymbol{\alpha})=T(\boldsymbol{\alpha};H,X,Y)$ by

$T(\boldsymbol{\alpha})=\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{1\leq x\leq X}\displaystyle\sum_{1\leq y\leq
Y}e(hg(x,y))\biggr{|}.$

######  Lemma 3.1.

Let $k$ and $l$ be natural numbers satisfying $1\leq l\leq k-2$ and $k\geq 3,$
and consider real numbers $\alpha_{k}$ and $\alpha_{j}\ (0\leq j\leq l)$.
Suppose that $\alpha_{k}$ and $\alpha_{l}$ have rational approximations
$|\alpha_{k}-a_{1}/q_{1}|\leq q_{1}^{-2}$ and
$|\alpha_{l}-a_{2}/q_{2}|<q_{2}^{-2}$ with $(q_{1},a_{1})=(q_{2},a_{2})=1,$
respectively. Suppose that $X$ and $H$ are positive real numbers sufficiently
large in terms of $k$. Then, for each $\epsilon>0$, one has

$T(\boldsymbol{\alpha})\ll
H(XY)^{1+\epsilon}\left(\frac{1}{q_{1}}+\frac{1}{X}+\frac{q_{1}}{HX^{k}}\right)^{2^{1-k}}\left(\frac{1}{q_{2}}+\frac{1}{Y}+\frac{q_{2}}{HX^{l}Y^{k-l}}\right)^{2^{1-k}}$
(3.2)

###### Proof.

By Hölder’s inequality, one obtains

$T(\boldsymbol{\alpha})\ll\biggl{(}\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{1\leq y\leq
Y}1\biggr{)}^{1-2^{-l}}\biggl{(}\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{1\leq y\leq Y}\biggl{|}\displaystyle\sum_{1\leq x\leq
X}e(hg(x,y))\biggr{|}^{2^{l}}\biggr{)}^{2^{-l}}.$

Then by Weyl differencing (see, for example, the proof of Vaughan [References,
Lemma 2.4]), we obtain

$\displaystyle
T(\boldsymbol{\alpha})\ll(HY)^{1-2^{-l}}\biggl{(}X^{2^{l}-(l+1)}\sum_{1\leq
h\leq H}\sum_{1\leq y\leq Y}\sum_{|h_{1}|\leq X}\cdots\sum_{|h_{l}|}\sum_{x\in
I_{l}{(\boldsymbol{h}})}e(h\Delta_{l}^{x}(g(x,y),\mathbf{h}))\biggr{)}^{2^{-l}}$

where $I_{l}({\boldsymbol{h}})$ is an interval of integers contained in
$[-X,X].$ Note here that

$\Delta_{l}^{x}(g(x,y);\mathbf{h})=\Delta_{l}^{x}(\alpha_{k}x^{k})+\Delta_{l}^{x}(\alpha_{l}x^{l}y^{k-l})$

and that $\Delta_{l}^{x}(\alpha_{l}x^{l}y^{k-l})$ does not depend on $x.$
Hence, one obtains

$\displaystyle
T(\boldsymbol{\alpha})\ll(HY)^{1-2^{-l}}X^{1-(l+1)2^{-l}}V(\alpha_{k},\alpha_{l})^{2^{-l}},$
(3.3)

where

$V(\alpha_{k},\alpha_{l})=\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{|h_{1}|\leq X}\cdots\displaystyle\sum_{|h_{l}|\leq
X}|V_{1}(\alpha_{k};h,\mathbf{h})||V_{2}(\alpha_{l};h,\mathbf{h})|$

in which

$V_{1}(\alpha_{k};h,\mathbf{h})=\displaystyle\sum_{x\in
I_{l}}e(h\Delta_{l}^{x}(\alpha_{k}x^{k},\mathbf{h}))$

and

$V_{2}(\alpha_{l};h,\mathbf{h})=\displaystyle\sum_{1\leq y\leq
Y}e(h\Delta_{l}^{x}(\alpha_{l}x^{l}y^{k-l},\mathbf{h})).$

By applying Hölder’s inequality again,

$\displaystyle V(\alpha_{k},\alpha_{l})\ll\biggl{(}\sum_{1\leq h\leq
H}\sum_{|h_{1}|\leq X}\cdots\sum_{|h_{l}|\leq
X}1\biggr{)}^{1-2^{2-k+l}}W_{1}(\alpha_{k})^{2^{1-k+l}}W_{2}(\alpha_{l})^{2^{1-k+l}}$
(3.4)

where

$W_{1}(\alpha_{k})=\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{|h_{1}|\leq X}\cdots\displaystyle\sum_{|h_{l}|\leq
X}|V_{1}(\alpha_{k},h,\mathbf{h})|^{2^{k-l-1}},$
$W_{2}(\alpha_{l})=\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{|h_{1}|\leq X}\cdots\displaystyle\sum_{|h_{l}|\leq
X}|V_{2}(\alpha_{l};h,\mathbf{h})|^{2^{k-l-1}}.$

By applying classical Weyl differencing arguments, one has

$\displaystyle W_{1}(\alpha_{k})\ll X^{2^{k-l-1}-(k-l)}\sum_{1\leq h\leq
H}\sum_{|h_{1}|\leq X}\cdots\sum_{|h_{k-1}|\leq X}\biggl{|}\sum_{x\in
I_{k-1}(\boldsymbol{h})}e(c_{k}hh_{1}\cdots h_{k-1}x)\biggr{|}$

where $c_{k}$ is a constant depending on $k,$ and $I_{k-1}(\boldsymbol{h})$ is
an interval of integers contained in $[-X,X]$. Thus, for each $\epsilon>0,$
one has

$\displaystyle W_{1}(\alpha_{k})\ll
HX^{2^{k-l-1}+l+\epsilon}\left(\frac{1}{q_{1}}+\frac{1}{X}+\frac{q_{1}}{HX^{k}}\right).$

Furthermore, one has

$\displaystyle W_{2}(\alpha_{l})\ll Y^{2^{k-l-1}-(k-l)}\sum_{1\leq h\leq
H}\sum_{|h_{1}|\leq X}\cdots\sum_{|h_{l}|\leq X}\sum_{|h_{l+1}|\leq
Y}\cdots\sum_{|h_{k-1}|\leq Y}\biggl{|}\sum_{y\in
J_{k-1}(\boldsymbol{h})}e(d_{k}hh_{1}\cdots h_{k-1}y)\biggr{|}$

where $d_{k}$ is a constant depending on $k,$ and $J_{k-1}(\boldsymbol{h})$ is
an interval of integers contained in $[-Y,Y].$ Hence, for each $\epsilon,$ one
has

$W_{2}(\alpha_{l})\ll
HX^{l}Y^{2^{k-l-1}+\epsilon}\left(\frac{1}{q_{2}}+\frac{1}{Y}+\frac{q_{2}}{HX^{l}Y^{k-l}}\right).$

Substituting these estimates for $W_{1}(\alpha_{k})$ and $W_{2}(\alpha_{l})$
into (3.4) and from there into (3.3), one concludes that

$T(\boldsymbol{\alpha})\ll
H(XY)^{1+\epsilon}\left(\frac{1}{q_{1}}+\frac{1}{X}+\frac{q_{1}}{HX^{k}}\right)^{2^{1-k}}\left(\frac{1}{q_{2}}+\frac{1}{Y}+\frac{q_{2}}{HX^{l}Y^{k-l}}\right)^{2^{1-k}}$

∎

In order to describe Lemma 3.2, we introduce the set of smooth numbers

$\mathcal{A}(Y,R)=\\{1\leq n\leq Y|\ p\ \textrm{prime}\ \textrm{and}\
p|n\Rightarrow p\leq R\\}.$

The following lemma provides an upper bound for the exponential sum in
$(\ref{11'})$ with $A=[1,X]\cap{\mathbb{Z}}$ and $B=\mathcal{A}(Y,R)$ where
$R=Y^{\eta}$ with sufficiently small $\eta>0$. To describe the following
lemma, it is convenient to define an exponential sum
$S(\boldsymbol{\alpha})=S(\boldsymbol{\alpha};H,X,Y,R)$ by

$S(\boldsymbol{\alpha})=\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{1\leq x\leq
X}\displaystyle\sum_{y\in\mathcal{A}(Y,R)}e(hg(x,y))\biggr{|}$

######  Lemma 3.2.

Let $k$ and $l$ be natural numbers satisfying $1\leq l\leq\lfloor k/2\rfloor$
and $k\geq 3$, and consider real numbers $\alpha_{k}$ and $\alpha_{j}\ (0\leq
j\leq l)$. Suppose that $X,\ Y$ and $H$ are positive real numbers sufficiently
large in terms of $k$ with $Y\leq X$. Let $Z$ be a real number such that

$1\leq Z\leq\min\\{(HX^{l})^{1/(2(k-l)},Y\\},$

and let $N$ be a real number with $Z^{\frac{1}{2}(k-l)}\leq N\leq
HX^{l}Y^{k-l}Z^{-\frac{1}{2}(k-l)}$. Suppose that there exist
$q\in{\mathbb{N}}$ and $a\in{\mathbb{Z}}$ satisfying $q\leq N,(q,a)=1$ and
$|q\alpha_{l}-a|\leq N^{-1}.$ Then, for each $\epsilon>0,$ one has

$\displaystyle
S(\boldsymbol{\alpha})\ll\frac{H(XY)^{1+\epsilon}}{(q+HX^{l}Y^{k-l}|q\alpha_{l}-a|)^{1/(2^{l+1}(k-l))}}+H(XY)^{1+\epsilon}Z^{-1/2^{l+2}}.$
(3.5)

To prove Lemma 3.2, we shall use following two propositions.

######  Proposition 3.3.

Let $f$ be an arithmetic function. Suppose that $Y,M_{1},\ldots,M_{r}$ and $R$
satisfy $1\leq R<Y$, $1\leq M_{i}<Y\ (1\leq i\leq r)$ and $M_{1}\cdots
M_{r}<Y.$ Then, one has

$\displaystyle\sum_{y\in\mathcal{A}(Y,R)}f(y)\ll
F(Y,M_{1},\ldots,M_{r},R)+G(Y,M_{1},\ldots,M_{r},R),$

where

$\displaystyle\qquad\quad
F(Y,M_{1},\ldots,M_{r},R)=\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq
M_{r}R}\displaystyle\sum_{u\in\mathcal{A}(Y/(M_{1}\cdots
M_{r}),R)}\left|f(v_{1}\cdots v_{r}u)\right|$ and $\displaystyle\qquad\quad
G(Y,M_{1},\ldots,M_{r},R)=\displaystyle\sum_{j=1}^{r}M_{1}\cdots
M_{j}R^{j-1}\sup_{1\leq y\leq M_{j}}|f(y)|.$

###### Proof.

Observe that if $y\in\mathcal{A}(Y,R)$ and $y>M_{1},$ then there is a divisor
$v_{1}$ of $y$ satisfying $M_{1}<v_{1}\leq M_{1}R$. Moreover, one then has
$y/v_{1}\in\mathcal{A}(Y/M_{1},R).$ Hence

$\displaystyle\displaystyle\sum_{y\in\mathcal{A}(Y,R)}f(y)$
$\displaystyle\ll\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\displaystyle\sum_{u\in\mathcal{A}(Y/M_{1},R)}|f(v_{1}u)|+\displaystyle\sum_{y\leq
M_{1}}|f(y)|$ $\displaystyle\ll\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\displaystyle\sum_{u\in\mathcal{A}(Y/M_{1},R)}|f(v_{1}u)|+M_{1}\sup_{1\leq
y\leq M_{1}}|f(y)|.$

Inductively applying this relation, the conclusion follows.∎

Proposition 3.4 provides the upper bound for a certain type of exponential
sums in many variables. To prove Proposition 3.4, we follow the argument in
the proof of [References, Lemma 4.3]. To describe the following Proposition,
it is convenient to define an exponential sum
$\Xi(\beta)=\Xi(\beta;\boldsymbol{V},U,L)$ by

