# Sample path properties of multidimensional integral with respect to
stochastic measure

Boris Manikin1 and  Vadym Radchenko1 1Department of Mathematical Analysis,
Taras Shevchenko National University of Kyiv, 01601 Kyiv, Ukraine.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The integral with respect to a multidimensional stochastic measure, for which
we assume only $\sigma$-additivity in probability, is studied. The continuity
and differentiability of its realizations are established.

###### Key words and phrases:

Stochastic measure, trajectories of random functions, multivariable Haar
functions, Besov space

###### 2020 Mathematics Subject Classification:

Primary: 60H05; Secondary: 60G60, 60G17

##  1\. Introduction

The main purpose of this article is studying the regularity property of the
integral of a deterministic function with respect to stochastic measure $\mu$,
where $\mu$ is defined on Borel subsets of $[0,1]^{d}$. In basic statements of
the paper, for $\mu$ we assume only $\sigma$-additivity in probability. We
study the regularity with respect to a parameter and with respect to the upper
limit of the integral. Case $d=1$ was considered in [20, Section 2.3], but
methods of [20] do not work for $d\geq 2$. In our paper, we study the integral
using the Fourier – Haar expansion of the integrand.

Sample path behaviour of various stochastic processes was studied in many
works; we mention only some of them. The conditions for sample boundedness and
continuity of $\alpha$-stable processes were established in [22, Chapter 10].
Hölder continuity of harmonizable operator scaling stable random field is
given by Corollary 5.5 [5]. In Theorem 4.1 from [18] the uniform modulus of
continuity for some self-similar $S\alpha S$ random field was obtained, while
the upper bound of modulus of continuity of stable random fields is given by
Proposition 5.1 and Corollary 5.3 [6]. These results, for example, imply
Hölder continuity of mentioned fields.

Conditions of continuity of Gaussian random fields may be found in Theorems
3.3.3, 3.4.1 and 3.4.3 [1]. Hölder continuity of centered Gaussian random
fields under certain conditions is proved in [28, Theorem 4.2]. Theorem 2.1
[29] gives an estimate of modulus of continuity of general random fields on
metric space, these results are applied to harmonizable stable random fields.

In our article, we consider stochastic processes which are integrals with
respect to a general stochastic measure. The definition of these measures,
their properties and examples as well as information about Fourier – Haar
series may be found in Section 2. The continuity and differentiability of
sample paths of parameter-dependent integral are established in Section 3. The
continuity of realizations of the integral as a function of the upper limit is
studied in Section 4.

##  2\. Preliminaries

###  2.1. Stochastic measures

In this subsection, we give basic information concerning stochastic measures
in a general setting. In statements of Sections 3 and 4, this set function is
defined on Borel subsets of $[0,1]^{d}$.

Let ${\sf{L}}_{0}={\sf{L}}_{0}(\Omega,{\mathcal{F}},{{\sf{P}}})$ be the set of
all real-valued random variables defined on the complete probability space
$(\Omega,{\mathcal{F}},{{\sf{P}}})$. Convergence in ${\sf{L}}_{0}$ means the
convergence in probability. Let ${{\sf{X}}}$ be an arbitrary set and
${\mathcal{B}}$ a $\sigma$-algebra of subsets of ${{\sf{X}}}$.

######  Definition 1.

A $\sigma$-additive mapping $\mu:\ {\mathcal{B}}\to{\sf{L}}_{0}$ is called
stochastic measure (SM).

In other words,

$\displaystyle\mu(A\cup B)=\mu(A)+\mu(B)\text{ for }A\cap B=\emptyset,$
$\displaystyle\mu(A_{n})\stackrel{{\scriptstyle{\sf{P}}}}{{\to}}0\text{ for
}A_{n}\downarrow\emptyset.$ (1)

We do not assume the moment existence or martingale properties for SM. We can
say that $\mu$ is an ${\sf{L}}_{0}$–valued measure.

We note the following examples of SMs in multidimensional spaces. All
measures, if nothing else is mentioned, are considered on Borel
$\sigma$-algebra of sets in $[0,1]^{d}$.

######  Example 1.

Let $\mu$ be a random orthogonal measure — it is defined, for example, in [9,
Section 5.3] and [14, Section 2.3] — with finite structural function. Then
$\mu$ is an SM in the sense of Definition 1; condition (1) follows from [9,
Section 5.3, Theorem 1(d)]. For example, the set function
$\mu(A)=\int_{[0,1]^{d}}{\mathbf{1}}_{A}(t)\,{d}W(t)$, where we take multiple
Wiener-Itô integral (see its properties, for example, in [17, Section 1.1]),
is a random orthogonal measure with structural function
${\sf{m}}(A)=d!\lambda_{d}(A)$, where $\lambda_{d}$ is the $d$-dimensional
Lebesgue measure.

######  Example 2.

Let $\mu$ be an independently scattered $\alpha$-stable random measures for
$\alpha\in(0,2]$ with finite control measure. According to the definition —
see [22, Definition 3.3.1] — for disjoint Borel sets $\\{A_{n}:n\geq 1\\}$ the
following holds:

$\mu(\cup_{n=1}^{\infty}A_{n})=\sum_{n=1}^{\infty}\mu(A_{n})\ \text{a. s.}$

Therefore, $\mu$ is an SM.

######  Example 3.

Consider the set function $\mu:\mathcal{B}([0,1]^{d})\to{\sf{L}}_{0}$ such
that

$\mu(A)=\int_{\mathbb{R}^{d}}{\mathbf{1}}_{A}(t)\,{d}Z_{H}^{q}(t),$ (2)

where $Z_{H}^{q}$ is the Hermite sheet (see, for example, [7] or [27, Section
4]), $H=(H_{1},\dots,H_{d})$, $1/2<H_{i}<1$. Under these conditions, integral
in (2) is well defined, and $\mu$ is an SM in the sense of Definition 1; the
statement (1) can be proved using the representation of $\mu$ via Wiener
integral (see [27, (4.12)]). [7, Proposition 3] implies the Hölder continuity
of $\mu$.

If $q=1$ then the Hermite sheet coincides with the fractional Brownian sheet.

Theorem 10.1.1 of [15] states the sufficient conditions under which the
product of one-dimensional SMs generates an SM.

SMs may be used for study of stochastic dynamical systems (see [3]).

For deterministic measurable functions $f:{\sf{X}}\to\mathbb{R}$, an integral
of the form $\int_{{\sf{X}}}f\,d\mu$ is studied in [15, Chapter 7], [20,
Chapter 1]. In particular, every bounded measurable $f$ is integrable w. r. t.
any $\mu$. An analogue of the Lebesgue dominated convergence theorem holds for
this integral (see [15, Proposition 7.1.1] or [20, Theorem 1.5]).

Below we will use the following statement for SMs defined on arbitrary
$\sigma$-algebra $\mathcal{B}$.

######  Lemma 1.

(Lemma 3.1[19]) Let $\mu$ be an SM on $({\sf{X}},\mathcal{B})$, $f_{k}:\
{{\sf{X}}}\to\mathbb{R},\ k\geq 1$, be measurable functions such that
$\hat{f}(x)=\sum_{k=1}^{\infty}|{f_{k}}(x)|$ is integrable w. r. t. $\mu$.
Then

$\sum_{k=1}^{\infty}\Bigl{(}\int_{{\sf{X}}}f_{k}\,d\mu\Bigr{)}^{2}<\infty\quad\mbox{{~{}}a.~{}s.}$

###  2.2. Besov spaces

We recall the definition of Besov spaces $B^{\alpha}_{p,p}([0,1]^{d})$
following [8] and [12]. Other approaches to the definition and properties of
these spaces are considered in [4], [21, Sections 2.1 and 2.4], [25], [26,
Sections 2 and 5]. The equivalence of different definitions is discussed in
[8], [16, Section 17.2], [21, Section 2.3].

For functions $f\in{\sf{L}}_{p}([0,1]^{d})$, we consider the value

$\|f\|_{B^{\alpha}_{p,p}([0,1]^{d})}=\|f\|_{{\sf{L}}_{p}([0,1]^{d})}+\Bigl{(}\int_{0}^{1}(\omega_{p}(f,r))^{p}r^{-\alpha
p-1}\,dr\Bigr{)}^{1/p},$

where $\omega_{p}$ denotes the ${\sf{L}}_{p}$-modulus of continuity,

$\displaystyle\omega_{p}(f,r)$ $\displaystyle=$ $\displaystyle\sup_{|h|\leq
r}\Bigl{(}\int_{I_{h}}|f(x+h)-f(x)|^{p}\,dx\Bigr{)}^{1/p},$
$\displaystyle{\textrm{where}}\quad I_{h}$ $\displaystyle=$
$\displaystyle\\{x\in[0,1]^{d}:\ x+h\in[0,1]^{d}\\},$

and $|h|$ denotes the Euclidean norm in $\mathbb{R}^{d}$. Then

$B^{\alpha}_{p,p}([0,1]^{d})=\\{f\in{\sf{L}}_{p}([0,1]^{d}):\
\|f\|_{B^{\alpha}_{p,p}([0,1]^{d})}<+\infty\\},$

and $\|\cdot\|_{B^{\alpha}_{p,p}([0,1]^{d})}$ is a norm in this space.

Let $\mu$ be an SM defined on the Borel $\sigma$-algebra of $[0,1]^{d}$. For
$x=(x_{1},x_{2},\dots,x_{d})\in[0,1]^{d}$ set

$\mu(x)=\mu\Bigl{(}\prod_{i=1}^{d}[0,x_{i}]\Bigr{)}.$

In the sequel, we will refer to the following assumption.

######  Assumption A 1.

There exists a real-valued finite measure ${\sf{m}}$ on
$({\sf{X}},\mathcal{B})$ with the following property: if a measurable function
$g:{\sf{X}}\to\mathbb{R}$ satisfies $\int_{{\sf{X}}}g^{2}\,d{\sf{m}}<+\infty$
then $g$ is integrable w. r. t. $\mu$ on ${\sf{X}}$.

For orthogonal measure, we can take its structural measure as ${\sf{m}}$. For
an $\alpha$-stable random measure A1 holds for the control measure ${\sf{m}}$
(see (3.4.1) [22]). For an SM from Example 3 we can take measure ${\sf{m}}$
such that

${\sf{m}}(A)=\int_{A}dv\int_{[0,1]^{d}}\prod_{i=1}^{d}|u_{i}-v_{i}|^{2H_{i}-2}du,$

where $u$ and $v$ are $d$-dimensional vectors with the components $u_{i}$,
$v_{i}$.

We have the following statement concerning the Besov regularity of SMs defined
on Borel subsets of $[0,1]^{d}$.

######  Theorem 1.

(Theorem 5.1 [19]) Let the random function

$\mu(x)=\mu\Bigl{(}\prod_{i=1}^{d}[0,x_{i}]\Bigr{)},\ x\in[0,1]^{d},$

has continuous paths and Assumption A1 holds.

Then for any $1\leq p<+\infty,\ 0<\alpha<\min\\{1/p,1/2\\}$ the realization
$\mu(x)$, $x\in[0,1]^{d}$ with probability 1 belongs to the Besov space
$B^{\alpha}_{p,p}([0,1]^{d})$.

###  2.3. Haar functions

In order to construct a version of the stochastic integral, we use the
approximation of the integrand with Fourier–Haar series.

