Too much spacing of the paragraph before a long equation in Align environment

Question

I am typing my paper and I get a very awkward page before an Align environment which contains a long equation (there are too much spacing in the first page, the second page contains only that long equations). Does any one to help me to clear spacing appear in first page?

Here is my code

\documentclass[a4paper,12pt,intlimits,oneside]{amsart}

%\usepackage{showkeys}

\usepackage{enumerate}
\usepackage{amsfonts,amsmath}
%\usepackage{amsthm}showkeys,

\usepackage{latexsym,amssymb,amsthm}



\textwidth14cm \textheight21cm \evensidemargin.2cm
\oddsidemargin.2cm


\begin{document} 
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{example}{Example}[section]
\numberwithin{equation}{section}
\newtheorem{Theorem}{Theorem}[section]
\newtheorem{Lemma}{Lemma}[section]
\newtheorem{Proposition}{Proposition}[section]
\newtheorem{Remark}{Remark}[section]
\newtheorem{Corollary}{Corollary}[section]
\newtheorem{Definition}{Definition}[section]
\newtheorem{Example}{Example}[section]

In order to prove the converse of the theorem, we first need the following lemma.
\begin{lemma}\label{mainlem1} Let $w\in {\mathcal W}_\alpha$, $\alpha>-d$ and $\varepsilon>0$. Then the function 
\[
f_{p,\varepsilon}(x)=
\begin{cases}
0\qquad\qquad\quad\;\;\text{if}\quad|x|\leq1&\\
|x|^{-\frac{d+\alpha}p-\varepsilon}\qquad\text{if}\quad|x|\geq1,
\end{cases}
\]
belongs to $L^p_w\left(\mathbb R^d\right)$ and $\|f_{p,\varepsilon}\|_{p,w}=\left(\frac{w(S_d)}{p\varepsilon}\right)^{1/p}$.
\end{lemma}
Since the proof of the lemma is straightforward, we omit it. Now we shall prove the theorem. Let $\varepsilon$ be  an arbitrary positive number and  for each $k=1,\ldots,m$ we set $\varepsilon_k=\frac{p\varepsilon}{p_k}$ and
\[
f_{p_k,\varepsilon_k}(x)=
\begin{cases}
0\qquad\qquad\qquad\;\text{if}\quad|x|\leq1&\\
|x|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\qquad\text{if}\quad|x|\geq1.
\end{cases}
\]
Lemma \ref{mainlem1} implies that $f_{p_k,\varepsilon_k}\in L^{p_k}_{\omega_k}\left(\mathbb R^d\right)$ and 
\[
\|f_{p_k,\varepsilon_k}\|_{p_k,\omega_k}=\left(\frac{\omega_k(S_d)}{p_k\varepsilon_k}\right)^{1/p_k}=\left(\frac{\omega_k(S_d)}{p\varepsilon }\right)^{1/p_k}>0,
\]
for each $k=1,\ldots,m$. For each $x\in\mathbb R^d$ which $|x|\geq1$, let
\begin{align*}
S_x=\bigcap\limits_{k=1}^m\{t\in[0,1]^n:\;|s_k(t)x|>1\}.
\end{align*}
From the assumption 
$|s_k(t_1,\ldots,t_n)|\geq \min\{t_1^\beta,\ldots,t_n^\beta\}$ a.e $t=(t_1,\ldots,t_n)\in[0,1]^n$, there exists a null subset $E$ of $[0,1]^n$ so that $S_x$ contains $\left[1/|x|^{1/\beta},1\right]^n\setminus E$. From (\ref{sec2eq1}), we have 
%\leavevmode\vspace{\dimexpr-\abovedisplayskip-\baselineskip\relax}
\begin{align*}
\;&\|U^{m,n}_{\psi,\overrightarrow{s}}\left(f_{p_1,\varepsilon_1},\ldots,f_{p_m,\varepsilon_m}\right)\|_{L^p_\omega\left(\mathbb R^d\right)}^p=\int_{\mathbb R^d}\left|\int_{[0,1]^n}\left(\prod\limits_{k=1}^mf_k\left(s_k(t)x\right)\right)\psi(t)dt\right|^p\omega(x)dx\\
=\;& \int_{\mathbb R^d}\prod_{k=1}^m|x|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\omega(x)\left|\int_{S_x}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|^pdx\\
\geq\;& \int_{|x|\geq \varepsilon^{-\beta}}|x|^{-(d+\alpha+\varepsilon p)}\omega(x)\left|\,\int_{[\varepsilon,1]^n}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|^pdx\\
=\;& \varepsilon^{p\beta\varepsilon}\left(\int_{|x|\geq1}|x|^{-(d+\alpha+\varepsilon p)}\omega(x)dx\right)\left|\,\int_{[\varepsilon,1]^n}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|^p \\
=\;& \varepsilon^{p\beta\varepsilon}\left(\frac{\omega(S_d)}{\varepsilon p}\right)\left|\,\int_{[\varepsilon,1]^n}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|^p \\
\geq\;& \varepsilon^{p\beta\varepsilon}\prod_{k=1}^m\|f_{p_k,\epsilon_k}\|_{p_k,\varepsilon_k}^{p/p_k}\left|\,\int_{[\varepsilon,1]^n}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|^p. \\
\end{align*}
This implies that 
\begin{align*}
\left\|U^{m,n}_{\psi,\overrightarrow{s}}\right\|_{L^{p_1}_{\omega_1}\left(\mathbb R^d\right)\times\cdots\times L^{p_m}_{\omega_m}\left(\mathbb R^d\right)\to L^{p}_{\omega}\left(\mathbb R^d\right)}\geq  \varepsilon^{\beta\varepsilon}\cdot\left|\,\int_{[\varepsilon,1]^n}\prod_{k=1}^m|s_k(t)|^{-\frac{d+\alpha_k}{p_k}-\varepsilon_k}\psi(t)dt\right|.
\end{align*} 


\end{document}