$\Xi(\beta)=\displaystyle\sum_{v_{1},\ldots,v_{r}}\displaystyle\sum_{u_{1},u_{2}}\biggl{|}\displaystyle\sum_{1\leq
x\leq L}e(\beta(v_{1}\cdots v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})x)\biggr{|},$

where the summation is over integers satisfying

$1\leq v_{i}\leq V_{i}\ (1\leq i\leq r)\ \textrm{and}\ U/2<u_{1},u_{2}\leq U.$
(3.6)

######  Proposition 3.4.

Let $k$ and $l$ be natural numbers satisfying $1\leq
l\leq\lfloor\frac{k}{2}\rfloor$ and $k\geq 3.$ Suppose that $L$ is a positive
real number sufficiently large in terms of $k$. Let $U,V_{1},\ldots,V_{r}$ be
positive numbers satisfying

$\displaystyle 1\leq U\leq\tfrac{1}{2}(LV_{1}^{k-l}\cdots
V_{r}^{k-l})^{1/(2(k-l)-1)}$ (3.7) $\displaystyle 1\leq
V_{r}<\tfrac{1}{2}L^{1/(k-l)}U^{-\frac{1}{2(k-l)}}$ $\displaystyle 1\leq
V_{i}<\tfrac{1}{2}L^{1/(k-l)}V_{i+1}\cdots V_{r}U^{-1}\ (1\leq i\leq r-1).$

Let $\beta\in{\mathbb{R}}$, $Q=V_{1}^{k-l}\cdots V_{r}^{k-l}U^{k-l}L$ and $N$
be a real number with $U^{\frac{1}{2}(k-l)}\leq N\leq QU^{-\frac{1}{2}(k-l)}$.
Suppose that there exist $q\in{\mathbb{N}}$ and $a\in{\mathbb{Z}}$ satisfying
$q\leq N,(q,a)=1$ and $|q\beta-a|\leq N^{-1}.$ Then, for each $\epsilon>0,$
one has

$\displaystyle\Xi(\beta)\ll\frac{V_{1}\cdots
V_{r}U^{2}L^{1+\epsilon}}{(q+Q|q\beta-a|)^{1/(k-l)}}+V_{1}\cdots
V_{r}L^{1+\epsilon}U^{3/2},$ (3.8)

where the implicit constant may depend on $r.$

###### Proof.

By summing over $x$ in the inner sum of $\Xi(\beta)$, one has

$\Xi(\beta)\ll\displaystyle\sum_{v_{1},\ldots,v_{r}}\displaystyle\sum_{u_{1},u_{2}}\frac{L}{1+L\|\beta(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})\|},$ (3.9)

where the summations are over $v_{1},\ldots,v_{r},u_{1},u_{2}$ satisfying
$(\ref{lab3.6}).$ Plainly, one may restrict the summation to satisfy

$U/2\leq u_{2}<u_{1}\leq U\ \textrm{and}\ \|\beta(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})\|<L^{-1}U^{1/2}.$ (3.10)

Indeed, the contribution on the right hand side of (3.9) of the remaining
summands is $O(V_{1}\cdots V_{r}LU^{3/2}).$ For given
$u_{1},u_{2},v_{1},\ldots,v_{r}$, we may choose $n_{0}$ so that

$\|\beta(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})\|=|\beta(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-n_{0}|.$

Let $\mathcal{R}_{r-1}=(LV_{r}^{k-l}U^{-1/2})^{1/2}$. By Dirichlet’s
approximation theorem, there exist $q_{r-1}\in{\mathbb{N}}$ and $\
a_{r-1}\in{\mathbb{Z}}$ with $q_{r-1}\leq\mathcal{R}_{r-1}$ and
$(q_{r-1},a_{r-1})=1$ such that

$|\beta(v_{1}\cdots
v_{r-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})q_{r-1}-a_{r-1}|<1/\mathcal{R}_{r-1}.$
(3.11)

If $a_{r-1}=0$, then $q_{r-1}=1.$ Thus, on recalling $(\ref{con3.7})$, it
follows from (3.10) and (3.11) that

$\left|\frac{n_{0}}{v_{r}^{k-l}}-\frac{a_{r-1}}{q_{r-1}}\right|v_{r}^{k-l}q_{r-1}<\frac{\mathcal{R}_{r-1}U^{1/2}}{L}+\frac{V_{r}^{k-l}}{\mathcal{R}_{r-1}}<1.$

Thus, one has $q_{r-1}n_{0}=v_{r}^{k-l}a_{r-1}$. If $a_{r-1}=0$, then
$q_{r-1}=1.$ Hence, in all cases, $q_{r-1}|v_{r}^{k-l}.$ Let
$q_{r-1}=q_{1,r-1}q_{2,r-1}^{2}\cdots q_{k-l,r-1}^{k-l}$ where $q_{k-l,r-1}$
is maximal and $q_{1,r-1},\ldots,q_{k-l-1,r-1}$ are squarefree and coprime in
pairs. Then, $q_{1,r-1}\cdots q_{k-l,r-1}|v_{r}$. Thus, by writing

$v_{r}=q_{0,r-1}q_{1,r-1}\cdots q_{k-l,r-1},$

the bound (3.9) may be replaced by

$\displaystyle\sum_{v_{1},\ldots,v_{r-1}}\sum_{u_{1},u_{2}}\sum_{q_{0,r-1}\leq
V_{r}/(q_{1,r-1}\cdots q_{k-l,r-1})}\frac{L}{Z_{0}}+O(V_{1}\cdots
V_{r}LU^{3/2}),$ (3.12)

where

$Z_{0}=1+L(q_{0,r-1}\cdots q_{k-l,r-1})^{k-l}|\beta(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-1}/q_{r-1}|.$

Since we have $q_{r-1}\leq(q_{1,r-1}q_{2,r-1}\cdots q_{k-l,r-1})^{k-l}$, by
applying the upper bound (1) from Appendix A, the inner sum of $(\ref{22''})$
is

$\displaystyle\ll\frac{LV_{r}}{q_{1,r-1}\cdots
q_{k-l,r-1}(1+LV_{r}^{k-l}|\beta(v_{1}\cdots
v_{r-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-1}/q_{r-1}|)^{1/(k-l)}}$ (3.13)
$\displaystyle\ll\frac{LV_{r}}{(q_{r-1}+LV_{r}^{k-l}|q_{r-1}\beta(v_{1}\cdots
v_{r-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-1}|)^{1/(k-l)}}.$

Thus, by substituting the bound (3.13) into the inner sum in (3.12), the first
term in (3.12) is

$\ll\displaystyle\sum_{v_{1},\ldots,v_{r-1}}\displaystyle\sum_{u_{1},u_{2}}\frac{LV_{r}}{(q_{r-1}+LV_{r}^{k-l}|q_{r-1}\beta(v_{1}\cdots
v_{r-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-1}|)^{1/(k-l)}}$ (3.14)

We analyse the expression (3.14) by an inductive arguments similar to the
argument first described. At the $i$th step of this argument, with $1\leq
i\leq r,$ one finds that

$\Xi(\beta)\ll B_{i}+O(V_{1}\cdots V_{r}L^{1+\epsilon}U^{3/2}),$ (3.15)

where

$B_{i}=\displaystyle\sum_{v_{1},\ldots,v_{r-i}}\displaystyle\sum_{u_{1},u_{2}}\frac{LV_{r}V_{r-1}\cdots
V_{r-i+1}(\log L)^{i}}{(q_{r-i}+LV_{r}^{k-l}\cdots
V_{r-i+1}^{k-l}|q_{r-i}\beta(v_{1}\cdots
v_{r-i})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-i}|)^{1/(k-l)}}$ (3.16)

in which $q_{r-i}\in{\mathbb{Z}}$ and $a_{r-i}\in{\mathbb{N}}$ satisfy
$(q_{r-i},a_{r-i})=1,$

$q_{r-i}\leq\mathcal{R}_{r-i}=\left\\{\begin{array}[]{l}L^{1/2}(V_{r}\cdots
V_{r-i+1})^{(k-l)/2}\ \ \ \ \ \ \ \ \ \ \textrm{when}\ \ i\geq 2\\\
L^{1/2}V_{r}^{(k-l)/2}U^{-1/4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\textrm{when}\ \ i=1.\end{array}\right.$

and

$|\beta(v_{1}\cdots
v_{r-i})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})q_{r-i}-a_{r-i}|<1/\mathcal{R}_{r-i}.$

The case $i=1$ obviously follows by (3.11), (3.12) and (3.14). Assume that
$(\ref{eq3.15})$ holds for a particular $i$ with $i\leq r-1.$ We shall show
that (3.15) holds for $i+1.$ Plainly, one may restrict the summation in
$(\ref{eq3.16})$ to satisfy

$q_{r-i}+LV_{r}^{k-l}\cdots V_{r-i+1}^{k-l}|q_{r-i}\beta(v_{1}\cdots
v_{r-i})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-i}|<U^{\frac{1}{2}(k-l)}.$ (3.17)

Indeed, the contribution of the remaining summands is $O(V_{1}\cdots
V_{r}LU^{3/2}).$ Let $\mathcal{R}_{r-i-1}=L^{1/2}(V_{r}\cdots
V_{r-i})^{(k-l)/2}.$ By Dirichlet’s approximation theorem, there exist
$q_{r-i-1}\in{\mathbb{N}},a_{r-i-1}\in{\mathbb{Z}}$ with
$q_{r-i-1}\leq\mathcal{R}_{r-i-1}$ and $(q_{r-i-1},a_{r-i-1})=1$ such that

$|\beta(v_{1}\cdots
v_{r-i-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})q_{r-i-1}-a_{r-i-1}|<1/\mathcal{R}_{r-i-1}.$
(3.18)

Notice that if $a_{r-i-1}=0$ then $q_{r-i-1}=1.$ Thus, on recalling
$(\ref{con3.7})$, it follows from $(\ref{25})$ and $(\ref{26})$ that

$\left|\frac{a_{r-i}}{q_{r-i}v_{r-i}^{k-l}}-\frac{a_{r-i-1}}{q_{r-i-1}}\right|q_{r-i}v_{r-i}^{k-l}q_{r-i-1}<\frac{\mathcal{R}_{r-i-1}U^{\frac{k-l}{2}}}{LV_{r}^{k-l}\cdots
V_{r-i+1}^{k-l}}+\frac{U^{\frac{k-l}{2}}V_{r-i}^{k-l}}{\mathcal{R}_{r-i-1}}<1.$

Thus, one has $q_{r-i}v_{r-i}^{k-l}a_{r-i-1}=q_{r-i-1}a_{r-i}.$ If $a_{r-i}=0$
then $q_{r-i}=1.$ Hence, in all cases, $q_{r-i}|q_{r-i-1}.$ By writing
$q^{\prime}_{r-i-1}=q_{r-i-1}/q_{r-i},$ one has

$v_{r-i}^{k-l}a_{r-i-1}=q^{\prime}_{r-i-1}a_{r-i}.$

Since $(q^{\prime}_{r-i-1},a_{r-i-1})=1,$ one has
$q_{r-i-1}^{\prime}|v_{r-i}^{k-l}.$ Similarly, one has

$q_{r-i-1}^{\prime}=q_{1,r-i-1}q_{2,r-i-1}^{2}\cdots q_{k-l,r-i-1}^{k-l},$

where $q_{k-l,r-i-1}$ is maximal and $q_{1,r-i-1},\ldots,q_{k-l-1,r-i-1}$ are
squarefree and coprime in pairs. Then, $q_{1,r-i-1}q_{2,r-i-1}\cdots
q_{k-l,r-i-1}|v_{r-i}$. Thus, by writing $v_{r-i}=q_{0,r-i-1}\tau$ with

$\tau=q_{1,r-i-1}\cdots q_{k-l,r-i-1},$

the equation (3.16) may be replaced by

$\displaystyle\sum_{v_{1},\ldots,v_{r-i-1}}\displaystyle\sum_{u_{1},u_{2}}\displaystyle\sum_{q_{0,r-i-1}<V_{r-i}/\tau}\frac{LV_{r}\cdots
V_{r-i+1}(\log L)^{i}}{Z}+O(V_{1}\cdots V_{r}L^{1+\epsilon}U^{3/2}),$ (3.19)

where

$Z^{k-l}=q_{r-i}\biggl{(}1+L(V_{r}\cdots
V_{r-i+1})^{k-l}\bigl{(}q_{0,r-i-1}\tau)^{k-l}\bigl{|}\beta(v_{1}\cdots
v_{r-i-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-\frac{a_{r-i-1}}{q_{r-i-1}}\bigr{|}\biggr{)}.$