We follow the definition of one-dimensional Haar functions $\chi_{n}(x)$,
$x\in[0,1]$, from [13, Section 3]. For $n=2^{j}+i$, $1\leq i\leq 2^{j}$,
$j\geq 0$ we write

$\begin{split}\Delta_{n}=\Delta_{j}^{i}=\bigl{(}(i-1)2^{-j},i2^{-j}\bigr{)},\quad\overline{\Delta}_{n}=\bigl{[}(i-1)2^{-j},i2^{-j}\bigr{]},\\\
\Delta_{1}=\Delta_{0}^{0}=(0,1),\quad\overline{\Delta}_{1}=[0,1],\\\
\Delta_{n}^{+}=\bigl{(}\Delta_{j}^{i}\bigr{)}^{+}=\bigl{(}(i-1)2^{-j},(2i-1)2^{-j-1}\bigr{)}=\Delta_{j+1}^{2i-1},\\\
\Delta_{n}^{-}=\bigl{(}\Delta_{j}^{i}\bigr{)}^{-}=\bigl{(}(2i-1)2^{-j-1},i2^{-j}\bigr{)}=\Delta_{j+1}^{2i}.\end{split}$
(3)

Now let $\chi_{1}=1$, and

$\chi_{n}(x)=\left\\{\begin{array}[]{l}0,\quad
x\notin\overline{\Delta}_{n},\\\ 2^{j/2},\quad x\in\Delta_{n}^{+},\\\
-2^{j/2},\quad x\in\Delta_{n}^{-}.\end{array}\right.$

for $2^{j}+1\leq n\leq 2^{j+1}$. In discontinuity points and at the endpoints
of $[0,1]$ we define

$\begin{split}\chi_{n}(x)=\bigl{(}\chi_{n}(x-)+\chi_{n}(x+)\bigr{)}/2,\quad
x\in(0,1),\\\
\chi_{n}(0)=\chi_{n}(0+),\quad\chi_{n}(1)=\chi_{n}(1-).\end{split}$ (4)

For $g\in{\sf{L}}_{1}([0,1])$, we define the Fourier – Haar coefficients and
sums in a standard way:

$\displaystyle
c_{n}(g)=\int_{[0,1]}g(x)\chi_{n}(x)\,dx=2^{j/2}\Bigl{(}\int_{\Delta_{n}^{+}}g(x)\,dx-\int_{\Delta_{n}^{-}}g(x)\,dx\Bigr{)},$
$\displaystyle S_{N}(g,x)=\sum_{n=1}^{N}c_{n}(g)\chi_{n}(x).$

We will approximate $g$ taking the Fourier – Haar sums for values $N=2^{k}$,
$k\in\mathbb{Z}_{+}$.

By (III.8), (III.8′) [13], for $x\neq i2^{-k}$, $0\leq i\leq 2^{k}$, holds

$S_{2^{k}}(g,x)=\sum_{i=1}^{2^{k}}2^{k}\mathbf{1}_{\Delta_{k}^{i}}(x)\int_{\Delta_{k}^{i}}g(t)\,dt,$
(5)

and

$\begin{split}S_{2^{k}}\Bigl{(}g,\frac{i}{2^{k}}\Bigr{)}=2^{k-1}\Bigl{(}\int_{\Delta_{k}^{i}}g(t)\,dt+\int_{\Delta_{k}^{i+1}}g(t)\,dt\Bigr{)},\quad
i=1,2,\dots 2^{k}-1,\\\
S_{2^{k}}(g,0)=2^{k}\int_{\Delta_{k}^{1}}g(t)\,dt,\quad
S_{2^{k}}(g,1)=2^{k}\int_{\Delta_{k}^{2^{k}}}g(t)\,dt.\end{split}$ (6)

For functions $f:[0,1]^{d}\to\mathbb{R}$, we will use approximation by
multivariate Fourier – Haar sums.

The multivariate Haar functions are defined with equalities

$\chi^{(d)}_{n_{1},\dots,n_{d}}(x)=\chi_{n_{1}}(x_{1})\chi_{n_{2}}(x_{2})\dots\chi_{n_{d}}(x_{d}),$

where $\chi_{n_{s}}$ are one-dimensional Haar functions,
$x=(x_{1},\dots,x_{d})\in[0,1]^{d}$. Consider Fourier – Haar coefficients

$c^{(d)}_{n_{1},\dots,n_{d}}(f)=\int_{[0,1]^{d}}f(x)\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,dx,$

and sums

$S^{(d)}_{2^{k}}(f,x)=\sum_{n_{1},\dots,n_{d}\in\\{1,2,3,\dots
2^{k}\\}}c^{(d)}_{n_{1},\dots,n_{d}}(f)\chi^{(d)}_{n_{1},\dots,n_{d}}(x).$

In the sequel, we denote the set of elements $(n_{1},\dots,n_{d})$ such that
$1\leq n_{s}\leq 2^{k}$ for all $s,1\leq s\leq d$ by $\mathbb{N}_{k}$. In
other words, $\mathbb{N}_{k}=\\{1,2,3,\dots,2^{k}-1,2^{k}\\}^{d}$.

We need the following statement on the uniform convergence of multivariate
Fourier – Haar sums for continuous functions. For $d=1$ this fact may be
found, for example, in [13, Theorem III.2]. A similar statement for
multivariate periodic functions was proved in [2].

######  Lemma 2.

If $f\in\mathbb{C}([0,1]^{d})$ then

$\sup_{x\in[0,1]^{d}}|S^{(d)}_{2^{k}}(f,x)-f(x)|\to 0,\quad k\to\infty.$

###### Proof.

For bounded functions $u:[0,1]^{d}\to\mathbb{R}$ we use the following notation
of the uniform modulus of continuity

$\omega(u,r)=\sup_{x,x+h\in[0,1]^{d},\ |h|\leq r}|u(x+h)-u(x)|.$

From (5) and (6) (see also (III.15)[13]) it follows that for $d=1$ holds

$|S_{2^{k}}(g,x)-g(x)|\leq\omega(g,2^{-k}).$ (7)

For $t=(t_{1},\dots t_{d})$ we get

$\displaystyle S^{(d)}_{2^{k}}(f,x)=\sum_{1\leq n_{1}\leq
2^{k}}\dots\sum_{1\leq n_{d}\leq
2^{k}}\chi_{n_{1}}(x_{1})\dots\chi_{n_{d}}(x_{d})$
$\displaystyle\times\int_{[0,1]^{d}}f(t)\chi_{n_{1}}(t_{1})\chi_{n_{2}}(t_{2})\dots\chi_{n_{d}}(t_{d})\,dt.$

We consider the operators

$P_{k,j}(u)=S_{2^{k}}(u,x_{j}),$

where we take one-dimensional Fourier – Haar sums of the function
$u(x_{1},\dots,\ x_{d})$ in coordinate $x_{j}$. Then

$S^{(d)}_{2^{k}}(f,x)=P_{k,d}(P_{k,d-1}(\dots P_{k,1}(f)\dots)).$

Estimate (7) implies

$|P_{k,j}(u)(x)-u(x)|\leq\omega(u,2^{-k}),$ (8)

and it is easy to see that for integral $m\geq 1$

$\omega(P_{k,j}(u),m2^{-k})\leq\omega(u,(m+2)2^{-k}).$ (9)

We have

$\displaystyle|S^{(d)}_{2^{k}}(f,x)-f(x)|\leq|P_{k,1}(f)(x)-f(x)|$
$\displaystyle+\sum_{l=1}^{d-1}|P_{k,l+1}(P_{k,l}(\dots
P_{k,1}(f)\dots))-P_{k,l}(P_{k,l-1}(\dots P_{k,1}(f)\dots))|,$

where

$\displaystyle|P_{k,l+1}(P_{k,l}(\dots
P_{k,1}(f)\dots))-P_{k,l}(P_{k,l-1}(\dots P_{k,1}(f)\dots))|$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:pfomega}}}{{\leq}}\omega(P_{k,l}(\dots
P_{k,1}(f)\dots),2^{-k})\stackrel{{\scriptstyle\eqref{eq:puomega}}}{{\leq}}\omega(f,(2l+1)2^{-k}).$

Therefore,

$|S^{(d)}_{2^{k}}(f,x)-f(x)|\leq\sum_{l=0}^{d-1}\omega(f,(2l+1)2^{-k}).$ (10)

For continuous $f$ the right hand side of (10) tends to 0 as $k\to\infty$,
that implies the statement of our lemma. ∎

######  Remark 1.

We obtained estimate (10) for all bounded $f\in{\sf{L}}_{1}([0,1]^{d})$, not
necessarily continuous.

##  3\. Parameter dependent integral

In this section, we study the properties of random function

$\eta(z)=\int_{[0,1]^{d}}f(z,x)\,d\mu(x),\ z\in Z,$ (11)

Note that parameter stochastic integral w.r.t. Brownian sheet was considered
in [23], [24].

If $f$ is continuous in $x$ then Lemma 2 and analogue of the dominated
convergence theorem ([20, Theorem 1.5]) imply that for each fixed $z$

$\int_{[0,1]^{d}}f(z,x)\,d\mu(x)={\rm
p}\lim_{k\to\infty}\int_{[0,1]^{d}}S^{(d)}_{2^{k}}(f,x)\,d\mu(x).$

Therefore,

$\tilde{\eta}(z)=\int_{[0,1]^{d}}S^{(d)}_{1}(f,x)\,d\mu(x)+\sum_{k=1}^{\infty}\int_{[0,1]^{d}}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x).$
(12)

is the version of $\eta(z)$.

In (12) we have integrals of simple functions, and each integral is equal to
respective linear combination of values of $\mu$. We fix the same version of
$\mu$ for all these values.

We will use notation $e_{i}=(0,\dots,0,1,0,\dots,0)$, where 1 stays on $i$-th
place, $e_{i}\in\mathbb{R}^{d}$.

######  Lemma 3.

Let the function $f(z,x):Z\times[0,1]^{d}\to\mathbb{R}$ be continuously
differentiable $l$ times on $[0,1]^{d}$ for each $z$, $l\geq 0$, and

$\Bigl{|}\dfrac{{\partial}^{r}f(z,x)}{{\partial}x_{s_{1}}\dots\partial
x_{s_{r}}}\Bigr{|}\leq C_{f},\ x\in[0,1]^{d}$ (13)

for some constant $C_{f}>0$, which is independent of $z$, and all $0\leq r\leq
l$, $s_{1},\dots,s_{r}\subset\\{1,\dots,d\\}$. Moreover, let $l$-th
derivatives be Hölder continuous with an exponent $\alpha>0$:

$\Bigl{|}\dfrac{\partial^{l}f(z,x^{(1)})}{\partial x_{s_{1}}\dots\partial
x_{s_{l}}}-\dfrac{\partial^{l}f(z,x^{(2)})}{\partial x_{s_{1}}\dots\partial
x_{s_{l}}}\Bigr{|}\leq C_{f}|x^{(1)}-x^{(2)}|^{\alpha}.$ (14)

If $l+\alpha>d/2$, then the version (12) satisfies the inequality

$|\tilde{\eta}(z)|\leq C_{f}C^{(d)}_{\mu}(\omega),$

where $C^{(d)}_{\mu}(\omega)$ denotes a random constant that depends only of
dimension $d$ and SM $\mu$, $C^{(d)}_{\mu}(\omega)<\infty$ a.s.

Moreover, series (12) converge absolutely and uniformly, i. e.

$\sum_{k=1}^{\infty}\sup_{z\in
Z}\Bigl{|}\int_{[0,1]^{d}}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x)\Bigr{|}\leq
C_{f}C^{(d)}_{\mu}(\omega).$ (15)

###### Proof.

We have that

$\int_{[0,1]^{d}}S^{(d)}_{2^{k}}(f,x)\,d\mu(x)=\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}c^{(d)}_{n_{1},\dots,n_{d}}(f)\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x).$
(16)

For each $n_{s}\geq 2$, take $j_{s}$ such that $2^{j_{s}}+1\leq n_{s}\leq
2^{j_{s}+1}$, $j_{s}\geq 0$. Denote by $\\{j^{(k)},1\leq k\leq d\\}$ the
permutation of $\\{j_{k},1\leq k\leq d\\}$ such that

$j^{(1)}\leq j^{(2)}\leq\dots\leq j^{(d)}.$

If $n_{s_{1}}=\dots=n_{s_{i}}=1$, we set $j^{(1)}=\dots=j^{(i)}=0$. We rewrite
the expression for $c^{(d)}_{n_{1},\dots,n_{d}}(f)$ in the following way:

$\begin{split}c^{(d)}_{n_{1},\dots,n_{d}}(f)=\int_{[0,1]^{d}}f(z,t)\chi^{(d)}_{n_{1},\dots,n_{d}}(t)\,dt=\int_{[0,1]^{d}}f(z,t)\prod_{1\leq
s\leq d}\chi_{n_{s}}(t_{s})\,dt\\\ =2^{\sum_{1\leq s\leq d,n_{s}\geq
2}j_{s}/2}\int_{[0,1]^{d}}f(z,t)\prod_{1\leq s\leq d,n_{s}\geq
2}\Bigl{(}\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s})-\mathbf{1}_{\Delta_{n_{s}}^{-}}(t_{s})\Bigr{)}\,dt\\\
=2^{\sum_{1\leq s\leq d,n_{s}\geq 2}j_{s}/2}\int_{[0,1]^{d}}f(z,t)\\\
\times\prod_{1\leq s\leq d,n_{s}\geq
2}\Bigl{(}\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s})-\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s}-2^{-j_{s}-1})\Bigr{)}\,dt\\\
\stackrel{{\scriptstyle(*)}}{{=}}2^{\sum_{1\leq s\leq d,n_{s}\geq
2}j_{s}/2}\\\
\times\int_{[0,1]^{d}}\sum_{\varepsilon_{s}\in\\{0,1\\},n_{s}\geq
2}(-1)^{\varepsilon_{1}+\dots+\varepsilon_{d}}f(z,t_{1}+\varepsilon_{1}2^{-j_{1}-1},\dots,t_{d}+\varepsilon_{d}2^{-j_{d}-1})\\\
\times\prod_{1\leq s\leq d,n_{s}\geq
2}\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s})\,dt=2^{\sum_{1\leq s\leq d,n_{s}\geq
2}j_{s}/2}\\\ \times\int_{\prod_{1\leq s\leq d,n_{s}\geq
2}\Delta_{n_{s}}^{+}}\sum_{\varepsilon_{s}\in\\{0,1\\},n_{s}\geq
2}(-1)^{\varepsilon_{1}+\dots+\varepsilon_{d}}\\\ \times
f(z,t_{1}+\varepsilon_{1}2^{-j_{1}-1},\dots,t_{d}+\varepsilon_{d}2^{-j_{d}-1})\,dt.\end{split}$
(17)

In equality (*) we opened the brackets in the product

$\prod_{1\leq s\leq d,n_{s}\geq
2}\Bigl{(}\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s})-\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s}-2^{-j_{s}-1})\Bigr{)}$

and used the change of the variables $t_{s}-2^{-j_{s}-1}\to t_{s}$ in
$\mathbf{1}_{\Delta_{n_{s}}^{+}}(t_{s}-2^{-j_{s}-1})$.

The last sum in (17) can be represented as the sum of at most $2^{d-1}$
summands of the form

$f(z,t^{*})-f(z,t^{*}+2^{-j^{(d)}-1}e_{s_{d}})=\int_{0}^{2^{-j^{(d)}-1}}\frac{\partial
f(z,t^{*}+h_{s_{d}})}{\partial t_{s_{d}}}dh_{s_{d}},\qquad j^{(d)}=j_{s_{d}}.$

Now we repeat the same thoughts $l-1$ times and use the properties of a
function $f$, which leads to the following inequality

$\displaystyle|c^{(d)}_{n_{1},\dots,n_{d}}(f)|\leq
C_{f}2^{d-l-1}2^{\sum_{1\leq s\leq d,n_{s}\geq
2}-j_{s}/2}2^{-(j^{(d)}+\dots+j^{(d-l+1)}+\alpha j^{(d-l)})}.$ (18)

Further, we have the estimate

$\begin{split}\sum_{k=1}^{\infty}\sup_{z\in
Z}\Bigl{|}\int_{[0,1]^{d}}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x)\Bigr{|}\\\
\leq\sum_{k=1}^{\infty}\sup_{z\in
Z}\Bigl{|}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}c^{(d)}_{n_{1},\dots,n_{d}}(f)\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)\Bigr{|}\\\
\stackrel{{\scriptstyle\eqref{estc}}}{{\leq}}2^{d}C_{f}\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}2^{\sum_{1\leq
s\leq d,n_{s}\geq 2}-j_{s}/2}2^{-(j^{(d)}+\dots+j^{(d-l+1)}+\alpha
j^{(d-l)})}\\\
\times\Bigl{|}\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)\Bigr{|}\\\
\leq
2^{d}C_{f}\biggl{(}\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}2^{\beta\sum_{1\leq
s\leq d,n_{s}\geq 2}j_{s}}2^{-2(j^{(d)}+\dots+j^{(d-l+1)}+\alpha
j^{(d-l)})}\biggr{)}^{1/2}\\\
\times\biggl{(}\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}2^{(-\beta-1)\sum_{1\leq
s\leq d,n_{s}\geq
2}j_{s}}\Bigl{(}\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)\Bigr{)}^{2}\biggr{)}^{1/2}\\\
:=2^{d}C_{f}P_{1}^{1/2}P_{2}^{1/2}.\end{split}$

For a sufficiently small $\beta>0$ we have the following estimates for
$P_{1}$:

$\displaystyle P_{1}=\sup_{k\geq
1}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}2^{\beta\sum_{1\leq s\leq
d,n_{s}\geq 2}j_{s}}2^{-2(j^{(d)}+\dots+j^{(d-l+1)}+\alpha j^{(d-l)})}$
$\displaystyle=\sup_{k\geq
1}\sum_{A\subset\\{1,\dots,d\\}}\sum_{\begin{subarray}{c}0\leq j_{i}\leq k,\\\
i\in A\end{subarray}}2^{(\beta+1)\sum_{i\in
A}j_{i}-2(j^{(d)}+\dots+j^{(d-l+1)}+\alpha j^{(d-l)})}$ $\displaystyle\leq
2^{d}\sup_{k\geq 1}\sum_{0\leq j_{i}\leq
k}2^{(\beta+1-2(l+\alpha)/d)\sum_{i=1}^{d}j_{i}}=2^{d}C.$ (19)

Here we denoted the set of indexes $i$, for which $n_{s}=1$, as $A$. It is
left to prove that for each fixed $\beta>0$ $P_{2}\leq C_{\mu}^{(d)}(\omega)$.
As follows from Lemma 1, it is sufficient to show that

$\tilde{P}_{2}=\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}2^{(-\beta/2-1/2)\sum_{1\leq
s\leq d,n_{s}\geq 2}j_{s}}|\chi^{(d)}_{n_{1},\dots,n_{d}}(x)|\leq C^{(d)},$
(20)

where by $C^{(d)}$ we denote positive constant that depends only on $d$. The
inequality (20) holds true, as is proved below.

$\displaystyle\tilde{P}_{2}=\sup_{k\geq
1}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}2^{(-\beta/2-1/2)\sum_{1\leq
s\leq d,n_{s}\geq 2}j_{s}}|\chi_{n_{1}}(x_{1})\dots\chi_{n_{d}}(x_{d})|$
$\displaystyle\leq\sup_{k\geq
1}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}2^{(-\beta/2)\sum_{1\leq s\leq
d,n_{s}\geq 2}j_{s}}$
$\displaystyle\times\prod_{s=1}^{d}\Bigl{(}\mathbf{1}_{\Delta_{n_{s}}}(x_{s})+\frac{1}{2}\mathbf{1}_{\\{(i-1)2^{-j_{s}}\\}}(x_{s})+\frac{1}{2}\mathbf{1}_{\\{i2^{-j_{s}}\\}}(x_{s})\Bigr{)}$
$\displaystyle\stackrel{{\scriptstyle(*)}}{{\leq}}\sup_{k\geq
1}\sum_{A\subset\\{1,\dots,d\\}}\sum_{\begin{subarray}{c}0\leq j_{s}\leq k,\\\
s\in A\end{subarray}}2^{(-\beta/2)\sum_{s\in A}j_{s}}\leq 2^{d}\sup_{k\geq
1}\sum_{0\leq j_{s}\leq k}2^{(-\beta/2)\sum_{s=1}^{d}j_{s}}=C^{(d)}.$

Here in inequality (*) we used that for each set $j_{1},\dots,j_{d}$, for each
$x$, $x_{s}\neq i2^{-j_{s}}$, exists exactly one set $(n_{1},\dots,n_{d})$
such that $\chi^{(d)}_{n_{1},\dots,n^{d}}(x)\neq 0$ and $2^{j_{s}}+1\leq
n_{s}\leq 2^{j_{s}+1}$. If we have coordinates $x_{s}=i2^{-j_{s}}$, we take
into account that $\mathbf{1}_{\\{i2^{-j_{s}}\\}}(x_{s})$ has the coefficient
$1/2$, as follows from (4). Now the statement of the lemma is a consequence of
(19) and (20). ∎

######  Remark 2.

Assume that constants in inequalities (13) and (14) depend not only on $f$,
but also on $z$. Then the series (12) converges a. s. for each fixed $z\in Z$
and the following analogue of (15):

$\sum_{k=1}^{\infty}\Bigl{|}\int_{[0,1]^{d}}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x)\Bigr{|}\leq
C_{f,z}C^{(d)}_{\mu}(\omega).$

This statement is proved similarly to Lemma 3. We just refer to $C_{f,z}$
everywhere instead of $C_{f}$.

The following statement gives the conditions of the continuity with respect to
parameter $z\in Z$.

######  Theorem 2.

Let $Z$ be a metric space, and the function
$f(z,x):Z\times[0,1]^{d}\to\mathbb{R}$ be such that all paths $f(\cdot,x)$ are
continuous on $Z$ for each $x$, $f(z,\cdot)$ is continuously differentiable
$l$ times on $[0,1]^{d}$ for each $z$, $l\geq 0$, and

$\Bigl{|}\dfrac{{\partial}^{r}f(z,x)}{{\partial}x_{s_{1}}\dots\partial
x_{s_{r}}}\Bigr{|}\leq C_{f},\ x\in[0,1]^{d}$

for some constant $C_{f}>0$ (independent of $z$) and all $0\leq r\leq l$,
$s_{1},\dots,s_{r}\subset\\{1,\dots,d\\}$. Let $l$-th derivatives be Hölder
continuous with the exponent $\alpha$ and

$\Bigl{|}\dfrac{\partial^{l}f(z,x^{(1)})}{\partial x_{s_{1}}\dots\partial
x_{s_{l}}}-\dfrac{\partial^{l}f(z,x^{(2)})}{\partial x_{s_{1}}\dots\partial
x_{s_{l}}}\Bigr{|}\leq C_{f}|x^{(1)}-x^{(2)}|^{\alpha}.$

If, in addition, $l+\alpha>d/2$, then the random function $\eta$ defined by
(11) has a version (12) with continuous paths on $Z$ a. s.

###### Proof.

From (17) it follows that each $c^{(d)}_{n_{1},\dots,n_{d}}$ is continuous
function of variable $z$, and (16) implies that
$\int_{[0,1]^{d}}S^{(d)}_{2^{k}}(f,x)\,d\mu(x)$ is continuous for all $k\geq
1$, $\omega\in\Omega$.