By applying the upper bound (2) from Appendix A, the inner sum in (3.19) is

$\displaystyle\ll\frac{LV_{r}\cdots V_{r-i}(\log
L)^{i+1}}{q_{r-i}^{1/(k-l)}\tau(1+L(V_{r}\cdots
V_{r-i})^{k-l}|\beta(v_{1}\cdots
v_{r-i-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-i-1}/q_{r-i-1}|)^{1/(k-l)}}.$

Since we have $q^{\prime}_{r-i-1}\leq\tau^{k-l}$ and
$q_{r-i-1}=q^{\prime}_{r-i-1}q_{r-i}$, we find that this expression is

$\displaystyle\leq\frac{LV_{r}\cdots V_{r-i}(\log
L)^{i+1}}{q_{r-i-1}^{1/(k-l)}(1+L(V_{r}\cdots V_{r-i})^{k-l}|\beta(v_{1}\cdots
v_{r-i-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-i-1}/q_{r-i-1}|)^{1/(k-l)}}$
(3.20) $\displaystyle=\frac{LV_{r}\cdots V_{r-i}(\log
L)^{i+1}}{(q_{r-i-1}+L(V_{r}\cdots V_{r-i})^{k-l}|q_{r-i-1}\beta(v_{1}\cdots
v_{r-i-1})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})-a_{r-i-1}|)^{1/(k-l)}}.$

Hence, by substituting the bound (3.20) into the inner sum (3.19), the first
term in (3.19) is seen to be $O(B_{i+1}).$ Thus, this confirms that the bound
(3.15) holds for $i+1$. Therefore, one infers by induction that

$\Xi(\beta)\ll\Xi(\beta)+O(V_{1}\cdots V_{r}L^{1+\epsilon}U^{3/2}),$ (3.21)

where

$\Xi_{1}(\beta)=\displaystyle\sum_{u_{1},u_{2}}\frac{LV_{r}\cdots V_{1}(\log
L)^{r}}{(q_{0}+LV_{r}^{k-l}\cdots
V_{1}^{k-l}|q_{0}\beta(u_{1}^{k-l}-u_{2}^{k-l})-a_{0}|)^{1/(k-l)}}$

in which ($q_{0},a_{0}$)=1, $q_{0}\leq\mathcal{R}_{0}=L^{1/2}(V_{r}\cdots
V_{1})^{(k-l)/2}$ and
$|q_{0}\beta(u_{1}^{k-l}-u_{2}^{k-l})-a_{0}|<1/\mathcal{R}_{0}.$ By Cauchy’s
inequality, one has

$\Xi_{1}(\beta)\ll UL^{1+\epsilon/2}V_{r}\cdots V_{1}\Xi_{2}(\beta)^{1/2},$
(3.22)

where

$\Xi_{2}(\beta)=\displaystyle\sum_{u_{1},u_{2}}\frac{1}{(q_{0}+LV_{r}^{k-l}\cdots
V_{1}^{k-l}|q_{0}\beta(u_{1}^{k-l}-u_{2}^{k-l})-a_{0}|)^{2/(k-l)}}.$

Plainly, one may also restrict the summation to satisfy

$q_{0}+LV_{r}^{k-l}\cdots
V_{1}^{k-l}|q_{0}\beta(u_{1}^{k-l}-u_{2}^{k-l})-a_{0}|<U^{\frac{1}{2}(k-l)}.$
(3.23)

Indeed, the contribution from the summation in (3.22) arising from remaining
terms is $O(V_{1}\cdots V_{r}L^{1+\epsilon}U^{3/2}).$ We put

$j=(u_{1},u_{2}),\ \ \ n=u_{2}/j,\ \ \ m=(u_{1}-u_{2})/j$

so that

$j\leq U,\ \ m\leq U/j,\ \ U/(2j)<n<n+m\leq U/j,\ \ (n,n+m)=1.$

Now, $q_{0}$ and $a_{0}$ will depend on $n,m,j.$ Let
$S=((U/j)^{k-l-1}LV_{r}^{k-l}\cdots V_{1}^{k-l})^{1/2}$. Then for fixed $j$
and $m$, by Dirichlet’s approximation theorem, there exists $c\in{\mathbb{Z}}$
and $s\in{\mathbb{N}}$ such that

$(c,s)=1,\ \ \ s\leq S,\ \ \ \textrm{and}\ \ \ |\beta
j^{k-l}m-c/s|\leq(sS)^{-1}.$

Notice again that if $c=0,$ then $s=1$. Let $D=((n+m)^{k-l}-n^{k-l})/m.$ Then

$D=\frac{k-l}{m}\displaystyle\int_{n}^{n+m}x^{k-l-1}dx,$

and so

$(k-l)(U/(2j))^{k-l-1}\leq D\leq(k-l)(U/j)^{k-l-1}.$

Thus, from (3.23), one has $q_{0}\leq U^{\frac{1}{2}(k-l)}$, and

$\left|\beta
j^{k-l}m-\frac{a_{0}}{q_{0}D}\right|\leq\frac{U^{\frac{1}{2}(k-l)}}{q_{0}DLV_{r}^{k-l}\cdots
V_{1}^{k-l}}$

Therefore, since $U\leq\tfrac{1}{2}(LV_{1}^{k-l}\cdots
V_{r}^{k-l})^{1/(2(k-l)-1)},$ one has

$|q_{0}Dc-
sa_{0}|=\left|\frac{c}{s}-\frac{a_{0}}{q_{0}D}\right|sq_{0}D\leq\frac{U^{\frac{1}{2}(k-l)}D}{S}+\frac{U^{\frac{1}{2}(k-l)}S}{LV_{r}^{k-l}\cdots
V_{1}^{k-l}}<1.$

Thus, we see that $cq_{0}D=a_{0}s.$ Hence $q_{0}|s.$ Let $s_{1}=s/q_{0}.$ Then
$cD=a_{1}s_{1}.$ Hence $c|a_{1}$ and $s_{1}|D.$ Therefore, as ($n,\ n+m$)=1,
we have $(n(n+m),s_{1})=1.$ Thus, one finds that

$\Xi_{2}(\beta)\ll\displaystyle\sum_{j\leq U}\displaystyle\sum_{m\leq
U/j}\displaystyle\sum_{s_{1}|s}\Xi_{3}(\beta)$ (3.24)

where

$\Xi_{3}(\beta)=\displaystyle\sum_{n}\frac{(s_{1}/s)^{2/(k-l)}}{(1+LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/(2j))^{k-l-1}|\beta j^{k-l}m-c/s|)^{2/(k-l)}}$

in which the innermost sum is over $n$ satisfying

$n\leq U/j,\ \ \ (n(n+m),s_{1})=1,\ \ \ s_{1}|D.$ (3.25)

The third condition $s_{1}|D$ implies that

$(n+m)^{k-l}\equiv n^{k-l}\ (\textrm{mod}\ s_{1}m).$ (3.26)

Since we have $(n,n+m)=1$ and $(n,s_{1})=1$, one finds that $(n,s_{1}m)=1$.
Thus, there exists $n_{0}$ such that $nn_{0}\equiv 1\ (\textrm{mod}\ s_{1}m).$
Hence, the congruence $(\ref{eq3.26})$ is equivalent to

$(1+mn_{0})^{k-l}\equiv 1\ (\textrm{mod}\ s_{1}m).$

Notice that the congruence $y^{k-l}\equiv 1\ (\textrm{mod}\ s_{1}m)$ has $v$
solutions modulo $s_{1}m$, say $g_{1},\ldots,g_{v}$, where
$v\ll(s_{1}m)^{\epsilon}.$ Hence $1+mn_{0}\equiv g_{i}\ (\textrm{mod}\
s_{1}m)$ for some $1\leq i\leq v$. Thus, there are at most $v$ choices for
$n_{0}$, and so for $n,$ modulo $s_{1}.$ Then, on noting that $s_{1}m$ is
bounded above by powers of $L$ which may depend on $r$, we see that

$\Xi_{3}(\beta)\ll\left(\frac{U}{js_{1}}+1\right)\frac{L^{\epsilon/4}(s_{1}/s)^{2/(k-l)}}{(1+LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/(2j))^{k-l-1}|\beta j^{k-l}m-c/s|)^{2/(k-l)}},$ (3.27)

where the implicit may depend on $r.$ By examining separately the contribution
arising from the terms $U/(js_{1})$ and $1$ in the first factor of
$(\ref{39'})$, we see from (3.24) that

$\Xi_{2}(\beta)\ll A+O(L^{\epsilon/2}U),$

where

$A=\displaystyle\sum_{j\leq U}\displaystyle\sum_{m\leq
U/j}\frac{L^{\epsilon/4}Uj^{-1}}{(s+LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/(2j))^{k-l-1}|\beta j^{k-l}ms-c|)^{2/(k-l)}}.$

Plainly, we may restrict the inner sum in $A$ to those $m$ satisfying

$s+LV_{r}^{k-l}\cdots V_{1}^{k-l}(U/(2j))^{k-l-1}|\beta
j^{k-l}ms-c|<U^{\frac{1}{2}(k-l)}.$

Indeed, the contribution on $A$ of the remaining sum is $O(L^{\epsilon}U)$,
and thus the right hand side in (3.22) is $O(V_{1}\cdots
V_{r}L^{1+\epsilon}U^{3/2}).$ Let $T=(LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/j)^{k-l})^{1/2}$. Then for fixed $j$, by Dirichlet’s
approximation theorem there exists $d$ and $t$ with

$(d,t)=1,\ \ \ t\leq T\ \ \ \textrm{and}\ \ \ |\beta
j^{k-l}-d/t|\leq(tT)^{-1}.$

Notice again that if $d=0$, then $t=1$. Then, on recalling
$U\leq\tfrac{1}{2}(LV_{1}^{k-l}\cdots V_{r}^{k-l})^{1/(2(k-l)-1)}$, for fixed
$j$ and $m$ one has

$\left|\frac{c}{ms}-\frac{d}{t}\right|tms\leq\frac{U^{\frac{1}{2}(k-l)}U/j}{T}+\frac{TU^{\frac{1}{2}(k-l)}}{LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/(2j))^{k-l-1}}<1.$

Thus, we see that $ct=dms,$ and so $s|t$. Let $t_{1}=t/s$. Then $ct_{1}=dm.$
Thus $t_{1}|m.$ Let $m_{1}=m/t_{1}.$ Therefore, the summation $A$ is bounded
above by

$\displaystyle\sum_{j\leq
U}\displaystyle\sum_{t_{1}|t}(t_{1}/t)^{2/(k-l)}A_{1},$

where

$A_{1}=\displaystyle\sum_{m_{1}\leq
U/(jt_{1})}\frac{L^{\epsilon/4}Uj^{-1}}{(1+LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/(2j))^{k-l-1}m_{1}t_{1}|\beta j^{k-l}-d/t|)^{2/(k-l)}}.$

Hence, by applying the upper bound (3) from Appendix A, the innermost sum is

$A_{1}\ll\frac{L^{\epsilon/4}U^{2}j^{-2}t_{1}^{-1}\log
X}{(1+LV_{r}^{k-l}\cdots V_{1}^{k-l}(U/j)^{k-l}|\beta
j^{k-l}-d/t|)^{2/(k-l)}}$

Hence, one finds that

$\Xi_{2}(\beta)\ll L^{\epsilon/2}\left(U+\displaystyle\sum_{j\leq
U}\frac{U^{2}j^{-2}}{(t+LV_{r}^{k-l}\cdots V_{1}^{k-l}(U/j)^{k-l}|\beta
j^{k-l}t-d|)^{2/(k-l)}}\right),$ (3.28)

and one may restrict the summation over $j$ to satisfy

$t+LV_{r}^{k-l}\cdots V_{1}^{k-l}(U/j)^{k-l}|\beta
j^{k-l}t-d|<\frac{1}{2}(U/j)^{(k-l)/2}.$ (3.29)