By (15), $\int_{[0,1]^{d}}S^{(d)}_{2^{k}}(f,x)\,d\mu(x)$ converges to
$\tilde{\eta}(z)$ uniformly a.s. as $k\to\infty$, that implies the statement
of our theorem ∎

######  Theorem 3.

Let the function $f(z,x):Z\times[0,1]^{d}\to\mathbb{R}$, $Z=[a,b]$, be
continuous differentiable $l+1$ times on $Z\times[0,1]^{d}$ while

$\Bigl{|}\dfrac{\partial^{r+1}f(z,x)}{\partial{z}\partial
x_{s_{1}}\dots\partial x_{s_{r}}}\Bigr{|}\leq C_{f},\ x\in[0,1]^{d},$

for all $0\leq r\leq l$, $s_{1},\dots,s_{r}\subset\\{1,\dots,d\\}$. Moreover,
let $l+1$-th derivatives be Hölder continuous with the exponent $\alpha>0$:

$\Bigl{|}\dfrac{\partial^{l+1}f(z,x^{(1)})}{\partial z\partial
x_{s_{1}}\dots\partial x_{s_{l}}}-\dfrac{\partial^{l+1}f(z,x^{(2)})}{\partial
z\partial x_{s_{1}}\dots\partial x_{s_{l}}}\Bigr{|}\leq
C_{f}|x^{(1)}-x^{(2)}|^{\alpha}.$

If, in addition, $l+\alpha>d/2$, then paths of a random function
$\tilde{\eta}$, which is defined by (12), have bounded derivatives on $Z$:

$\dfrac{d\tilde{\eta}(z)}{dz}=\int_{[0,1]^{d}}\dfrac{\partial f(z,x)}{\partial
z}\,d\mu(x).$

###### Proof.

Lemma 3 implies that

$\displaystyle\int_{[0,1]^{d}}\dfrac{\partial f(z,x)}{\partial
z}\,d\mu(x)=\int_{[0,1]^{d}}S^{(d)}_{1}\Bigl{(}\frac{\partial f}{\partial
z},x\Bigr{)}\,d\mu(x)$
$\displaystyle+\sum_{k=1}^{\infty}\int_{[0,1]^{d}}\Bigl{(}S^{(d)}_{2^{k}}\Bigl{(}\frac{\partial
f}{\partial z},x\Bigr{)}-S^{(d)}_{2^{k-1}}\Bigl{(}\frac{\partial f}{\partial
z},x\Bigr{)}\Bigr{)}\,d\mu(x),$

where the series in the right hand side converges uniformly. According to
equalities

$\displaystyle\int_{[0,1]^{d}}{S^{(d)}_{2^{k}}}\Bigl{(}\frac{\partial
f}{\partial z},x\Bigr{)}\,d\mu(x)$
$\displaystyle=\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}c^{(d)}_{n_{1},\dots,n_{d}}\Bigl{(}\frac{\partial
f}{\partial
z}\Bigr{)}\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)$
$\displaystyle\stackrel{{\scriptstyle\eqref{eq:estcn}}}{{=}}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}}\frac{\partial
c^{(d)}_{n_{1},\dots,n_{d}}(f)}{\partial
z}\int_{[0,1]^{d}}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)$
$\displaystyle=\frac{d}{dz}\int_{[0,1]^{d}}{S^{(d)}_{2^{k}}}(f,x)\,d\mu(x)$

and Remark 2, we can differentiate the series in (12) and obtain that

$\displaystyle\dfrac{d\tilde{\eta}(z)}{dz}=\frac{d}{dz}\int_{[0,1]^{d}}S^{(d)}_{1}(f,x)\,d\mu(x)$
$\displaystyle+\sum_{k=1}^{\infty}\frac{d}{dz}\int_{[0,1]^{d}}\bigl{(}S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x)\bigr{)}\,d\mu(x)=\int_{[0,1]^{d}}\dfrac{\partial
f(z,x)}{\partial z}\,d\mu(x),$

which finishes the proof. ∎

##  4\. Integral as a function of upper limit

For continuous function $f(x):[0,1]^{d}\to\mathbb{R}$ and
$y=(y_{1},\dots,y_{d})\in[0,1]^{d}$, an we consider the random function

$\xi(y)=\int_{\prod_{s=1}^{d}[0,y_{s}]}f(x)\,d\mu(x).$ (21)

Lemma 2 implies that

$\begin{split}\tilde{\xi}(y)=\int_{\prod_{s=1}^{d}[0,y_{s}]}S^{(d)}_{1}(f,x)\,d\mu(x)\\\
+\sum_{k=1}^{\infty}\int_{\prod_{s=1}^{d}[0,y_{s}]}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x).\end{split}$
(22)

is the version of $\xi(y)$.

We refer to the following assumptions on SM $\mu$.

######  Assumption A 2.

Random function $\mu(x)=\mu\Bigl{(}\prod_{i=1}^{d}[0,x_{i}]\Bigr{)},\
x\in[0,1]^{d}$, has continuous paths.

In the following condition on the uniform modulus of continuity we take the
continuous version of $\mu$.

######  Assumption A 3.

If $d\geq 2$ then
$\sum_{k=1}^{\infty}k^{d-2}\omega(\mu,2^{-k})<\infty\quad\textrm{a.\ s.}$

It is easy to see that A3 holds for $\mu(x)$ with Hölder continuous paths. It
also holds if, for example, $\omega(\mu,\tau)\leq C|\ln\tau|^{-\epsilon}$,
$\epsilon>d-1$.

######  Example 4.

Denote

$q(\tau)=|\ln\tau|^{-\gamma},\,\gamma>d-1/2,\qquad\mathcal{K}(z)=\sqrt{\dfrac{dq^{2}(z)}{dz}}.$

We introduce the following stochastic process:

$B^{q}(x)=\int_{\prod_{s=1}^{d}[0,x_{s}]}\prod_{s=1}^{d}\mathcal{K}(x_{s}-y_{s})dW(y),\
x=(x_{1},\dots,x_{d})$

where $W$ is a $d$-dimensional Wiener process. Theorem 3.1 in [11] implies the
existence of rectangle $[t,T]=\prod_{i=1}^{d}[t_{i},T_{i}]\subset[0,1]^{d}$
such that

$\lim_{\varepsilon\to 0+}\sup_{\begin{subarray}{c}x,\bar{x}\in[t,T]\\\
\delta_{x,\bar{x}}\leq\varepsilon\end{subarray}}\frac{|B^{q}(x)-B^{q}(\bar{x})|}{\delta_{x,\bar{x}}\sqrt{\ln\Bigl{(}\frac{D}{q^{-}1(\delta_{x,\bar{x}})}\Bigr{)}}}=C\
\text{a. s.}$ (23)

Here $\delta_{x,\bar{x}}=\|B^{q}(x)-B^{q}(\bar{x})\|_{L^{2}(\Omega)}\leq
Cq(|x-\bar{x}|)$, $D$ is a diameter of $[t,T]$. From (23) it follows that for
all $x,\bar{x}\in[t,T]$

$\displaystyle|B^{q}(x)-B^{q}(\bar{x})|\leq C\delta_{x,\bar{x}}\sqrt{\ln
D+\delta_{x,\bar{x}}^{-1/\gamma}}$
$\displaystyle\stackrel{{\scriptstyle\gamma>1/2}}{{\leq}}C|\ln|x-\bar{x}||^{-\gamma}\sqrt{\ln
D+|\ln|x-\bar{x}||}$ $\displaystyle\leq C|\ln|x-\bar{x}||^{1/2-\gamma}.$

Therefore, $\omega_{[t,T]}(B^{q},\tau)\leq C|\ln\tau|^{1/2-\gamma}$.

Now we are ready to formulate the main result of the section.

######  Theorem 4.

Let Assumptions A2 and A3 hold, and the function $f(x):[0,1]^{d}\to\mathbb{R}$
be continuously differentiable $d$ times on $[0,1]^{d}$.

Then, for the random function $\xi$ defined by (21), version (22) has
continuous paths on $[0,1]^{d}$ a. s.

###### Proof.

For version (22), we have that

$\tilde{\xi}(y)=\lim_{k\to\infty}\int_{\prod_{s=1}^{d}[0,y_{s}]}S^{(d)}_{2^{k}}(f,x)\,d\mu(x).$
(24)

Here $S^{(d)}_{2^{k}}(f,x)$ is a simple function. By our assumption, SM $\mu$
has a continuous paths. Therefore, for each $k$, the random function of
variable $y$

$\int_{\prod_{s=1}^{d}[0,y_{s}]}S^{(d)}_{2^{k}}(f,x)\,d\mu(x)$

has a continuous paths. We will prove that the convergence in (24) is uniform
in $y\in[0,1]^{d}$ a.s., and this will imply the continuity of $\tilde{\xi}$.

Our aim is to show that

$\sum_{k=1}^{\infty}\sup_{y\in[0,1]^{d}}\Bigl{|}\int_{\prod_{s=1}^{d}[0,y_{s}]}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x)\Bigr{|}\leq\tilde{C}^{(d)}_{f,\mu}(\omega)$

for some random constant $\tilde{C}^{(d)}_{f,\mu}(\omega)<\infty$ a. s. that
may depends on $d,f,\mu$.

Recall that paths of $\mu(x_{1},\dots,x_{d})$ are continuous, therefore for
any cut of the set $\prod_{s=1}^{d}[y_{s1},y_{s2}]\subset[0,1]^{d}$ we have

$\mu\Bigl{(}\prod_{s=1}^{d}[y_{s1},y_{s2}]\cap\\{x_{s}=a\\}\Bigr{)}=0.$ (25)

Thus, we obtain

$\begin{split}\int_{\prod_{s=1}^{d}[0,y_{s}]}(S^{(d)}_{2^{k}}(f,x)-S^{(d)}_{2^{k-1}}(f,x))\,d\mu(x)\\\
=\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}c^{(d)}_{n_{1},\dots,n_{d}}(f)\int_{\prod_{s=1}^{d}[0,y_{s}]}\chi^{(d)}_{n_{1},\dots,n_{d}}(x)\,d\mu(x)\\\
\stackrel{{\scriptstyle\eqref{eq:muzero}}}{{=}}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}c^{(d)}_{n_{1},\dots,n_{d}}(f)2^{(j_{1}+\dots+j_{d})/2}\\\
\times\int_{\prod_{s=1}^{d}[0,y_{s}]}\prod_{1\leq s\leq d,n_{s}\geq
2}\Bigl{(}\mathbf{1}_{\Delta_{n_{s}}^{+}}(x_{s})-\mathbf{1}_{\Delta_{n_{s}}^{-}}(x_{s})\Bigr{)}\,d\mu(x)\\\
=\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}c^{(d)}_{n_{1},\dots,n_{d}}(f)2^{(j_{1}+\dots+j_{d})/2}\\\
\times\sum_{\varepsilon_{s}\in\\{+,-\\}}\mu\Bigl{(}\prod_{1\leq s\leq
d}(\Delta_{n_{s}})^{\varepsilon_{s}}\cap\prod_{1\leq s\leq
d}[0,y_{s}]\Bigr{)}\\\ :=A_{k1}(y)+A_{k2}(y).\end{split}$

Here $A_{k1}(y)$ is the sum of terms with
$(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}$,
$\varepsilon_{s}\in\\{+,-\\}$ such that

$\prod_{1\leq s\leq
d}(\Delta_{n_{s}})^{\varepsilon_{s}}\subset\prod_{s=1}^{d}[0,y_{s}]$

(here for $n_{s}=1$ we take only $\varepsilon_{s}=+$ and
$(\Delta_{n_{s}})^{\varepsilon_{s}}=(0,1)$). Taking into account the
definition of $(\Delta_{n_{s}})^{+}$ and $(\Delta_{n_{s}})^{-}$ in (3), we get
that in $A_{k1}(y)$

$\prod_{1\leq s\leq
d}(\Delta_{n_{s}})^{\varepsilon_{s}}\cap\prod_{s=1}^{d}[0,y_{s}]=\prod_{1\leq
s\leq d}(\Delta_{n_{s}})^{\varepsilon_{s}}=\prod_{1\leq s\leq
d}\Delta_{m_{s}}$

for some $(m_{1},\dots,m_{d})\in\mathbb{N}_{k+1}\setminus\mathbb{N}_{k}$.
Therefore,

$\begin{split}\sum_{k=1}^{\infty}\sup_{y\in[0,1]^{d}}|A_{k1}(y)|\\\
\leq\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}|c^{(d)}_{n_{1},\dots,n_{d}}(f)|2^{(j_{1}+\dots+j_{d})/2}\sum_{\varepsilon_{s}\in\\{+,-\\}}\Bigl{|}\mu\Bigl{(}\prod_{1\leq
s\leq d}(\Delta_{n_{s}})^{\varepsilon_{s}}\Bigr{)}\Bigr{|}\\\
\stackrel{{\scriptstyle\eqref{estc}}}{{\leq}}C\sum_{k=1}^{\infty}\sum_{(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}}2^{-(j_{1}+\dots+j_{d})}\sum_{\varepsilon_{s}\in\\{+,-\\}}\Bigl{|}\mu\Bigl{(}\prod_{1\leq
s\leq d}(\Delta_{n_{s}})^{\varepsilon_{s}}\Bigr{)}\Bigr{|}\\\ \leq
C\sum_{k=1}^{\infty}\sum_{(m_{1},\dots,m_{d})\in\mathbb{N}_{k+1}\setminus\mathbb{N}_{k}}2^{d}2^{-(j_{1}^{\prime}+\dots+j_{d}^{\prime})}\Bigl{|}\mu\Bigl{(}\prod_{1\leq
s\leq d}\Delta_{m_{s}}\Bigr{)}\Bigr{|}.\end{split}$

Here $j^{\prime}_{s}$ are taken such that $2^{j^{\prime}_{s}}+1\leq m_{s}\leq
2^{j^{\prime}_{s}+1}$, and $j^{\prime}_{s}=j_{s}+1$ for respective $m_{s}$.
From estimates in (19), (20) it follows that

$\sum_{k=1}^{\infty}\sup_{y\in[0,1]^{d}}|A_{k1}(y)|\leq\tilde{C}^{(d)}_{\mu}(\omega).$

for some random constant $\tilde{C}^{(d)}_{\mu}(\omega)$ that depends only of
$d$ and $\mu$. (It is easy to see that all estimates in (19), (20) remain
valid if we change $\chi^{(d)}_{n_{1},\dots,n_{d}}(x)$ to
$2^{(j_{1}+\dots+j_{d})/2}\prod_{1\leq s\leq
d}\mathbf{1}_{\Delta_{m_{s}}}(x_{s})$.)

Further, we estimate $A_{k2}(y)$, i. e. the sum of terms with
$(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1}$ such that
$y_{s}\in\Delta_{n_{s}}$ for some $s$.

We get

$\displaystyle|A_{k2}(y)|\leq\sum_{\begin{subarray}{c}(n_{1},\dots,n_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1},\\\
\exists
y_{s}\in(\Delta_{n_{s}})^{\varepsilon_{s}}\end{subarray}}2^{-\sum_{1\leq s\leq
d,n_{s}\geq 2}j_{s}}\Bigl{|}\mu\Bigl{(}\prod_{1\leq s\leq
d}((\Delta_{n_{s}})^{\varepsilon_{s}}\cap[0,y_{s}])\Bigr{)}\Bigr{|}$
$\displaystyle=2^{d}\sum_{\begin{subarray}{c}(m_{1},\dots,m_{d})\in\mathbb{N}_{k}\setminus\mathbb{N}_{k-1},\\\
\exists y_{s}\in\Delta_{m_{s}}\end{subarray}}2^{-\sum_{1\leq s\leq d,m_{s}\geq
3}j^{\prime}_{s}}\Bigl{|}\mu\Bigl{(}\prod_{1\leq s\leq
d}(\Delta_{m_{s}}\cap[0,y_{s}])\Bigr{)}\Bigr{|}$ $\displaystyle\leq
2^{d}\omega(\mu,2^{-k-1})\sum_{\begin{subarray}{c}(m_{1},\dots,m_{d})\in\mathbb{N}_{k+1}\setminus\mathbb{N}_{k},\\\
\exists y_{s}\in\Delta_{m_{s}}\end{subarray}}2^{-\sum_{1\leq s\leq d,m_{s}\geq
3}j^{\prime}_{s}}:=D_{k+1}.$

Now we check the convergence of the series $\sum_{k=1}^{\infty}D_{k}$ with the
help of the sum

$\sum_{\begin{subarray}{c}(m_{1},\dots,m_{d})\in\mathbb{N}_{k},\\\ \exists
y_{s}\in\Delta_{m_{s}}\end{subarray}}2^{-\sum_{1\leq s\leq d,m_{s}\geq
3}j^{\prime}_{s}}:=T_{k}.$

Notice that for each fixed set $(j^{\prime}_{1},\dots,j^{\prime}_{d})$ there
exist at most

$2^{\sum_{1\leq s\leq d,m_{s}\geq 3}j^{\prime}_{s}}-\prod_{1\leq s\leq
d,m_{s}\geq 3}\bigl{(}2^{j^{\prime}_{s}}-1\bigr{)}$

sets $(m_{1}\dots,m_{d})$, for which $\exists y_{s}\in\Delta_{m_{s}}$.
Denoting the set of indexes $s$, which satisfy the equality $n_{s}=1$, as $A$,
we obtain that

$\displaystyle\sup_{y}T_{k}\leq\sum_{A\subset\\{1,\dots,d\\}}\sum_{\begin{subarray}{c}1\leq
j_{i}^{\prime}\leq k,\\\ i\in A\end{subarray}}\biggl{(}1-\prod_{i\in
A}\bigl{(}1-2^{-j^{\prime}_{i}}\bigr{)}\biggr{)}$
$\displaystyle=\sum_{A\subset\\{1,\dots,d\\}}\sum_{\begin{subarray}{c}1\leq
j_{i}^{\prime}\leq k,\\\ i\in
A\end{subarray}}\sum_{\begin{subarray}{c}B\subset A,\\\
B\neq\emptyset\end{subarray}}(-1)^{|B|+1}2^{-\sum_{i\in B}j^{\prime}_{i}}$
$\displaystyle\leq\sum_{A\subset\\{1,\dots,d\\}}\sum_{\begin{subarray}{c}B\subset
A,\\\ B\neq\emptyset\end{subarray}}(-1)^{|B|+1}\sum_{\begin{subarray}{c}1\leq
j_{i}^{\prime}\leq k,\\\ i\in A\end{subarray}}2^{-\sum_{i\in
B}j^{\prime}_{i}}$
$\displaystyle\stackrel{{\scriptstyle(*)}}{{=}}\sum_{u=1}^{d}\sum_{v=1}^{u}(-1)^{v+1}\left(\begin{array}[]{c}d\\\
u\end{array}\right)\left(\begin{array}[]{c}u\\\
v\end{array}\right)\sum_{\begin{subarray}{c}1\leq j^{\prime}_{i}\leq k,\\\
1\leq i\leq u\end{subarray}}2^{-\sum_{1\leq i\leq v}j^{\prime}_{i}}$
$\displaystyle=\sum_{u=1}^{d}\sum_{v=1}^{u}(-1)^{v+1}\left(\begin{array}[]{c}d\\\
u\end{array}\right)\left(\begin{array}[]{c}u\\\
v\end{array}\right)(1-2^{-k})^{v}k^{u-v}$
$\displaystyle=\sum_{u=1}^{d}\left(\begin{array}[]{c}d\\\
u\end{array}\right)(k^{u}-(k-1+2^{-k})^{u})$
$\displaystyle=(k+1)^{d}-(k+2^{-k})^{d}.$

Here in $(*)$ we used the fact that sum
$(-1)^{|B|+1}\sum_{\begin{subarray}{c}1\leq j_{i}^{\prime}\leq k,\\\ i\in
A\end{subarray}}2^{-\sum_{i\in B}j^{\prime}_{i}}$ depends only on $|A|$ and
$|B|$. Therefore,

$\displaystyle D_{k}\leq
2^{d}\omega(\mu,2^{-k})\bigl{(}(k+1)^{d}-(k+2^{-k})^{d}-k^{d}+(k-1+2^{-k+1})\bigr{)}$
$\displaystyle=2^{d}\omega(\mu,2^{-k})\bigl{(}(k+1)^{d}-2k^{d}+(k-1)^{d}+O(2^{-\beta
k})\bigr{)},$

where $0<\beta<1$. Applying Assumption A3 we get the statement of the theorem.
∎

If $\mu(x)$ have a Hölder continuous paths with exponent $\gamma>0$, then the
statement of Theorem 4 follows from Theorem 16 [10]. Moreover, Theorem 16 [10]
states that $\xi$ has the version that is Hölder continuous with the same
exponent $\gamma>0$. In [10], integral is considered in the Young sense, and
its value coincides with value of our integral with respect to Hölder
continuous $\mu$.

######  Theorem 5.

Let Assumptions A1, A2 and A3 hold, and the function
$f(x):[0,1]^{d}\to\mathbb{R}$ be continuously differentiable $d$ times on
$[0,1]^{d}$.

Then, for the random function $\xi$ defined by (21), for any $1\leq
p<+\infty,\ 0<\alpha<\min\\{1/p,1/2\\}$ version (22) with probability 1
belongs to the Besov space $B^{\alpha}_{p,p}([0,1]^{d})$.

###### Proof.

The statement follows from Theorems 1 and 4. ∎

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# Some QCH Kähler surfaces with zero scalar curvature.

Włodzimierz Jelonek  Institute of Mathematics
Cracow University of Technology
Warszawska 24
31-155 Krakòw, POLAND<EMAIL_ADDRESS>
###### Abstract.

In this paper we prove that some well known Kähler surfaces with zero scalar
curvature are QCH Kähler. We prove that family of generalized Taub-Nut Kähler
surfaces parametrized by $k\in[-1,1]$ is of orthotoric type for $k\in(-1,1)$
and of Calabi type for $k\in\\{-1,1\\}$.

###### Key words and phrases:

Kähler surface, holomorphic sectional curvature, quasi constant holomorphic
sectional curvature, QCH Kähler surface, orthotoric Kähler surface

###### 2010 Mathematics Subject Classification:

53C55,53C25,53B35

##  1\. Introduction

In the paper we study toric Kähler surfaces with zero scalar curvature (see
[W]). Kähler surfaces are an object of intense study ( see [A-C-G], [D], [Do],
[D-T], [G-R], [W]). We assume that $(M,g,J)$ is a Kähler surface admmiting two
linearly independent holomorphic Killing commuting vector fields
$X_{1},X_{2}$, i.e.
$\mathfrak{h}=\operatorname{span}\\{X_{1},X_{2}\\}\subset\mathfrak{hol}(M)$.
Let $\phi_{1},\phi_{2}$ be potentials of the fields $X_{1},X_{2}$ i.e.
$X_{i}=J\nabla\phi_{i}$. The moment map
$(\phi_{1},\phi_{2}):M\rightarrow\Sigma\subset\mathbb{R}^{2}$ produces the
Riemannian projection onto the metric polytope. On the toric Kähler surfaces
with zero scalar curvature the volumetric normal coordinates $(x,y)$ on
$\sigma$ are defined. We have
$x=\sqrt{|X_{1}|^{2}|X_{2}|^{2}-<X_{1},X_{2}>^{2}}$ which is a harmonic
function on $\Sigma$ and $y=*dx$.

QCH Kähler surfaces (see [J-1]-[J-7]) are Kähler surfaces $(M,g,J)$ admitting
a global, $2$-dimensional, $J$-invariant distribution $\mathcal{D}$ having the
following property: The holomorphic curvature $K(\pi)=R(X,JX,JX,X)$ of any
$J$-invariant $2$-plane $\pi\subset T_{x}M$, where $X\in\pi$ and $g(X,X)=1$,
depends only on the point $x$ and number
$|X_{\mathcal{D}}|=\sqrt{g(X_{\mathcal{D}},X_{\mathcal{D}})}$, where
$X_{\mathcal{D}}$ is the orthogonal projection of $X$ onto $\mathcal{D}$ (see
[J-1]). Every QCH Kähler surface admits an opposite almost Hermitian structure
$I$ such that the Ricci tensor $\rho$ of $(M,g,J)$ is $I$-invariant and
$(M,g,I)$ satisfies the second Gray condition:

$(G2)R(X,Y,Z,W)-R(IX,IY,Z,W)=R(IX,Y,IZ,W)+R(IX,Y,Z,IW).$

Here we restrict to QCH Kähler surfaces for which the structure $I$ is
integrable. These are essentialy generalized Calabi type Kähler surfaces and
generalized orthotoric type Kähler surfaces. If additionaly $I$ is conformally
Kähler then they are Calabi type and orthotoric type Kähler surfaces. The Lee
form $\theta$ of the Hermitian structure $I$ satisfies
$d\omega_{I}=2\theta\wedge\omega_{I}$ where $\omega_{I}(X,Y)=g(IX,Y)$ is the
Kähler form of $(M,g,I)$.

######  Definition 1.1.

A generalized Calabi type QCH Kähler surface is a QCH Kähler surface for which
$I$ is Hermitian, $|\nabla I|>0$ and one of the distributions
$\mathcal{D},\mathcal{D}^{\perp}$ coincides with
$\mathcal{D}_{0}=\operatorname{span}\\{E_{3},E_{4}\\}=\operatorname{span}\\{\theta^{\sharp},I\theta^{\sharp}\\}=\operatorname{Null}I$
and is integrable (see [J-4],[J-6]).

######  Definition 1.2.

A generalized orthotoric QCH Kähler surface is a QCH Kähler surface with
integrable structure $I$ with $|\nabla I|>0$ and such that both distributions
$\mathcal{D},\mathcal{D}^{\perp}$ satisfy
$\mathcal{D}\cap\mathcal{D}_{0}=\mathcal{D}^{\perp}\cap\mathcal{D}_{0}=\\{0\\}$.