Let $\overline{\mathcal{R}}=Q^{1/2}$. Then by Dirichlet’s approximation
theorem, there exists $\overline{q}$ and $\overline{a}$ such that

$(\overline{q},\overline{a})=1,\ \ \ \overline{q}\leq\overline{\mathcal{R}}\ \
\ \textrm{and}\ \ \
|\overline{q}\beta-\overline{a}|\leq\overline{\mathcal{R}}^{-1}.$

Since $t$ is non-zero, one finds that $j\leq U.$ Therefore, on recalling

$U\leq\tfrac{1}{2}(LV_{1}^{k-l}\cdots V_{r}^{k-l})^{1/(2(k-l)-1)},$

when $j$ satisfies $(\ref{42})$, one has

$\left|\frac{\overline{a}}{\overline{q}}-\frac{d}{j^{k-l}t}\right|j^{k-l}t\overline{q}<\frac{(U/j)^{(k-l)/2}j^{k-l}}{2\overline{\mathcal{R}}}+\frac{\overline{\mathcal{R}}(U/j)^{(k-l)/2}}{2LV_{r}^{k-l}\cdots
V_{1}^{k-l}(U/j)^{k-l}}\leq 1.$

Thus, we see that $j^{k-l}t\overline{a}=\overline{q}d$. Hence
$t|\overline{q}.$ Let $\overline{q}^{\prime}=\overline{q}/t.$ Then
$j^{k-l}\overline{a}=\overline{q}^{\prime}d.$ Hence
$\overline{q}^{\prime}|j^{k-l}$. Let
$\overline{q}^{\prime}=q_{1}^{\prime}q_{2}^{\prime 2}\cdots q_{k-l}^{\prime
k-l}$ where $q_{k-l}^{\prime}$ is maximal and
$q_{1}^{\prime},\ldots,q_{k-l-1}^{\prime}$ are square-free and coprime in
pairs. Then $q_{1}^{\prime}\cdots q_{k-l}^{\prime}|j$, so the summation over
$j$ in (3.28) is

$\ll\displaystyle\sum_{w}\displaystyle\sum_{t,q_{1}^{\prime},\ldots,q_{k-l}^{\prime}}\frac{U^{2}(wq_{1}^{\prime}\cdots
q_{k-l}^{\prime})^{-2}t^{-2/(k-l)}}{(1+LV_{r}^{k-l}\cdots
V_{1}^{k-l}U^{k-l}|\beta-\overline{a}/\overline{q}|)^{2/(k-l)}},$ (3.30)

where the inner sum is over $t,q_{1}^{\prime},\ldots,q_{k-l}^{\prime}$
satisfying $\overline{q}=tq_{1}^{\prime}q_{2}^{\prime 2}\cdots q_{k-l}^{\prime
k-l}$. Thus, on recalling $Q=(V_{1}\cdots V_{r}U)^{k-l}L$, since we have
$\overline{q}^{2/(k-l)}\leq t^{2/(k-l)}q_{1}^{\prime}q_{2}^{\prime 2}\cdots
q_{k-l}^{\prime k-l}$, the bound (3.30) is

$\ll\frac{U^{2}L^{\epsilon/2}}{(\overline{q}+Q|\beta\overline{q}-\overline{a}|)^{2/(k-l)}}.$

When
$\overline{q}+Q|\overline{q}\beta-\overline{a}|\geq\frac{1}{2}U^{(k-l)/2}$ we
are done, so we may suppose that
$\overline{q}+Q|\overline{q}\beta-\overline{a}|\leq\frac{1}{2}U^{(k-l)/2}.$
Thus,

$\left|\frac{\overline{a}}{\overline{q}}-\frac{a}{q}\right|q\overline{q}<\frac{U^{(k-l)/2}N}{2Q}+\frac{U^{(k-l)/2}}{2N}\leq
1,$

since by assumption, $U^{\frac{1}{2}(k-l)}\leq N\leq QU^{-\frac{1}{2}(k-l)}.$
Hence, one has $q=\overline{q},a=\overline{a}.$ Therefore, we complete the
proof. ∎

We now turn to prove Lemma 3.2. To prove this lemma, we exploit Proposition
3.3 and Proposition 3.4.

###### Proof of Lemma 3.2.

Note that

$S(\boldsymbol{\alpha})\leq\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{y\in\mathcal{A}(Y,R)}\biggl{|}\displaystyle\sum_{1\leq
x\leq X}e(hg(x,y))\biggr{|}.$ (3.31)

We shall first apply Proposition 3.3 to the right hand side in (3.31). Let
$M_{1},\ldots,M_{r}$ be real numbers with $M_{1},\ldots,M_{r}\geq 1$
satisfying $M_{1}\cdots M_{r}=Y/Z$. We will define each of
$M_{1},\ldots,M_{r}$ later for applications of Proposition 3.4. For now, we
temporarily assume that such $M_{1},\ldots,M_{r}$ exist. Then, by applying
Proposition 3.3 with $R=Y^{\eta}$ and

$f(y)=\displaystyle\sum_{1\leq x\leq X}e(hg(x,y)),$

one finds from (3.31) that

$S(\boldsymbol{\alpha})\ll S_{1}(\boldsymbol{\alpha})+\displaystyle\sum_{1\leq
h\leq H}\displaystyle\sum_{j=1}^{r}M_{1}\cdots M_{j}R^{j-1}\sup_{1\leq y\leq
M_{j}}|f(y)|,$ (3.32)

where

$S_{1}(\boldsymbol{\alpha})=\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq M_{r}R}\\\
\displaystyle\sum_{u\in\mathcal{A}(Z,R)}\left|\displaystyle\sum_{1\leq x\leq
X}e\left(hg(x,v_{1}\ldots v_{r}u)\right)\right|.$

Since by $Y/(M_{1}\cdots M_{r})=Z$ and $\sup_{1\leq y\leq M_{i}}|f(y)|\leq X$
for all $h$, it follows trivially that

$\displaystyle\sum_{1\leq h\leq H}\displaystyle\sum_{j=1}^{r}M_{1}\cdots
M_{j}R^{j-1}\sup_{1\leq y\leq M_{i}}|f(y)|\ll
H(XY)^{1+\epsilon}Z^{-1/2^{l+2}}.$

Then, it suffices to bound $S_{1}(\boldsymbol{\alpha})$. By applying Hölder’s
inequality, we see that

$S_{1}(\boldsymbol{\alpha})\leq(HYR^{r})^{1-2^{-l}}S_{2}(\boldsymbol{\alpha})^{2^{-l}},$
(3.33)

where

$S_{2}(\boldsymbol{\alpha})=\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{u\in\mathcal{A}(Z,R)}\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq
M_{r}R}\biggl{|}\displaystyle\sum_{1\leq x\leq X}e\left(hg(x,v_{1}\cdots
v_{r}u)\right)\biggr{|}^{2^{l}}.$

By applying Weyl differencing, we have

$\displaystyle S_{2}(\boldsymbol{\alpha})\ll
X^{2^{l}-l-1}\displaystyle\sum_{1\leq h\leq
H}\displaystyle\sum_{u\in\mathcal{A}(Z,R)}\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq
M_{r}R}S_{3}(\boldsymbol{\alpha}),$ (3.34)

where

$S_{3}(\boldsymbol{\alpha})=\displaystyle\sum_{|h_{1}|\leq
X}\cdots\displaystyle\sum_{|h_{l}|\leq X}\displaystyle\sum_{x\in
I_{l}(\mathbf{h})}e\left(h\Delta_{l}^{x}(g(x,v_{1}\cdots
v_{r}u),\mathbf{h})\right)$

in which $I_{l}(\mathbf{h})$ is an interval of integers contained in $[-X,X]$.
Note here that

$\Delta_{l}^{x}(g(x,v_{1}\cdots
v_{r}u),\mathbf{h})=\Delta_{l}^{x}(\alpha_{k}x^{k})+\Delta_{l}^{x}(\alpha_{l}x^{l}(v_{1}\cdots
v_{r}u)^{k-l})$

and $\Delta_{l}^{x}(\alpha_{l}x^{l}(v_{1}\cdots v_{r}u)^{k-l})$ does not
depend on $x.$ Thus, by splitting $\mathcal{A}(Z,R)$ into dyadic intervals
$[Z_{0}/2^{i},Z_{0}/2^{i-1}]$ with $Z_{0}\leq Z$, for any $\epsilon>0$, one
infers from (3.34) by applying a standard divisor estimate that

$S_{2}(\boldsymbol{\alpha})\ll
X^{2^{l}-l+\epsilon}\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq
M_{r}R}\displaystyle\sum_{1\leq n\leq
l!HX^{l}}\biggl{|}\displaystyle\sum_{u}e\left(\alpha_{l}(v_{1}\cdots
v_{r}u)^{k-l}n\right)\biggr{|}$

where the innermost sum is over $u\in\mathcal{A}(Z,R)$ and $Z_{0}/2\leq u\leq
Z_{0}$ for some $Z_{0}$ with $Z_{0}\leq Z$. Since we have $M_{1}\cdots
M_{r}=Y/Z$, by applying Cauchy’s inequality, we deduce that

$S_{2}(\boldsymbol{\alpha})\ll
X^{2^{l}-l+\epsilon}(HX^{l}(Y/Z)R^{r})^{1/2}S_{4}(\boldsymbol{\alpha})^{1/2},$
(3.35)

where

$S_{4}(\boldsymbol{\alpha})=\displaystyle\sum_{M_{1}<v_{1}<M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}<M_{r}R}\displaystyle\sum_{1\leq
n\leq l!HX^{l}}\left|\displaystyle\sum_{u}e\left(\alpha_{l}(v_{1}\cdots
v_{r}u)^{k-l}n\right)\right|^{2}.$

By squaring out and change the order of summations, we see that

$S_{4}(\boldsymbol{\alpha})\leq\displaystyle\sum_{M_{1}<v_{1}\leq
M_{1}R}\cdots\displaystyle\sum_{M_{r}<v_{r}\leq M_{r}R}\
\displaystyle\sum_{Z_{0}/2\leq u_{1},u_{2}\leq
Z_{0}}\left|\displaystyle\sum_{1\leq n\leq
l!HX^{l}}e\left(\alpha_{l}n(v_{1}\cdots
v_{r})^{k-l}(u_{1}^{k-l}-u_{2}^{k-l})\right)\right|.$ (3.36)

We define $M_{1},\ldots,M_{r}$ here so that we apply Proposition 3.4. We set
$r=1/\eta$, and define

$a=\biggl{\lfloor}\frac{1}{\eta}\biggl{(}1-\frac{\log Z}{\log
Y}\biggr{)}\biggr{\rfloor},$

which satisfies $Y^{a\eta}\leq Y/Z.$ Then, define $M_{r},\ldots,M_{1}$ by

$M_{i}=\left\\{\begin{array}[]{l}Y^{\eta}\ \ \ \ \ \ \ \ \ \ \textrm{when}\ \
i>r-a\\\ Y/(ZY^{a\eta})\ \textrm{when}\ \ i=r-a\\\ 1\ \ \ \ \ \ \ \ \ \ \ \ \
\textrm{when}\ \ 1\leq i<r-a.\end{array}\right.$

Since $Y/Z=M_{1}\cdots M_{r}$, this choice for $M_{1},\ldots,M_{r}$ is in
accordance with the hypotheses of Proposition 3.3.