An orthotoric type Kähler surface is a Kähler surface $(M,g,J)$ with a metric

$g=(\xi-\eta)(\frac{1}{F(\xi)}d\xi^{2}-\frac{1}{G(\eta)}d\eta^{2})+\frac{1}{\xi-\eta}(F(\xi)(dt+\eta
dz)^{2}-G(\eta)(dt+\xi dz)^{2}.$

These are Kähler surfaces with zero scalar curvature if and only if
$F(\xi)=A\xi^{2}+a\xi+b,G(\eta)=A\xi^{2}+c\eta+d$.

A QCH Kähler surface with integrable $I$ is of generalized Calabi type if and
only if $d(\omega_{J}-\omega_{I})=\phi\wedge(\omega_{J}-\omega_{I})$ or
$d(\omega_{J}+\omega_{I})=\phi\wedge(\omega_{J}+\omega_{I})$ for some 1-form
$\phi$.

######  Definition 1.3.

We say that a QCH Kähler surface $(M,g,J)$ is of orthotoric or of Calabi type
if there exists an open and dense subset $U\subset M$ such that
$(U,g_{|U},J_{|U})$ is of orthotoric or of Calabi type Kähler surface.

.

In the paper we prove that all Kähler surfaces investigated by Weber in [W]
are QCH Kähler surfaces of orthotoric or Calabi type. The generalized Taub-Nut
Kähler surfaces are orthotoric for $k\in(-1,1)$ and Calabi type for
$k\in\\{-1,1\\}$. The exceptional half plane surface is of Calabi type. We
also prove that the Burns metric (see [L], [B]) is of Calabi type. If
$(M,g,J)$ is not hyperkähler then the opposite Hermitian structure $I$ has the
Kähler form proportional to the Ricci form $\rho$ of $(M,g,J)$, which explains
why $\rho$ is an eigenvalue of the Weyl tensor $W^{-}$ and why $W^{-}$ is
degenerate which was noticed by Weber in [W].

##  2\. Generalized Taub-Nut surfaces

The generalized Taub-Nut surfaces were first discribed by Donaldson in [Do](
see also [W]). For the generalized Taub-Nut surfaces we have
$M=\mathbb{C}^{2}$ and (see [W])

$\displaystyle\phi_{1}=\frac{1}{\sqrt{2}}(-y+\sqrt{x^{2}+y^{2}})+\frac{\alpha}{2}x^{2}$
$\displaystyle\phi_{2}=\frac{1}{\sqrt{2}}(y+\sqrt{x^{2}+y^{2}})+\frac{\beta}{2}x^{2}$

and

$g_{\Sigma}=\frac{1+2M(ky+\sqrt{x^{2}+y^{2}})}{\sqrt{x^{2}+y^{2}}}(dx\otimes
dx+dy\otimes dy).$

Note that for $k=0$ we get standard Taub-Nut hyperkähler surface.

######  Theorem 2.1.

Let $(M,g,J)$ be a generalized Taub-Nut Kähler surface with $k\in(-1,1)$. Then
$(M,g,J)$ is an orthotoric type QCH Kähler surface with a metric (on an open
and dense subset)

$g=(\xi-\eta)(\frac{1}{F(\xi)}d\xi^{2}-\frac{1}{G(\eta)}d\eta^{2})+\frac{1}{\xi-\eta}(F(\xi)(dt+\eta
dz)^{2}-G(\eta)(dt+\xi dz)^{2}.$

where $F(\xi)=a\xi+b,G(\eta)=c\eta+d$, and $a,c>0$, $da-bc>0$. We also have
$M=\frac{(c+a)a^{2}c^{2}}{4(da-bc)^{2}}$, $k=\frac{a-c}{a+c}$

###### Proof.

Let $F(\xi)=a\xi+b,G(\eta)=c\eta+d$ where $a,c>0$ and
$\xi\geq\xi_{0}=-\frac{b}{a},\eta\leq\eta_{0}=-\frac{d}{c}$. Assume that
$-\frac{b}{a}+\frac{d}{c}>0$ which means $da-bc>0$. Consider a Kähler surface
$(M,g,J)$ with a metric

$g=(\xi-\eta)(\frac{1}{F(\xi)}d\xi^{2}-\frac{1}{G(\eta)}d\eta^{2})+\frac{1}{\xi-\eta}(F(\xi)(dt+\eta
dz)^{2}-G(\eta)(dt+\xi dz)^{2}.$

The fields $X_{1}=\frac{\partial}{\partial t},X_{2}=\frac{\partial}{\partial
z}$ are holomorphic Killing with potentials
$\phi_{1}=\xi+\eta,\phi_{2}=\xi\eta$ respectively. Note that
$g(X_{1},X_{1})=\frac{1}{\xi-\eta}(F-G),g(X_{2},X_{2})=\frac{1}{\xi-\eta}(F\eta^{2}-G\xi^{2}),g(X_{1},X_{2})=\frac{1}{\xi-\eta}(\eta
F-\xi G)$.

Let
$\widetilde{x}=\frac{2}{a}\sqrt{a\xi+b},\widetilde{y}=\frac{2}{c}\sqrt{-(c\xi+d)}$
be isothermal coordinates on $M$,
$\widetilde{z}=\widetilde{x}+i\widetilde{y}$. Then the metric on $\Sigma$ is
$g_{\Sigma}=(\xi-\eta)(d\widetilde{x}^{2}+d\widetilde{y}^{2})$.

The area of a parallelogram defined by $X_{1},X_{2}$ is
$\mathcal{V}=\sqrt{-FG}$. Let us introduce the volumetric coordinates
$x=\sqrt{-FG}$ and $y$ such that $dy=*dx$. Then
$y=\frac{c}{2}\xi+\frac{a}{2}\eta+e$ where $e=\frac{cb}{2a}+\frac{ad}{2c}$ we
choose in such a way that $(\xi_{0},\eta_{0})$ has a new coordinates $(0,0)$.
If $z=x+iy$ then $z=\frac{ac}{8}\widetilde{z}^{2}$. ZHence
$dx^{2}+dy^{2}=a^{2}c^{2}\frac{|\widetilde{z}|^{2}}{16}(d\widetilde{x}^{2}+d\widetilde{y}^{2})$.

Consequently
$g_{\Sigma}=\frac{16(\xi-\eta)}{a^{2}c^{2}|z|^{2}}(dx^{2}+dy^{2})=\frac{4(\xi-\eta)}{c^{2}F-a^{2}G}(dx^{2}+dy^{2})$.

Note that $\sqrt{x^{2}+y^{2}}=(\frac{c}{2}\xi-\frac{a}{2}\eta+\gamma)$ where
$\gamma=\frac{bc^{2}-da^{2}}{2ac}$. Define a Lie subalgebra
$\mathfrak{h}=span\\{X_{1},X_{2}\\}\subset\mathfrak{hol}(M)$ of the Lie
algebra of holomorphic Killing vector fields on $M$. If we change a basis of
this algebra $Y_{1}=AX_{1}+BY_{2},Y_{2}=CX_{1}+DX_{2}$ then for potentials
$\psi_{1},\psi_{2}$ of fields $Y_{1},Y_{2}$ we get
$\psi_{1}=A\phi_{1}+B\phi_{2},\psi_{2}=C\phi_{1}+D\phi_{2}$ (up to a
constant). We also have $det(Y_{1},Y_{2})=|AD-BC|det(X_{1},X_{2})$. Hence the
appropriate volumetric coordinates satisfy $z_{1}=|AD-BC|z$.

Note that

$\displaystyle\xi+\eta+\frac{b}{a}+\frac{d}{c}=\frac{1}{c}(y+\sqrt{x^{2}+y^{2}})-\frac{1}{a}(-y+\sqrt{x^{2}+y^{2}}),$
$\displaystyle\xi+\frac{b}{a}=\frac{1}{c}(y+\sqrt{x^{2}+y^{2}}),\eta+\frac{d}{c}=-\frac{1}{a}(-y+\sqrt{x^{2}+y^{2}}),$
$\displaystyle\xi\eta-\frac{bd}{ac}=-\frac{1}{ac}x^{2}+\frac{ad}{c}(y+\sqrt{x^{2}+y^{2}})-\frac{bc}{a}(-y+\sqrt{x^{2}+y^{2}})$

Now we find how to change the basis to obtain the moment mappings from the
Weber work.

We have
$\psi_{1}=\frac{1}{\sqrt{2}}(y^{\prime}+\sqrt{(x^{\prime})^{2}+(y^{\prime})^{2}})+A(x^{\prime})^{2},\psi_{2}=\frac{1}{\sqrt{2}}(-y^{\prime}+\sqrt{(x^{\prime})^{2}+(y^{\prime})^{2}})+B(x^{\prime})^{2}$
in a new volumetric coordinates corresponding to $\psi_{1},\psi_{2}$.

Hence $x^{\prime}=\alpha x,y^{\prime}=\alpha y$.

$\psi_{2}=\frac{1}{\sqrt{2}}(y^{\prime}+\sqrt{(x^{\prime})^{2}+(y^{\prime})^{2}})+A(x^{\prime})^{2}=\alpha\frac{1}{\sqrt{2}}(y+\sqrt{x^{2}+y^{2}})+\alpha^{2}Ax^{2}=\alpha\frac{c}{\sqrt{2}}\xi+\alpha\frac{ad}{\sqrt{2}c}-\alpha^{2}Aab\xi\eta-\alpha^{2}Ada\xi-\alpha^{2}Abc\eta-\alpha^{2}Adb=(\frac{1}{\sqrt{2}}\alpha
c-\alpha^{2}Ada)\xi-\alpha^{2}Abc\eta-\alpha^{2}Aab\xi\eta+\alpha\frac{da}{\sqrt{2}c}-\alpha^{2}Adb$
has to be a linear combination of $\xi+\eta,\xi\eta$ and a constant.

It follows $A=\frac{c}{\alpha\sqrt{2}(da-bc)}$.

$\psi_{1}=\frac{1}{\sqrt{2}}(-y^{\prime}+\sqrt{(x^{\prime})^{2}+(y^{\prime})^{2}})+B(x^{\prime})^{2}=\alpha\frac{1}{\sqrt{2}}(-y+\sqrt{x^{2}+y^{2}})+\alpha^{2}Bx^{2}=\alpha\frac{a}{\sqrt{2}}\eta+\alpha\frac{cb}{\sqrt{2}a}-\alpha^{2}Bab\xi\eta-\alpha^{2}Bda\xi-\alpha^{2}Bbc\eta-\alpha^{2}Bdb=-\alpha^{2}Ada)\xi+(\alpha\frac{a}{\sqrt{2}}-\alpha^{2}Abc)\eta-\alpha^{2}Aab\xi\eta+\alpha\frac{da}{\sqrt{2}c}-\alpha^{2}Adb$
Similarly for $\psi_{2}$ we have $B=\frac{a}{\alpha\sqrt{2}(da-bc)}$.
Consequently

$\psi_{1}=\frac{\alpha bc^{2}}{\sqrt{2}(da-bc)}\phi_{1}+\frac{\alpha
ac^{2}}{\sqrt{2}(da-bc)}\phi_{2}+C_{1},$

$\psi_{2}=\frac{\alpha da^{2}}{\sqrt{2}(da-bc)}\phi_{1}+\frac{\alpha
a^{2}c}{\sqrt{2}(da-bc)}\phi_{2}+C_{2}$ where $C_{1},C_{2}$ are constant.

Hence the area of a parallelogram is
$\alpha\mathcal{V}=\frac{\alpha^{2}a^{2}c^{2}}{2(da-bc)}\mathcal{V}$ hence
$\alpha=\frac{2(da-bc)}{a^{2}c^{2}}$.

To show that generalized taub-Nut and orthotoric with $F,G$ are isometric we
have to prove that

$\frac{4(\xi-\eta)}{c^{2}F-a^{2}G}=\frac{\alpha(1+2\alpha
M(ky+\sqrt{x^{2}+y^{2}}))}{\sqrt{x^{2}+y^{2}}}$. We prove that
$k=\frac{a-c}{a+c}$ and $M=\frac{(c+a)a^{2}c^{2}}{4(da-bc)^{2}}$.

Note that $c^{2}F-a^{2}G=2ac\sqrt{x^{2}+y^{2}}$. Hence we have to show that
$4(\xi-\eta)=2ac\alpha(1+2\alpha M(ky+\sqrt{x^{2}+y^{2}}))$.