On recalling that $Z_{0}\leq Z\leq(HX^{l})^{1/(2(k-l))}$ and by observing
$Y^{\eta}\geq M_{r}\geq M_{r-1}\geq\cdots\geq M_{1},$ one finds that

$\displaystyle 1\leq
M_{r}R<\frac{1}{2}(HX^{l})^{1/(k-l)}Z_{0}^{-\frac{1}{2(k-l)}},$ (3.37)
$\displaystyle 1\leq
M_{i}R<\frac{1}{2}(HX^{l})^{1/(k-l)}(M_{i+1}R)\cdots(M_{r}R)Z_{0}^{-1}\
(i=1,\ldots,r-1).$ (3.38)

Furthermore, on recalling that $Z\leq(HX^{l})^{1/(2(k-l))}$, one deduces that

$Z_{0}\leq\tfrac{1}{2}(HX^{l}(M_{1}R)^{k-l}\cdots(M_{r}R)^{k-l})^{1/(2(k-l)-1)}.$
(3.39)

Thus, on setting $U=Z_{0}$ and $V_{i}=M_{i}R\ (1\leq i\leq r),$ we see that
the hypothese of Proposition 3.4 are satisfied. Then, since we have
$Y/Z=M_{1}\cdots M_{r}$, by applying Proposition 3.4 to (3.36) with
$L=l!HX^{l}$ and $V_{i}=M_{i}R\ (i=1,\ldots,r)$, we obtain that

$S_{4}(\boldsymbol{\alpha})\ll
HX^{l}(Y/Z)R^{r}Z_{0}^{2}X^{\epsilon}\biggl{(}\frac{1}{(q+(Z_{0}/Z)^{k-l}HX^{l}Y^{k-l}|q\alpha_{l}-a|)^{1/(k-l)}}+Z_{0}^{-1/2}\biggr{)}.$
(3.40)

Therefore, on substituting (3.40) into (3.35) and that into (3.33), we deduce
that

$S_{1}(\boldsymbol{\alpha})\ll
X^{\epsilon}HXYR^{r}(Z_{0}Z^{-1})^{2^{-l}}\left(\frac{1}{(q+(Z_{0}/Z)^{k-l}HX^{l}Y^{k-l}|q\beta-a|)^{1/(k-l)}}+Z_{0}^{-1/2}\right)^{2^{-l-1}}.$

Since we have $Z_{0}\leq Z,$ by choosing $\eta$ small in terms of $\epsilon,$
one concludes

$S_{1}(\boldsymbol{\alpha})\ll\frac{H(XY)^{1+\epsilon}}{(q+HX^{l}Y^{k-l}|q\beta-a|)^{1/(2^{l+1}(k-l))}}+H(XY)^{1+\epsilon}Z^{-2^{-l-2}}.$

Therefore, we are done.

∎

##  4\. Proof of Theorem 1.1

Our goal in this section is to prove Theorem 1.1. As we mentioned in section
2, we reduce the problem of bounding the fractional parts of polynomial in
Theorem 1.1 to the corresponding diagonal problems. Thus, we begin this
section by examining the corresponding diagonal problem.

######  Proposition 4.1.

Let $\alpha$, $\beta\in{\mathbb{R}}$ and $k\in{\mathbb{N}}$ with $k\geq 2$.
Then, for any $\epsilon>0$, there exists a real number $X(k,\epsilon)$ such
that whenever $X\geq X(k,\epsilon)$ and $Y\geq X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq
X^{-\sigma+\epsilon}Y^{-\sigma+\epsilon},$

where $\sigma=2^{1-k}.$

###### Proof.

Let $H=X^{\sigma-\epsilon}Y^{\sigma-\epsilon}.$ Suppose that there exists no
$x$ and $y$ satisfying

$\|\alpha x^{k}+\beta y^{k}\|\leq 1/H.$

By Dirichlet approximation theorem, there exits $q_{1},q_{2}\in{\mathbb{N}}$
and $a_{1},a_{2}\in{\mathbb{Z}}$ such that $(q_{1},a_{1})=(q_{2},a_{2})=1$ and

$\left|\alpha-\frac{a_{1}}{q_{1}}\right|<\frac{1}{q_{1}X^{k-1}H},\
\left|\beta-\frac{a_{2}}{q_{2}}\right|<\frac{1}{q_{2}Y^{k-1}H}.$

If $q_{1}\leq X$, then by observing that $|q_{1}\alpha-
a_{1}|<(HX^{k-1})^{-1},$ one has

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq\|\alpha
q_{1}^{k}\|\leq q_{1}^{k-1}\|\alpha
q_{1}\|<X^{k-1}\frac{1}{X^{k-1}H}\leq\frac{1}{H}.$

Similarly, if $q_{2}\leq Y,$ one has

$\min_{\begin{subarray}{c}1\leq x\leq X\\\ 1\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq\frac{1}{H}.$

These are contradiction to our assumption. Thus, we may assume that $q_{1}\geq
X$ and $q_{2}\geq Y$.

It follows from our assumption by Lemma 2.1 that

$XY\ll\displaystyle\sum_{1\leq h\leq
H}\bigl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\ 1\leq y\leq
Y\end{subarray}}e(h(\alpha x^{k}+\beta y^{k}))\bigr{|}.$ (4.1)

On writing

$\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\ 1\leq y\leq
Y\end{subarray}}e(h(\alpha x^{k}+\beta y^{k}))=\displaystyle\sum_{1\leq x\leq
X}e(h\alpha x^{k})\displaystyle\sum_{1\leq y\leq Y}e(h\beta y^{k}),$

by Cauchy’s inequality, the right hand side in (4.1) is

$\ll\left(\displaystyle\sum_{1\leq h\leq H}\biggl{|}\displaystyle\sum_{1\leq
x\leq X}e(h\alpha
x^{k})\biggr{|}^{2}\right)^{1/2}\left(\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{1\leq y\leq Y}e(h\beta
y^{k})\biggr{|}^{2}\right)^{1/2}$

Then, by applying Weyl differencing, the last expression is for every
$\delta>0$

$\ll
X^{1+\delta}Y^{1+\delta}H\left(\frac{1}{q_{1}}+\frac{1}{X}+\frac{q_{1}}{X^{k}H}\right)^{2^{1-k}}\left(\frac{1}{q_{2}}+\frac{1}{Y}+\frac{q_{2}}{Y^{k}H}\right)^{2^{1-k}}.$
(4.2)

Therefore, on recalling that $q_{1}\geq X$ and $q_{2}\geq Y,$ one finds that
(4.2) is

$\ll X^{1+\delta}Y^{1+\delta}X^{-\epsilon}Y^{-\epsilon}.$

By taking $\delta<\epsilon$, it contradicts $(\ref{35})$. Therefore, we are
forced to have

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq
X^{-\sigma+\epsilon}Y^{-\sigma+\epsilon}.$

∎

As sketched in section 2, we shall reduce Theorem 1.1 to this diagonal problem
through Lemma 3.1 and inductive arguments.

###### Proof of Theorem 1.1.

Let $H=X^{\sigma-\epsilon}$ with $\sigma=\frac{l+2}{l+1}2^{1-k}.$ There exists
$q_{k}$ and $a_{k}$ with $(q_{k},a_{k})=1$ such that $q_{k}\leq X^{k-1}H$ and

$|\alpha_{k}-a_{k}/q_{k}|\leq\frac{1}{q_{k}X^{k-1}H}.$ (4.3)

Suppose that

$\frac{1}{H}<\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|$
(4.4)

for all $0\leq x,y\leq X$ but $(x,y)=(0,0).$ Then, we may assume $q_{k}>X$.
Indeed, if one were to have $q_{k}\leq X$, by (4.3), one has

$\displaystyle\min_{\begin{subarray}{c}1\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|$
$\displaystyle\leq\min_{1\leq x\leq X}\|\alpha_{k}x^{k}\|\leq
q_{k}^{k-1}\|\alpha_{k}q_{k}\|\leq\frac{1}{H},$

which contradicts our assumption.

From our assumption (4.4), we derive analogues of (4.4) by an inductive
arguments. Specifically, at the $i$-th step, with $0\leq i\leq l$, we shall
prove that there exist coefficients
$\alpha_{l-i}^{(i)},\ldots,\alpha_{0}^{(i)}$, suitable integral multiples of
$\alpha_{l-i},\ldots,\alpha_{0}$, respectively, such that

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|$
(4.5)

for all $0\leq x\leq X$ and $0\leq y\leq Y_{i}=X^{1-i/(l+1)}$ but
$(x,y)=(0,0).$ The case $i=0$ obviously follows from $(\ref{4.6}).$ Assume
that $(\ref{4.77})$ holds for a particular $i\leq l-1.$ We shall show that
(4.5) holds for $i$ replaced by $i+1.$

By Lemma 2.1, the inequality (4.5) implies that

$XY_{i}\ll\displaystyle\sum_{1\leq h\leq H}\biggl{|}\displaystyle\sum_{1\leq
x\leq X}\displaystyle\sum_{1\leq y\leq
Y_{i}}e(h(\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}))\biggr{|}.$
(4.6)

Note that for any $\delta>0$, it follows from Dirichlet’s approximation
theorem that there exists $q\in{\mathbb{N}}$ such that

$\displaystyle q\leq X^{l-i+1}Y_{i}^{k-l+i}H^{-2^{k-1}+1}X^{-\delta}$ (4.7)
$\displaystyle\|q\alpha_{l-i}^{(i)}\|\leq\frac{X^{\delta}}{X^{l-i+1}Y_{i}^{k-l+i}H^{-2^{k-1}+1}}.$

Thus, on recalling $X<q_{k}<X^{k-1}H$, it follows by Lemma 3.1 that the right
hand side in (4.6) is

$\displaystyle\ll$ $\displaystyle
HX^{1+\epsilon}Y_{i}\biggl{(}\frac{1}{q_{k}}+\frac{1}{X}+\frac{q_{k}}{X^{k}H}\biggr{)}^{2^{1-k}}\biggl{(}\frac{1}{q}+\frac{1}{Y_{i}}+\frac{q}{X^{l-i}Y_{i}^{k-l+i}H}\biggr{)}^{2^{1-k}}$
$\displaystyle\ll$ $\displaystyle
HX^{1+\epsilon}Y_{i}X^{-2^{1-k}}(q^{-2^{1-k}}+Y_{i}^{-2^{1-k}}+X^{2^{1-k}}H^{-1-\delta}).$

Note that $(XY_{i})^{2^{1-k}}\geq H.$ Thus, combining this and the lower bound
of (4.6), one obtains

$XY_{i}\ll HX^{1+\epsilon}Y_{i}X^{-2^{1-k}}q^{-2^{1-k}}.$

This implies

$q\leq X^{1/(l+1)}.$

We apply the triangle inequality and put $y=qy_{1}$ with $0\leq y_{1}\leq
X^{1-(i+1)/(l+1)}$. Thus, we have $qy_{1}\leq Y_{i}$ and

$\displaystyle\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|$
(4.8)
$\displaystyle\leq\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|+\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|.$

Since we have $H=X^{\sigma-\epsilon}$ with $\sigma=\frac{l+2}{l+1}2^{1-k},$ by
applying the triangle inequality and (4.7), one has

$\displaystyle\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|$
$\displaystyle\leq x^{l-i}(qy_{1})^{k-l+i-1}y_{1}\|\alpha_{l-i}^{(i)}q\|$
(4.9) $\displaystyle\leq
X^{l-i}Y_{i}^{k-l+i-1}X^{1-(i+1)/(l+1)}\frac{X^{\delta}}{X^{l-i+1}Y_{i}^{k-l+i}H^{-2^{k-1}+1}}$
$\displaystyle\leq\frac{X^{\epsilon}}{HX}H^{2^{k-1}}X^{-1/(l+1)}$
$\displaystyle\leq\frac{X^{-\epsilon}}{H}.$

Hence, by substituting the bound (4.9) into (4.8) and the lower bound (4.5),
one obtains

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|$
(4.10)

for all $0\leq x\leq X$ and $0\leq y_{1}\leq Y_{i+1}$ but $(x,y_{1})=(0,0).$
By writing

$\alpha_{l-i-r}^{(i+1)}=\alpha_{l-i-r}^{(i)}q^{k-l+i+r}\ (1\leq r\leq l-i),$

one concludes from (4.10) that

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i+1)}x^{l-i-1}y_{1}^{k-l+i+1}+\cdots+\alpha_{0}^{(i+1)}y_{1}^{k}\|$
(4.11)

for all $0\leq x\leq X$ and $0\leq y_{1}\leq Y_{i+1}$ but $(x,y)=(0,0).$

Thus, one infers from induction that

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{0}^{(l)}y^{k}\|$

for all $0\leq x\leq X$ and $0\leq y\leq X^{1/(l+1)}$ but $(x,y)=(0,0),$ which
contradicts the conclusion of Proposition 4.1. Thus, we are forced to conclude
that

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\leq
1/H.$

∎

###### Proof of Corollary 1.2.

By change of variables,

$\displaystyle x=x_{1}-c_{k-1}y_{1}$ (4.12) $\displaystyle y=kc_{k}y_{1},$

one has

$\displaystyle\varphi(x,y)=\displaystyle\sum_{0\leq i\leq
k}c_{i}x^{i}y^{k-i}=\displaystyle\sum_{0\leq i\leq
k}c_{i}(x_{1}-c_{k-1}y_{1})^{i}(kc_{k}y_{1})^{k-i}=\displaystyle\sum_{\begin{subarray}{c}0\leq
i\leq k\\\ i\neq k-1\end{subarray}}c_{i}^{\prime}x_{1}^{i}y_{1}^{k-i},$ (4.13)

where each $c_{i}^{\prime}$ is obtained by the Binomial theorem. Furthermore,
notice from (4.12) that

$0\leq x_{1}\leq X/2\ \textrm{and}\ 0\leq y_{1}\leq
Y=\frac{X}{k(1+|c_{k}|)(1+|c_{k-1}|)}$

implies that

$|x|\leq X\ \textrm{and}\ |y|\leq X.$

Hence, one has

$\displaystyle\min_{\begin{subarray}{c}0\leq|x|,|y|\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}\varphi(x,y)+\alpha_{k-2}x^{k-2}y^{2}+\cdots+\alpha_{0}y^{k}\|$
$\displaystyle\leq$ $\displaystyle\ \min_{\begin{subarray}{c}0\leq x_{1}\leq
X/2\\\ 0\leq y_{1}\leq Y\\\
(x_{1},y_{1})\neq(0,0)\end{subarray}}\|\alpha_{k}x_{1}^{k}+\alpha_{k-2}^{\prime}x_{1}^{k-2}y_{1}^{2}+\cdots+\alpha_{0}^{\prime}y_{1}^{k}\|,$

where each $\alpha_{i}^{\prime}$ is obtained by (4.13) and the Binomial
theorem. Then, by applying Theorem 1.1 to the last expression, we conclude
that for any $\epsilon>0$, there exists $X(k,\epsilon)$ such that whenever
$X>X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq|x|,|y|\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}\varphi(x,y)+\alpha_{k-2}x^{k-2}y^{2}+\cdots+\alpha_{0}y^{k}\|\leq
X^{-\sigma+\epsilon},$

where $\sigma=k/(k-1)2^{1-k}.$

∎

##  5\. Proof of Theorem 1.3

Our goal in this section is to prove Theorem 1.3. We begin this section by
examining the corresponding diagonal problem. The following proposition is
useful for larger $k.$

######  Proposition 5.1.

Let $\alpha$, $\beta\in{\mathbb{R}}$ and $k\in{\mathbb{N}}$ with $k\geq$.
Then, for any $\epsilon>0$, there exists a real number $X(k,\epsilon)$ such
that whenever $X\geq X(k,\epsilon)$ and $Y\geq X(k,\epsilon)$ one has

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq
X^{-\sigma+\epsilon}Y^{-\sigma+\epsilon},$ (5.1)

where $\sigma=1/(k(k-1)).$

###### Proof.

Let $H=X^{\sigma-\epsilon}Y^{\sigma-\epsilon}$ with $\sigma=1/(k(k-1)).$
Suppose that we have

$\|\alpha x^{k}+\beta y^{k}\|>1/H$ (5.2)

for all $x$ and $y$. By Dirichlet’s approximation theorem, there exist
$q_{1},q_{2}\in{\mathbb{N}}$ and $a_{1},a_{2}\in{\mathbb{Z}}$ with
$(q_{1},a_{1})=(q_{2},a_{2})=1$ such that $q_{1},q_{2}\leq X^{k-1}H$ and

$\bigl{|}\alpha-\frac{a_{1}}{q_{1}}\bigr{|}\leq\frac{1}{q_{1}X^{k-1}H},\
\bigl{|}\beta-\frac{a_{2}}{q_{2}}\bigr{|}\leq\frac{1}{q_{2}X^{k-1}H}.$

By the same treatment in the proof of Proposition 4.1, we may assume that
$q_{1}$ and $q_{2}$ are greater than $X.$

By Dirichlet’s approximation theorem, there exists $r_{1}\in{\mathbb{N}}$ and
$b_{1}\in{\mathbb{Z}}$ with $(r_{1},b_{1})=1$ such that $|h\alpha-
b_{1}/r_{1}|\leq r_{1}^{-1}X^{1-k}$ and $r_{1}\leq X^{k-1}.$ Thus, by
[References, Theorem 5.2] depending on Vinogradov’s man value theorem, one has

$\displaystyle\sum_{1\leq x\leq X}e(h\alpha x^{k})\ll
X^{1+\epsilon/3}\biggl{(}\frac{1}{r_{1}}+\frac{1}{X}+\frac{r_{1}}{X^{k}}\biggr{)}^{\sigma},$

where $\sigma=1/(k(k-1)).$ Then, by the transference principle [References,
Lemma 14.1], one obtains

$\displaystyle\sum_{1\leq x\leq X}e(h\alpha x^{k})\ll
X^{1+\epsilon/3}\biggl{(}\frac{1}{r_{1}+X^{k}\|r_{1}h\alpha\|}+\frac{1}{X}+\frac{r_{1}+X^{k}\|r_{1}h\alpha\|}{X^{k}}\biggr{)}^{\sigma}.$

Since $r_{1}\leq X^{k-1}$, this bound is seen to be

$\displaystyle\sum_{1\leq x\leq X}e(h\alpha x^{k})\ll
X^{1-\sigma+\epsilon/3}+X^{1+\epsilon/3}\bigl{(}r_{1}+X^{k}\|r_{1}h\alpha\|\bigr{)}^{-\sigma}.$
(5.3)

Similarly, whenever $|h\beta-b_{2}/r_{2}|\leq r_{2}^{-1}Y^{1-k}$ and
$r_{2}\leq Y^{k-1}$ with $(r_{2},b_{2})=1,$ one obtains

$\displaystyle\sum_{1\leq y\leq y}e(h\beta y^{k})\ll
Y^{1-\sigma+\epsilon/3}+Y^{1+\epsilon/3}\bigl{(}r_{2}+Y^{k}\|r_{2}h\beta\|\bigr{)}^{-\sigma}.$
(5.4)

Notice here that $r_{1},\ r_{2}$ may depend on $h.$ Meanwhile, by Lemma 2.1,
it follows from our assumption $(\ref{5.22})$ that

$XY\ll\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\ 1\leq y\leq
Y\end{subarray}}e(h(\alpha x^{k}+\beta y^{k}))\biggr{|}.$ (5.5)

Then, by $(\ref{eq5.2})$ and $(\ref{eq5.3})$, we have

$\displaystyle\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\ 1\leq y\leq
Y\end{subarray}}e(h(\alpha x^{k}+\beta y^{k}))\biggr{|}$
$\displaystyle=\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq
X\end{subarray}}e(h(\alpha x^{k}))\biggr{|}\biggl{|}\displaystyle\sum_{1\leq
y\leq Y}e(h\beta y^{k})\biggr{|}$ $\displaystyle\ll\displaystyle\sum_{1\leq
h\leq H}(U_{1}(h)+U_{2}(h)+U_{3}(h)+U_{4}(h)),$

where

$\displaystyle U_{1}(h)=X^{1-\sigma+\epsilon/3}Y^{1-\sigma+\epsilon/3},$
$\displaystyle
U_{2}(h)=X^{1-\sigma+\epsilon/3}Y^{1+\epsilon/3}(r_{2}+Y^{k}\|r_{2}h\beta\|)^{-\sigma},$
$\displaystyle
U_{3}(h)=X^{1+\epsilon/3}Y^{1-\sigma+\epsilon/3}(r_{1}+X^{k}\|r_{1}h\alpha\|)^{-\sigma},$
$\displaystyle
U_{4}(h)=X^{1+\epsilon/3}Y^{1+\epsilon/3}(r_{1}+X^{k}\|r_{1}h\alpha\|)^{-\sigma}(r_{2}+Y^{k}\|r_{2}h\beta\|)^{-\sigma}$

in which $r_{1}$ and $r_{2}$ depend on $h.$

We shall show that

$\displaystyle\sum_{1\leq h\leq
H}(U_{1}(h)+U_{2}(h)+U_{3}(h)+U_{4}(h))\ll(XY)^{1-\epsilon/2}.$

First, since $H=X^{\sigma-\epsilon}Y^{\sigma-\epsilon},$ we have

$\displaystyle\sum_{1\leq h\leq H}U_{1}(h)\ll
X^{1-\epsilon/2}Y^{1-\epsilon/2}.$ (5.6)

Second, consider $\displaystyle\sum_{1\leq h\leq H}U_{2}(h).$ By Hölder’s
inequality, one has

$\displaystyle\sum_{1\leq h\leq H}U_{2}(h)$ $\displaystyle\leq
X^{1-\sigma+\epsilon/3}Y^{1+\epsilon/3}H^{1-\sigma}\left(\sum_{1\leq h\leq
H}(r_{2}+Y^{k}\|r_{2}h\beta\|)^{-1}\right)^{\sigma}$ (5.7)
$\displaystyle=X^{1-\sigma+\epsilon/3}Y^{1+\epsilon/3}H^{1-\sigma}\left(Y^{-k}\sum_{1\leq
h\leq
H}\min\left\\{\frac{Y^{k}}{r_{2}},\|r_{2}h\beta\|^{-1}\right\\}\right)^{\sigma}.$

Note that since $r_{2}\leq Y^{k-1}$, one has

$\displaystyle\displaystyle\sum_{1\leq h\leq
H}\min\left\\{\frac{Y^{k}}{r_{2}},\|r_{2}h\beta\|\right\\}$
$\displaystyle\leq\displaystyle\sum_{1\leq r\leq
Y^{k-1}}\displaystyle\sum_{1\leq h\leq
H}\min\left\\{\frac{Y^{k}}{r},\|rh\beta\|^{-1}\right\\}$
$\displaystyle\leq\displaystyle\sum_{1\leq r\leq
Y^{k-1}}\displaystyle\sum_{1\leq h\leq
H}\min\left\\{\frac{Y^{k}H}{rh},\|rh\beta\|^{-1}\right\\}.$

By a standard divisor estimate, this bound is seen to be

$\displaystyle\ll X^{\epsilon}\displaystyle\sum_{1\leq n\leq
Y^{k-1}H}\min\left\\{\frac{Y^{k}H}{n},\|n\beta\|\right\\}.$

By [References, Lemma 2.2], this bound is

$\ll
X^{\epsilon}Y^{k}H\left(\frac{1}{q_{2}}+\frac{1}{Y}+\frac{q_{2}}{Y^{k}H}\right)\leq
X^{\epsilon}Y^{k-1}H,$ (5.8)

where we have used inequalities $|\beta-a_{2}/q_{2}|\leq
q_{2}^{-1}X^{1-k}H^{-1}$ and $q_{2}>X.$ Hence, on substituting (5.8) into
(5.7), we find that

$\displaystyle\sum_{1\leq h\leq H}U_{2}(h)\ll
X^{1-\sigma+\epsilon/2}Y^{1-\sigma+\epsilon/2}H\ll
X^{1-\epsilon/2}Y^{1-\epsilon/2}.$ (5.9)

Third, consider $\displaystyle\sum_{1\leq h\leq H}U_{3}(h)$. By the same
treatment with just above, we obtain