$ky+\sqrt{x^{2}+y^{2}}=k(\frac{a}{2}\eta+\frac{c}{2}\xi+e)-\frac{a}{2}\eta+\frac{c}{2}\xi+\gamma=(k-1)\frac{a}{2}\eta+(k+1)\frac{c}{2}\xi+ke-\gamma$

and $(1-k)\frac{a}{2}=(k+1)\frac{c}{2}$ thus $k=\frac{a-c}{a+c}$ and
$\alpha^{2}ac^{2}M\frac{2a}{c+a}=2$ and $M\frac{4(da-
bc)^{2}}{a^{2}c^{2}(c+a)}=1$. It implies $M=\frac{(c+a)a^{2}c^{2}}{4(da-
bc)^{2}}$. Then $2\alpha M(ke+\gamma)=-1$ ∎

Now we introduce global coordinates on $M=\mathbb{C}^{2}$. Let us introduce
coordinates (compare [W])

$u=\sqrt{\frac{\alpha ca}{2}}\widetilde{x},v=\sqrt{\frac{\alpha
ca}{2}}\widetilde{y}$. Then

$g_{\Sigma}=\frac{2}{M}(1+(1+k)u^{2})+(1-k)v^{2})(du^{2}+dv^{2})$ and change
the coordinates $\theta_{i}$ by
$\theta_{i}^{\prime}=\frac{1}{\sqrt{2}}\theta_{i}$. Then (we denote
$\theta_{i}^{\prime}$ back by $\theta_{i}$)

$g=g_{\Sigma}+G^{ij}d\theta_{i}d\theta_{j}$ where

$\displaystyle
MG^{11}=\frac{2v^{2}((1+(1+k)u^{2})^{2}+(1+k)^{2}u^{2}v^{2}}{1+(1+k)u^{2}+(1-k)v^{2}},$
$\displaystyle
MG^{12}=MG^{21}=\frac{2u^{2}v^{2}(2+(1-k^{2})(u^{2}+v^{2})}{1+(1+k)u^{2}+(1-k)v^{2}}$
$\displaystyle
MG^{22}=\frac{2u^{2}((1+(1-k)v^{2})^{2}+(1-k)^{2}u^{2}v^{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$

.

Let us introduce a new coordinates
$x_{1}=v\cos\theta_{1},y_{1}=v\sin\theta_{1},x_{2}=u\cos\theta_{2},y_{1}=u\sin\theta_{2}$.
We get

$\frac{\partial}{\partial x_{1}}=\cos\theta_{1}\frac{\partial}{\partial
v}-\frac{1}{v}\sin\theta_{1}\frac{\partial}{\partial\theta_{1}},\frac{\partial}{\partial
y_{1}}=\sin\theta_{1}\frac{\partial}{\partial
v}+\frac{1}{v}\cos\theta_{1}\frac{\partial}{\partial\theta_{1}}$,

$\frac{\partial}{\partial x_{2}}=\cos\theta_{2}\frac{\partial}{\partial
u}-\frac{1}{u}\sin\theta_{2}\frac{\partial}{\partial\theta_{2}},\frac{\partial}{\partial
y_{2}}=\sin\theta_{2}\frac{\partial}{\partial
u}+\frac{1}{u}\cos\theta_{2}\frac{\partial}{\partial\theta_{2}}$.

Consequently

$Mg(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
x_{1}})=2(1+(1+k)u^{2})+2(1-k)x_{1}^{2}+\frac{2((k-1)y_{1}^{2}(1+(1+k)u^{2})+y_{1}^{2}(1+k)^{2}u^{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
y_{1}})=2(1-k)x_{1}y_{1}-2\frac{(1+k)^{2}u^{2}x_{1}y_{1}}{1+(1+k)u^{2}+(1-k)v^{2}}+\frac{2(1+(1+k)u^{2})(1-k)x_{1}y_{1}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial
y_{1}})=2(1+(1+k)u^{2})+2(1-k)y_{1}^{2}+\frac{2(1+(1+k)u^{2})(k-1)x_{1}^{2}+x_{1}^{2}(1+k)^{2}u^{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
x_{2}})=2y_{1}y_{2}\frac{2+(1-k^{2})(u^{2}+v^{2})}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
y_{2}})=-2y_{1}x_{2}\frac{2+(1-k^{2})(u^{2}+v^{2})}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial x_{2}},\frac{\partial}{\partial
x_{2}})=2(1+(1-k)v^{2})+2(1+k)x_{2}^{2}+\frac{2(-(k+1)y_{2}^{2}(1+(1-k)v^{2})+y_{2}^{2}(1-k)^{2}v^{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial x_{2}},\frac{\partial}{\partial
y_{2}})=2(1+k)x_{2}y_{2}-2\frac{(1-k)^{2}v^{2}x_{2}y_{2}}{1+(1-k)v^{2}+(1+k)u^{2}}+\frac{2(1+(1-k)v^{2})(1+k)x_{2}y_{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial y_{2}},\frac{\partial}{\partial
y_{2}})=2(1+(1-k)v^{2})+2(1+k)y_{2}^{2}+\frac{2(1+(1-k)v^{2})(-k-1)x_{2}^{2}+x_{2}^{2}(1-k)^{2}v^{2}}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial
y_{2}})=2x_{1}x_{2}\frac{2+(1-k^{2})(u^{2}+v^{2})}{1+(1+k)u^{2}+(1-k)v^{2}}$,

$Mg(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial
x_{2}})=-2x_{1}y_{2}\frac{2+(1-k^{2})(u^{2}+v^{2})}{1+(1+k)u^{2}+(1-k)v^{2}}$.

Since $u^{2}=x_{2}^{2}+y_{2}^{2},v^{2}=x_{1}^{2}+y_{2}^{2}$ it follows $g$ is
defined in coordinates $z_{1}=x_{1}+iy_{2},z_{2}=x_{2}+iy_{2}$ on the whole of
$\mathbb{C}^{2}$.

On $int(\Sigma)\times T^{2}$ we have coordinates $u,v,\theta_{1},\theta_{2}$
where $u>,v>0,\theta_{1}\in[0,2\pi),\theta_{2}\in[0,2\pi)$ (in fakt
$\theta_{i}$ are coordinates on $T^{2}=\mathbb{R}^{2}/2\pi\mathbb{Z}^{2}$. It
is easy to check that the maps $u,v,\theta_{1},\theta_{2}$ and
$x_{1},y_{1},x_{2},y_{2}$ agree on $int(\Sigma)\times T^{2}$ hence for
$u>0,v>0$ and $int(\Sigma)\times T^{2}$ in a new coordinates
$z_{1}=x_{1}+iy_{2},z_{2}=x_{2}+iy_{2}$ is
$(\mathbb{C}-\\{0\\})\times(\mathbb{C}-\\{0\\})$. The basis of killing vector
fields is $Y_{1}=x_{1}\frac{\partial}{\partial
y_{1}}-y_{1}\frac{\partial}{\partial
x_{1}},Y_{2}=x_{2}\frac{\partial}{\partial
y_{2}}-y_{2}\frac{\partial}{\partial x_{2}}$. The (real) surface
$L=\\{Y_{1}=0\\}$ in a new coordinates is the plane $x_{1}=y_{1}=0$. We shall
show that it coincides with one of the distributions $\mathcal{D},\mathcal{E}$
from the definition of $QCH$ Kähler surfaces. Let $TL$ be a tangent space of
$L$.

Let $T=\nabla Y_{1},S=\nabla Y_{2}$. Since on $L$ we have
$0=(L_{Y_{1}}-\nabla_{Y_{1}})I=[T,I]$ and $[T,J]=0$ it follows that $kerT=TL$
is $I,J$ ivariant. The equality $dimTL=2$ implies $I=J$ on $TL$ or $I=-J$ on
$TL$. Hence $TL$ coincides with $ker(IJ-id)$ or $ker(IJ+id)$ hence with
$\mathcal{D}$ or $\mathcal{E}$. Note that $R(Y_{1},Y_{2})=[T,S]$. Consequently
on $TL$ we have$[T,S]=0$. It implies $S(TL)\subset TL$. Thus $S=aI$ on $TL$
which means $\nabla_{Y_{2}}I=[S,I]=0$ on $L$. Therefore $Y_{2},JY_{2}\in TL$
and $Nul(I)=span(Y_{2},JY_{2})=TL$ where $Nul(I)=\\{X:\nabla_{X}I=0\\}$. Note
that at a point $(0,0,0,0)$ we have $\nabla I=0$ since
$||\nabla(\xi-\eta)||^{2}=\frac{1}{\xi-\eta}(F(\xi)-G(\eta)$ tends to $0$ when
$\xi\rightarrow\xi_{0},\eta\rightarrow\eta_{0}$.

We find the Kähler forms $\omega_{J},\omega_{I}$. We have (where again
$\theta_{i}^{\prime}=\theta_{i}$)

$\omega_{J}=\sqrt{2}(d\phi_{1}\wedge d\theta_{1}+d\phi_{2}\wedge d\theta_{2})$
where
$\phi_{1}=\frac{v^{2}}{\sqrt{2}M}(1+(1+k)u^{2}),\phi_{2}=\frac{u^{2}}{\sqrt{2}M}(1+(1-k)v^{2})$.
It is easy to show

$\frac{M}{2}\omega_{J}=(1+(1+k)u^{2})dx_{1}\wedge
dy_{1}+(1+(1-k)v^{2})dx_{2}\wedge
dy_{2}+(1+k)(x_{2}dx_{2}+y_{2}dy_{2})\wedge(x_{1}dy_{1}-y_{1}dx_{1})+(1-k)(x_{1}dx_{1}+y_{1}dy_{1})\wedge(x_{2}dy_{2}-y_{2}dx_{2})$.

Similarly

$\frac{M}{2}\omega_{I}=(1+(1+k)u^{2})dx_{1}\wedge
dy_{1}-(1+(1-k)v^{2})dx_{2}\wedge
dy_{2}-(1+k)(x_{2}dx_{2}+y_{2}dy_{2})\wedge(x_{1}dy_{1}-y_{1}dx_{1})+(1-k)(x_{1}dx_{1}+y_{1}dy_{1})\wedge(x_{2}dy_{2}-y_{2}dx_{2})$.

##  3\. Exceptional generalized Taub-Nut.

######  Theorem 3.1.

The exceptional generalized Taub-Nut surface is a QCH Kähler surface of Calabi
type.

###### Proof.

In an exceptional generalized Taub-Nut corresponding to $k=1$ we have
$g_{\Sigma}=2(1+2u^{2})(du^{2}+dv^{2})$,

$g=g_{\Sigma}+G^{ij}d\theta_{i}d\theta_{j}$ where

$MG^{11}=\frac{v^{2}((1+2u^{2})^{2}+4u^{2}v^{2}}{1+2u^{2}},MG^{12}=MG^{21}=\frac{2u^{2}v^{2}(u^{2}+v^{2})}{1+2u^{2}}$,

$MG^{22}=\frac{u^{2}}{1+2u^{2}}$. We also have
$\omega_{J}=\frac{2v}{\sqrt{2}}(1+2u^{2})dv\wedge
d\theta_{1}+\frac{4uv^{2}}{\sqrt{2}}du\wedge
d\theta_{1}+\frac{2u}{\sqrt{2}}du\wedge d\theta_{2}$,
$\omega_{I}=\frac{2v}{\sqrt{2}}(1+2u^{2})dv\wedge
d\theta_{1}-\frac{4uv^{2}}{\sqrt{2}}du\wedge
d\theta_{1}-\frac{2u}{\sqrt{2}}du\wedge d\theta_{2}$,

$I\frac{\partial}{\partial
u}=-\sqrt{2}\frac{(1+2u^{2})}{u}\frac{\partial}{\partial\theta_{2}},I\frac{\partial}{\partial
v}=-2\sqrt{2}v\frac{\partial}{\partial\theta_{2}}+\frac{\sqrt{2}}{v}\frac{\partial}{\partial\theta_{1}}$

and

$I\frac{\partial}{\partial\theta_{2}}=\frac{1}{\sqrt{2}}\frac{u}{1+2u^{2}}\frac{\partial}{\partial
u},I\frac{\partial}{\partial\theta_{1}}=\sqrt{2}\frac{u^{2}}{1+2u^{2}}\frac{\partial}{\partial
u}-\frac{\sqrt{2}}{2}v\frac{\partial}{\partial v}$.