$\displaystyle\sum_{1\leq h\leq H}U_{3}(h)$
$\displaystyle=X^{1+\epsilon/3}Y^{1-\sigma+\epsilon/3}H^{1-\sigma}\left(X^{-k}\sum_{1\leq
h\leq
H}\min\left\\{\frac{X^{k}}{r_{1}},\|r_{1}h\alpha\|^{-1}\right\\}\right)^{\sigma}$
(5.10) $\displaystyle\ll
X^{1+\epsilon/3}Y^{1-\sigma+\epsilon/3}H^{1-\sigma}\left(X^{\epsilon}H\left(\frac{1}{q_{1}}+\frac{1}{X}+\frac{q_{1}}{X^{k}H}\right)\right)^{\sigma}$
$\displaystyle\ll X^{1-\epsilon/2}Y^{1-\epsilon/2}.$

Finally, consider $\displaystyle\sum_{1\leq h\leq H}U_{4}(h).$ By Hölder’s
inequality, we have

$\displaystyle\sum_{1\leq h\leq
H}U_{4}(h)\ll(XY)^{1+\epsilon/3}H^{1-2\sigma}A^{\sigma}B^{\sigma},$ (5.11)

where

$A=X^{-k}\displaystyle\sum_{1\leq h\leq
H}\min\left\\{\frac{X^{k}}{r_{1}},\|r_{1}h\alpha\|^{-1}\right\\}$

and

$B=Y^{-k}\displaystyle\sum_{1\leq h\leq
H}\min\left\\{\frac{Y^{k}}{r_{2}},\|r_{2}h\beta\|^{-1}\right\\}.$

By the same treatment in the case $\displaystyle\sum_{1\leq h\leq H}U_{2}(h)$
and $\displaystyle\sum_{1\leq h\leq H}U_{3}(h)$, one infers from (5.11) that

$\displaystyle\sum_{1\leq h\leq
H}U_{4}(h)\ll(XY)^{1+\epsilon/3}H^{1-2\sigma}(Y^{\epsilon}HY^{-1})^{\sigma}(X^{\epsilon}HX^{-1})^{\sigma}\ll(XY)^{1-\epsilon/2}.$
(5.12)

Hence, by (5.6), (5.9), (5.10) and (5.12), we have

$\displaystyle\sum_{1\leq h\leq
H}(U_{1}(h)+U_{2}(h)+U_{3}(h)+U_{4}(h))\ll(XY)^{1-\epsilon/2}.$

This contradicts $(\ref{eq5.4})$ stemming from our assumption that

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|>1/H.$

Therefore, we are forced to conclude that

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq 1/H.$

∎

We shall reduce the problem in Theorem 1.3 to this diagonal problem by
exploiting the same argument in Theorem 1.1 with Lemma 3.1 replaced by Lemma
3.2.

###### Proof of Theorem 1.3.

Let $H=X^{\sigma-\epsilon}$ with $\sigma=\frac{2}{k(k-1)+\rho(k,l)}.$ Suppose
that

$1/H<\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\cdots+\alpha_{0}y^{k}\|$ (5.13)

for all $1\leq x,y\leq X.$ From (5.13), we shall derive a lower bound of
fractional parts of polynomial having fewer terms by an inductive arguments.
Specifically, at the $i$-th step, with $0\leq i\leq l,$ we shall show that for
all $1\leq x\leq X$ and $1\leq y\leq Y_{i}$ with

$Y_{i}=X^{1-(2^{l+1}(k-l)+2^{l}(k-l+1)+\cdots+2^{l-i+2}(k-l+i-1))\sigma},$
(5.14)

there exist coefficients $\alpha_{l-i}^{(i)},\ldots,\alpha_{0}^{(i)}$,
suitable integral multiples of $\alpha_{l-i},\ldots,\alpha_{0}$, respectively,
such that

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|.$
(5.15)

The case $i=0$ obviously follows from (5.13). Assume that (5.15) holds for a
particular $i\leq l-1.$ We shall show that (5.15) holds for $i$ replaced by
$i+1.$

Since $\mathcal{A}(Y_{i},R)\subseteq[1,Y_{i}]\cap{\mathbb{Z}}$, the assumption
(5.15) implies

$\frac{1}{H}\ll\min_{\begin{subarray}{c}1\leq x\leq X\\\
y\in\mathcal{A}(Y_{i},R)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|.$
(5.16)

Thus, by Lemma 2.1 and the fact that $|\mathcal{A}(Y_{i},R)|\asymp Y_{i},$ one
has

$XY_{i}\ll\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\
y\in\mathcal{A}(Y_{i},R)\end{subarray}}e(h(\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}))\biggr{|}$
(5.17)

By Dirichlet’s theorem, there exist $q\in{\mathbb{N}},a\in{\mathbb{Z}}$ with
$(q,a)=1$ such that

$q\leq HX^{l-i-2^{l-i+1}(k-l+i)\sigma+\eta}Y_{i}^{k-l+i}$

and

$\|q\alpha_{l-i}^{(i)}\|\leq\frac{X^{-\eta}X^{2^{l-i+1}(k-l+i)\sigma}}{HX^{l-i}Y_{i}^{k-l+i}}.$
(5.18)

Let $U=X^{2^{l-i+2}\sigma}.$ To apply Lemma 3.2 with $Z=U$ and $Y=Y_{i}$, we
verify here that $U$ and $Y_{i}$ satisfy the hypotheses of Lemma 3.2, namely,

$U\leq\min\\{(HX^{l-i})^{1/(2(k-l+i))},Y_{i}\\}.$

We first verify that $U\leq Y_{i}.$ Since we have $\rho(k,l)\leq k(k-1)$, one
has $2^{l+1}(k-l)\leq k(k-1),$ and thus

$2^{l}\leq k(k-1)/(2(k-l))\leq k-1,$ (5.19)

since by an obvious restriction $l\leq\lfloor\frac{k}{2}\rfloor.$ Note also
that

$\sigma=2/(k(k-1)+\rho(k,l))\leq\rho(k,l)^{-1},$ (5.20)

since $\rho(k,l)\leq k(k-1).$ Hence, on recalling (5.14), by
$(\ref{eqeq5.15})$ and $(\ref{eqeq5.16})$, one finds that

$\displaystyle U=X^{2^{l-i+2}\sigma}\leq X^{2^{l+2}\rho(k,l)^{-1}}\leq
X^{1-(\rho(k,l)-2^{2}(k-1))\rho(k,l)^{-1}}\leq
X^{1-(\rho(k,l)-2^{2}(k-1))\sigma}\leq Y_{i}.$ (5.21)

We turn to verify the other hypothesis. Since we have $\rho(k,l)\leq k(k-1)/7$
and by recalling the definition of $\rho(k,l)$, one has

$\sigma=\frac{2}{k(k-1)+\rho(k,l)}\leq\frac{1}{4\rho(k,l)}\leq\frac{1}{2^{l-i+3}(k-l+i)}.$

Hence, we have

$U=X^{2^{l-i+2}\sigma}\leq X^{1/(2(k-l+i))}$ (5.22)

Thus, from (5.21) and (5.22), we verified that

$U\leq\min\\{(HX^{l-i})^{1/(2(k-l+i))},Y_{i}\\}.$

Therefore, by applying Lemma 3.2 to the right hand side on $(\ref{eq5.14})$,
we obtain

$XY_{i}\ll\frac{H(XY_{i})^{1+\epsilon}}{(q+HX^{l-i}Y_{i}^{k-l+i}|q\alpha_{l}-a|)^{1/(2^{l-i+1}(k-l+i))}}+H(XY_{i})^{1+\epsilon}U^{-1/2^{l-i+2}}.$
(5.23)

This implies

$q\leq X^{2^{l-i+1}(k-l+i)\sigma}.$ (5.24)

By applying the triangle inequality and putting $y=qy_{1}$ with

$1\leq y_{1}\leq Y_{i}X^{-2^{l-i+1}(k-l+i)\sigma}=Y_{i+1},$

one finds that

$\displaystyle\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|$
(5.25)
$\displaystyle\leq\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|+\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|.$

By applying the triangle inequality and (5.18), one has

$\displaystyle\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|$
$\displaystyle\leq x^{l-i}(qy_{1})^{k-l+i-1}y_{1}\|\alpha_{l-i}^{(i)}q\|$
(5.26) $\displaystyle\leq
X^{l-i}Y_{i}^{k-l+i}X^{-2^{l-i+1}(k-l+i)\sigma}\frac{X^{-\eta}X^{2^{l-i+1}(k-l+i)\sigma}}{HX^{l-i}Y_{i}^{k-l+i}}$
$\displaystyle\leq\frac{X^{-\eta}}{H}.$

Thus, by $(\ref{eq5.11}),(\ref{eq5.30})$ and $(\ref{eq5.31}),$ one has

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|$
(5.27)

for all $1\leq x\leq X$ and $1\leq y_{1}\leq Y_{i+1}.$ By writing

$\alpha_{l-i-r}^{(i+1)}=\alpha_{l-i-r}^{(i)}q^{k-l+i+r}\ (1\leq r\leq l-i),$

one concludes from (5.27) that

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i+1)}x^{l-i-1}y_{1}^{k-l+i+1}+\cdots+\alpha_{0}^{(i+1)}y_{1}^{k}\|$

for all $1\leq x\leq X$ and $1\leq y_{1}\leq Y_{i+1}.$ Thus, this confirms the
inductive step.

Thus, one infers by induction that

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{0}^{(l)}y^{k}\|$

for all $1\leq x\leq X$ and $1\leq y\leq Y_{l}.$ This contradicts the
conclusion of Proposition 5.1. Thus, we are forced to conclude that

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\leq
1/H.$

∎

##  6\. Proof of Theorem 1.4

Our goal in this section is to prove Theorem 1.4. We begin this section by
examining the corresponding diagonal problem. To prove following Proposition
6.2, we require the minor arc estimates in [References, Corollary 2].
Throughout this section, we take $R=X^{\eta}$ with $\eta$ positive sufficient
small. We state here this corollary without proof as a proposition.

######  Proposition 6.1.

Let $\mathfrak{m}_{\lambda}$ denote the set of $\alpha\in{\mathbb{R}}$ such
that whenever $a\in{\mathbb{Z}},$ $q\in{\mathbb{N}},$ $(a,q)=1$ and
$|\alpha-a/q|\leq q^{-1}X^{\lambda-k},$ then $q>X^{\lambda}R$. Then there is a
natural number $k_{0}(\epsilon)$ with the following property. When $k\geq
k_{0}(\epsilon),$ there are real numbers $\lambda=\lambda(k)$, $\sigma(k)$ and
$C>0$ with

$\frac{\log\log k}{\log k}\ll 1-\lambda\ll\frac{\log\log k}{\log k}\ \
\textrm{and}\ \ \sigma(k)^{-1}=k(\log k+C\log\log k),$

and such that

$\displaystyle\sum_{x\in\mathcal{A}(X,R)}e(\alpha x^{k})\ll
X^{1-\sigma(k)+\epsilon}.$

###### Proof.

See [References, Corollary 2]. ∎

To state the following proposition, we exploit the exponent
$\lambda=\lambda(k)$ defined in Proposition 6.1.

######  Proposition 6.2.

Let $\alpha,\beta\in{\mathbb{R}}$ and $k\in{\mathbb{N}}$. Define

$\sigma=\frac{1}{k\log k+Ck\log\log k},\ \ \
\sigma_{0}=\frac{\sigma}{1-\lambda-\sigma}.$

Let $X$ and $Y$ be real numbers sufficiently large in terms of $k$ and
$\epsilon$ with $X^{\sigma_{0}}\leq Y\leq X.$ Then, one has

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq
X^{-\sigma+\epsilon}Y^{-\sigma+\epsilon}.$

###### Proof.

Let $H=X^{\sigma-\epsilon}Y^{\sigma-\epsilon}.$ Notice that
$\max\\{X^{\lambda-1},Y^{\lambda-1}\\}\leq 1/H.$ Suppose that there exists no
$x,y$ satisfying $\|\alpha x^{k}+\beta y^{k}\|\leq 1/H.$

Suppose first that there exist $q\in{\mathbb{N}},\ a\in{\mathbb{Z}},\
h\in[1,H]\cap{\mathbb{Z}}$ with $(q,a)=1$ such that $q\leq X^{\lambda}R$ and

$\bigl{|}h\alpha-\frac{a}{q}\bigr{|}\leq q^{-1}X^{\lambda-k}.$ (6.1)

Since we have $Y\leq X$, we find that $hq\leq HX^{\lambda}R\leq X$. Then, by
putting $y=qh$, it follows from $(\ref{62})$ that

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta
y^{k}\|\leq\|\alpha(qh)^{k}\|\leq(HX^{\lambda}R)^{k-1}\|\alpha qh\|\leq
X^{k-1}X^{\lambda-k}\leq 1/H,$

which contradicts our assumption.

Suppose next that there exist $q\in{\mathbb{N}},\ a\in{\mathbb{Z}},\
h\in[1,H]\cap{\mathbb{Z}}$ with $(q,a)=1$ such that $q\leq Y^{\lambda}R$ and

$\bigl{|}h\beta-\frac{a}{q}\bigr{|}\leq Y^{\lambda-k}.$ (6.2)

Since we have $X^{\sigma_{0}}\leq Y,$ we find that $qh\leq HY^{\lambda}R\leq
Y.$ Then, by putting $y=qh$, it follows from (6.2) that

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\gamma
y^{k}\|\leq\|\gamma(qh)^{k}\|\leq(HY^{\lambda}R)^{k-1}\|\gamma qh\|\leq
Y^{k-1}Y^{\lambda-k}\leq 1/H,$

which contradicts our assumption.

Thus, on recalling the definition of $\mathfrak{m}_{\lambda}$ in the statement
of proposition 6.1. We may assume that for all $h\in[1,H]\cap{\mathbb{Z}},$
one has $h\alpha,h\beta\in\mathfrak{m}_{\lambda}.$ Meanwhile, it follows from
our assumption by Lemma 2.1 that

$XY\ll\displaystyle\sum_{1\leq h\leq
H}\bigl{|}\displaystyle\sum_{\begin{subarray}{c}x\in\mathcal{A}(X,R)\\\
y\in\mathcal{A}(Y,R)\end{subarray}}e(h(\alpha x^{k}+\beta y^{k}))\bigr{|}.$

Therefore, on writing

$\bigl{|}\displaystyle\sum_{\begin{subarray}{c}x\in\mathcal{A}(X,R)\\\
y\in\mathcal{A}(Y,R)\end{subarray}}e(h(\alpha x^{k}+\beta
y^{k}))\bigr{|}=\bigl{|}\displaystyle\sum_{x\in\mathcal{A}(X,R)}e(h(\alpha
x^{k}))\bigr{|}\bigl{|}\displaystyle\sum_{y\in\mathcal{A}(Y,R)}e(h(\beta
y^{k}))\bigr{|},$

by applying Proposition 6.1, we find that

$\displaystyle\sum_{1\leq h\leq
H}\bigl{|}\displaystyle\sum_{\begin{subarray}{c}x\in\mathcal{A}(X,R)\\\
y\in\mathcal{A}(Y,R)\end{subarray}}e(h(\alpha x^{k}+\beta
y^{k}))\bigr{|}\ll(XY)^{1-\epsilon},$

which contradicts our assumption. Hence, we are forced to have

$\min_{\begin{subarray}{c}0\leq x\leq X\\\ 0\leq y\leq Y\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha x^{k}+\beta y^{k}\|\leq 1/H.$

∎

###### Proof fo Theorem 1.4.

Let $H=X^{\sigma-\epsilon}$ with $\sigma=2/(k\log k+\rho(k,l)+Ck\log\log k)$.
Suppose that

$1/H\leq\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\cdots+\alpha_{0}y^{k}\|$
(6.3)

for all $1\leq x,y\leq X.$ From (6.3), we shall derive a lower bound of
fractional parts of polynomial having fewer terms by an inductive arguments.
Specifically, at the $i$-th step, with $0\leq i\leq l,$ we shall show that for
all $1\leq x\leq X$ and $1\leq y\leq Y_{i}$ with

$Y_{i}=X^{1-(2^{l+1}(k-l)+2^{l}(k-l+1)+\cdots+2^{l-i+2}(k-l+i-1))\sigma},$

there exist coefficients $\alpha_{l-i}^{(i)},\ldots,\alpha_{0}^{(i)}$,
suitable integral multiples of $\alpha_{l-i},\ldots,\alpha_{0}$, respectively,
such that

$\frac{1}{H}\ll\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|.$
(6.4)

The case $i=0$ obviously follows from (6.3). Assume that (6.4) holds for a
particular $i\leq l-1.$ We shall show that (6.4) holds for $i$ replaced by
$i+1.$

Since $\mathcal{A}(Y_{i},R)\subseteq[1,Y_{i}]\cap{\mathbb{Z}},$ the assumption
(6.4) implies

$\frac{1}{H}\ll\min_{\begin{subarray}{c}1\leq x\leq X\\\
y\in\mathcal{A}(Y_{i},R)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|.$
(6.5)

Thus, by Lemma 2.1 and the fact that $|\mathcal{A}(Y_{i},R)|\asymp Y_{i},$ one
has

$XY_{i}\ll\displaystyle\sum_{1\leq h\leq
H}\biggl{|}\displaystyle\sum_{\begin{subarray}{c}1\leq x\leq X\\\
y\in\mathcal{A}(Y_{i},R)\end{subarray}}e(h(\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}))\biggr{|}$
(6.6)

By Dirichlet’s theorem, there exist $q\in{\mathbb{N}},a\in{\mathbb{Z}}$ with
$(q,a)=1$ such that

$q\leq HX^{l-i-2^{l-i+1}(k-l+i)\sigma+\eta}Y_{i}^{k-l+i}$

and

$\|q\alpha_{l-i}^{(i)}\|\leq\frac{X^{-\eta}X^{2^{l-i+1}(k-l+i)\sigma}}{HX^{l-i}Y_{i}^{k-l+i}}.$
(6.7)

Let $U=X^{2^{l-i+2}\sigma}.$ By the same treatment in the proof of Theorem 1.3
with $k(k-1)$ replaced by $k\log k$, we find that this $U$ and $Y_{i}$ satisfy
the hypotheses of Lemma 3.2. Therefore, by applying Lemma 3.2 to the right
hand side on (6.6), we obtain

$XY_{i}\ll\frac{H(XY_{i})^{1+\epsilon}}{(q+HX^{l-i}Y_{i}^{k-l+i}|q\alpha_{l}-a|)^{1/(2^{l-i+1}(k-l+i))}}+H(XY_{i})^{1+\epsilon}U^{-1/2^{l-i+2}}.$

This implies

$q\leq X^{2^{l-i+1}(k-l+i)\sigma}.$ (6.8)

By applying the triangle inequality and putting $y=qy_{1}$ with

$1\leq y_{1}\leq Y_{i}X^{-2^{l-i+1}(k-l+i)\sigma}=Y_{i+1},$

one finds that

$\displaystyle\|\alpha_{k}x^{k}+\alpha_{l-i}^{(i)}x^{l-i}y^{k-l+i}+\cdots+\alpha_{0}^{(i)}y^{k}\|$
(6.9)
$\displaystyle\leq\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|+\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|.$

By applying the triangle inequality and (6.7), one has

$\displaystyle\|\alpha_{l-i}^{(i)}x^{l-i}(qy_{1})^{k-l+i}\|$
$\displaystyle\leq x^{l-i}(qy_{1})^{k-l+i-1}y_{1}\|\alpha_{l-i}^{(i)}q\|$
(6.10) $\displaystyle\leq
X^{l-i}Y_{i}^{k-l+i}X^{-2^{l-i+1}(k-l+i)\sigma}\frac{X^{-\eta}X^{2^{l-i+1}(k-l+i)\sigma}}{HX^{l-i}Y_{i}^{k-l+i}}$
$\displaystyle\leq\frac{X^{-\eta}}{H}.$

Thus, by $(\ref{eq6.6}),(\ref{eq6.11})$ and $(\ref{eq6.12}),$ one has

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i)}x^{l-i-1}(qy_{1})^{k-l+i+1}+\cdots+\alpha_{0}^{(i)}(qy_{1})^{k}\|$
(6.11)

for all $1\leq x\leq X$ and $1\leq y_{1}\leq Y_{i+1}.$ By writing

$\alpha_{l-i-r}^{(i+1)}=\alpha_{l-i-r}^{(i)}q^{k-l+i+r}\ (1\leq r\leq l-i),$

one concludes from (6.11) that

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{l-i-1}^{(i+1)}x^{l-i-1}y_{1}^{k-l+i+1}+\cdots+\alpha_{0}^{(i+1)}y_{1}^{k}\|$

for all $1\leq x\leq X$ and $1\leq y_{1}\leq Y_{i+1}.$

Thus, one infers from induction that

$1/H\ll\|\alpha_{k}x^{k}+\alpha_{0}^{(l)}y^{k}\|$ (6.12)

for all $1\leq x\leq X$ and $1\leq y\leq Y_{l}.$ Since
$Y_{l}=X^{1-\rho(k,l)\sigma}$ and $\rho(k,l)\leq k\log k,$ a modicum
computation leads to the relation that $Y_{l}\geq X^{\sigma_{0}}.$ Therefore,
the inequality (6.12) contradicts the conclusion of Proposition 6.2. Thus, we
are forced to conclude that

$\min_{\begin{subarray}{c}0\leq x,y\leq X\\\
(x,y)\neq(0,0)\end{subarray}}\|\alpha_{k}x^{k}+\alpha_{l}x^{l}y^{k-l}+\alpha_{l-1}x^{l-1}y^{k-l+1}+\cdots+\alpha_{0}y^{k}\|\leq
1/H.$

∎

## Appendix A. Estimations in the proof of Proposition 3.4

Let $\alpha,\ N$ be positive real numbers, and $k$ and $l$ be natural numbers
with $k-l\geq 2$. Then, one has following:

$\displaystyle(1)\ \displaystyle\sum_{1\leq n\leq N}\frac{1}{1+\alpha
n^{k-l}}\ll\frac{N}{(1+N^{k-l}\alpha)^{1/(k-l)}}$ $\displaystyle(2)\
\displaystyle\sum_{1\leq n\leq N}\frac{1}{(1+\alpha
n^{k-l})^{1/(k-l)}}\ll\frac{N\log N}{(1+N^{k-l}\alpha)^{1/(k-l)}}$
$\displaystyle(3)\ \displaystyle\sum_{1\leq n\leq N}\frac{1}{(1+\alpha
n)^{2/(k-l)}}\ll\frac{N\log N}{(1+N\alpha)^{2/(k-l)}}$

###### Proof of (1).

We have

$\displaystyle\sum_{1\leq n\leq N}\frac{1}{1+\alpha
n^{k-l}}\ll\displaystyle\sum_{1\leq n\leq N}\min\\{1,\alpha^{-1}n^{-k+l}\\}\ll
N+\alpha^{-1/(k-l)},$

where $N$ and $\alpha^{-1/(k-l)}$ are derived from the contributions $1$ and
$\alpha^{-1}n^{-k+l}$, respectively. ∎

###### Proof of (2).

We have

$\displaystyle\sum_{1\leq n\leq N}\frac{1}{1+\alpha
n^{k-l}}\ll\displaystyle\sum_{1\leq n\leq
N}\min\\{1,\alpha^{-1/(k-l)}n^{-1}\\}\ll N+\alpha^{-1/(k-l)}\log N,$

where $N$ and $\alpha^{-1/(k-l)}\log N$ are derived from the contributions $1$
and $\alpha^{-1/(k-l)}n^{-1}$, respectively. ∎

###### Proof of (3).

We have

$\displaystyle\sum_{1\leq n\leq N}\frac{1}{1+\alpha
n^{k-l}}\ll\displaystyle\sum_{1\leq n\leq N}\min\\{1,(\alpha
n)^{-2/(k-l)}\\}\ll N+\frac{N\log N}{\alpha^{2/(k-l)}N^{2/(k-l)}},$

where $N$ and $\frac{N\log N}{\alpha^{2/(k-l)}N^{2/(k-l)}}$ are derived from
the contributions $1$ and $(\alpha n)^{-2/(k-l)}$, respectively. ∎

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