Hence $\frac{1}{2}(\omega_{J}-\omega_{I})=\frac{4uv^{2}}{\sqrt{2}}du\wedge
d\theta_{1}+\frac{2u}{\sqrt{2}}du\wedge
d\theta_{2},\phi=\frac{1}{2}(\omega_{J}+\omega_{I})=\frac{2v}{\sqrt{2}}(1+2u^{2})dv\wedge
d\theta_{1}$.

Consequently $d\phi=\frac{8vu}{\sqrt{2}}du\wedge dv\wedge
d\theta_{1}=\frac{4u}{1+2u^{2}}du\wedge\phi=d\ln(1+2u^{2})\wedge\phi$. It
follows that exceptional Taub-Nut is of Calabi type Calabi type. ∎

##  4\. Exceptional half plane.

For an eceptional half plane instanton we have $M=\mathbb{C}^{2}$. Recall that
$g=(1+x^{2})(dx^{2}+dy^{2})+G^{ij}d\theta_{i}d\theta_{j}$ where
$G^{11}=\frac{x^{2}}{1+x^{2}},G^{12}=G^{21}=\frac{2x^{2}y}{1+x^{2}},G^{22}=\frac{(1+x^{2})^{2}+4x^{2}y^{2}}{1+x^{2}}$
(see [W]).

######  Theorem 4.1.

An exceptional half plane is a QCH Kähler surface of Calabi type.

###### Proof.

First we give global coordinates on $M$. We introduce new coordinates
$x_{1}=x\cos\theta_{1},y_{1}=x\sin\theta_{1},x_{2}=y,y_{2}=\theta_{2}$. Since

$\frac{\partial}{\partial x_{1}}=\cos\theta_{1}\frac{\partial}{\partial
x}-\frac{1}{x}\sin\theta_{1}\frac{\partial}{\partial\theta_{1}},\frac{\partial}{\partial
x_{1}}=\frac{\partial}{\partial y},\frac{\partial}{\partial
y_{1}}=\frac{\partial}{\partial\theta_{2}}$.

it follows that

$g(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
x_{1}})=1+x_{1}^{2}-\frac{y_{1}^{2}}{1+x^{2}},g(\frac{\partial}{\partial
x_{1}},\frac{\partial}{\partial
y_{1}})=x_{1}y_{1}+\frac{x_{1}y_{1}}{1+x^{2}},g(\frac{\partial}{\partial
x_{1}},\frac{\partial}{\partial x_{2}})=0$,

$g(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial
y_{2}})=-\frac{2y_{1}x_{2}}{1+x^{2}},g(\frac{\partial}{\partial
x_{2}},\frac{\partial}{\partial x_{2}})=1+x^{2},g(\frac{\partial}{\partial
y_{2}},\frac{\partial}{\partial
y_{2}})=\frac{(1+x^{2})^{2}+4x^{2}x_{2}^{2}}{1+x^{2}}$.

The Kähler form of $(M,g,I)$ is
$\omega_{I}=\frac{1}{|Ric|}(-d\mathcal{R}^{1}\wedge
d\theta_{1}-d\mathcal{R}^{2}\wedge d\theta_{2})=xdx\wedge
d\theta_{1}-(1+x^{2})dy\wedge d\theta_{2}+2xydx\wedge d\theta_{2}$ Hence

$I\frac{\partial}{\partial
x}=\frac{(1+x^{2})}{x}\frac{\partial}{\partial\theta_{1}},I\frac{\partial}{\partial
y}=-\frac{\partial}{\partial\theta_{2}}+2y\frac{\partial}{\partial\theta_{1}}$

and

$I\frac{\partial}{\partial\theta_{2}}=-\frac{2xy}{1+x^{2}}\frac{\partial}{\partial
x}+\frac{\partial}{\partial
y},I\frac{\partial}{\partial\theta_{1}}=-\frac{x}{1+x^{2}}\frac{\partial}{\partial
x}$

The Kähler form of $(M,g,J)$ is $\omega_{J}=xdx\wedge
d\theta_{1}+(1+x^{2})dy\wedge d\theta_{2}+2xydx\wedge d\theta_{2}$. Since
$d\omega_{I}=\frac{4xdx}{1+x^{2}}\wedge\omega_{I}$, it follows that the Lee
form for $(M,g,I)$ is $\theta=d\ln(1+x^{2})$ i.e.
$d\omega_{I}=2\theta\wedge\omega_{I}$. One can easily check that
$N_{I}(X,Y)=0$ hence $I$ is integrable and $(M,g,I)$ is conformally Kähler.
Thus $(M,g,J)$ is a Kähler surface of Calabi type. The Killing fields are
$Y_{1}=x_{1}\frac{\partial}{\partial y_{1}}-y_{1}\frac{\partial}{\partial
x_{1}},Y_{2}=\frac{\partial}{\partial y_{2}}$.∎

Note that on $M$ acts a cylinder $\mathbb{R}\times S^{1}$ rather than a torus
$T^{2}$.

##  5\. The Burns metric.

Burns proved that the metric
$\omega=\partial\overline{\partial}(||z||^{2}+m\ln||z||^{2})$ on
$\mathbb{C}^{2}-\\{0\\}$ extends to the Kähler metric on the blow-up of
$\mathbb{C}^{2}$ at $0$. Let us introduce the coordinates $(u,z)$ on
$\widehat{\mathbb{C}^{2}}$ such that $z_{1}=z,z_{2}=uz$ on
$\mathbb{C}^{2}-\\{0\\}$. Wtedy
$||z||^{2}=|z_{1}|^{2}+|z_{2}|^{2}=|z|^{2}(1+|u|^{2})$.

$=\partial\overline{\partial}(||z||^{2})==\partial\overline{\partial}(|z|^{2}(1+|u|^{2})=\partial(zd\overline{z}+|u|^{2}zd\overline{z}+|z|^{2}ud\overline{u}=dz\wedge
d\overline{z}+|u|^{2}dz\wedge d\overline{z}+\overline{u}zdu\wedge
d\overline{z}+|z|^{2}du\wedge d\overline{u}+\overline{z}udz\wedge
d\overline{u}$.

$\partial\overline{\partial}(\ln||z||^{2})=\partial\overline{\partial}(\ln|z|^{2}(1+|u|^{2})=\partial\overline{\partial}(\ln(1+|u|^{2})=\partial(\frac{ud\overline{u}}{1+|u|^{2}}=\frac{du\wedge
d\overline{u}}{1+|u|^{2}}-\frac{u\overline{u}du\wedge
d\overline{u}}{(1+|u|^{2})^{2}}=\frac{du\wedge
d\overline{u}}{(1+|u|^{2})^{2}}$.

######  Theorem 5.1.

The space $\widehat{\mathbb{C}^{2}}$ with the Burns metric is a QCH Kähler
surface of Calabi type.

Hence $\omega=(1+|u|^{2})dz\wedge
d\overline{z}+(|z|^{2}+\frac{m}{(1+|u|^{2})^{2}})du\wedge
d\overline{u}+\overline{u}zdu\wedge d\overline{z}+u\overline{z}dz\wedge
d\overline{u}$.

One can also see that $\omega^{2}=2(|z|^{2}+\frac{m}{1+|u|^{2}})dz\wedge
d\overline{z}\wedge du\wedge d\overline{u}$ and the Ricci form is

$\rho=\partial\overline{\partial}\ln(|z|^{2}+\frac{m}{1+|u|^{2}})$.

Note that
$\overline{\partial}\ln(|z|^{2}+\frac{m}{1+|u|^{2}})=\frac{1}{|z|^{2}+\frac{m}{1+|u|^{2}}}(zd\overline{z}-\frac{mud\overline{u}}{(1+|u|^{2})^{2}})$.

We also have
$\partial\overline{\partial}\ln(|z|^{2}+\frac{m}{1+|u|^{2}})=\frac{1}{|z|^{2}+\frac{m}{1+|u|^{2}}}(dz\wedge
d\overline{z}-\frac{mdu\wedge
d\overline{u}}{(1+|u|^{2})^{2}}+\frac{2m|u|^{2}du\wedge
d\overline{u}}{(1+|u|^{2})^{3}})-\frac{1}{(|z|^{2}+\frac{m}{1+|u|^{2}})^{2}}(\overline{z}dz-\frac{m\overline{u}du}{(1+|u|^{2})^{2}})\wedge(zd\overline{z}-\frac{mud\overline{u}}{(1+|u|^{2})^{2}})$.

Consequently
$\rho=\frac{m}{(|z|^{2}+|z|^{2}|u|^{2}+m)^{2}}((1+|u|^{2})dz\wedge
d\overline{z}+\frac{(|z|^{2}+|z|^{2}|u|^{2}+m)(|u|^{2}-1)-m|u|^{2}}{(1+|u|^{2})^{2}}du\wedge
d\overline{u}+\overline{u}zdu\wedge d\overline{z}+u\overline{z}dz\wedge
d\overline{u})$.

Thus
$\omega\wedge\rho=\frac{m}{(|z|^{2}+|z|^{2}|u|^{2}+m)^{2}}(\frac{(|z|^{2}+|z|^{2}|u|^{2}+m)(|u|^{2}-1)-m|u|^{2}}{(1+|u|^{2})}+(|z|^{2}+\frac{m}{(1+|u|^{2})^{2}})(1+|u|^{2})-2|u|^{2}|z|^{2})du\wedge
d\overline{u}\wedge dz\wedge d\overline{z}=0$.

The diffeomorphism
$\phi:\mathbb{C}^{2}-\\{0\\}\rightarrow\mathbb{C}^{2}-\\{0\\}$,
$\phi(z)=\frac{z}{||z||^{2}}$, extends to the diffeomorphism
$\phi:\widehat{\mathbb{C}^{2}}\rightarrow\mathbb{CP}^{2}-\\{[1,0,0]\\}$( see
[L1]). Note that in maps
$\phi(u,z)=[|z|^{2}(1+|u|^{2}),z,uz]=[\overline{z}(1+|u|^{2}),1,u]$ and
$\phi(u,z)=[|z|^{2}(1+|u|^{2}),uz,z]=[\overline{z}(1+|u|^{2}),u,1]$. These are
smooth mappings and $\phi((0,[u_{1},u_{2}]))=[0,u_{1},u_{2}]$. Hence it is a
diffeomorphism.

If $z_{1},z_{2}$ are coordinates on $\mathbb{CP}^{2}$ then
$z_{1}=\overline{z}(1+|u|^{2}),z_{2}=u$ and
$dz_{1}=d\overline{z}(1+|u|^{2})+\overline{z}(ud\overline{u}+\overline{u}du),dz_{2}=du$.
Note that $u=z_{2},z=\frac{\overline{z_{1}}}{1+|z_{2}|^{2}}$. Consequently

$du=dz_{2},dz=\frac{\overline{dz_{1}}}{1+|z_{2}|^{2}}-\overline{z_{1}}\frac{z_{2}d\overline{z_{2}}+\overline{z_{2}}dz_{2}}{(1+|z_{2}|^{2})^{2}}.$

$\omega=(1+|u|^{2})\frac{\overline{dz_{1}}}{1+|z_{2}|^{2}}-\overline{z_{1}}\frac{z_{2}d\overline{z_{2}}+\overline{z_{2}}dz_{2}}{(1+|z_{2}|^{2})^{2}}\wedge(\frac{dz_{1}}{1+|z_{2}|^{2}}-z_{1}\frac{\overline{z_{2}}dz_{2}+z_{2}d\overline{z_{2}}}{(1+|z_{2}|^{2})^{2}})+(|\frac{\overline{z_{1}}}{1+|z_{2}|^{2}}|^{2}+\frac{m}{(1+|z_{2}|^{2})^{2}})dz_{2}\wedge
d\overline{z_{2}}+\overline{z_{2}}\frac{\overline{z_{1}}}{1+|z_{2}|^{2}}dz_{2}\wedge(\frac{dz_{1}}{1+|z_{2}|^{2}}-z_{1}\frac{\overline{z_{2}}dz_{2}+z_{2}d\overline{z_{2}}}{(1+|z_{2}|^{2})^{2}})+z_{2}\frac{z_{1}}{1+|z_{2}|^{2}}(\frac{\overline{dz_{1}}}{1+|z_{2}|^{2}}-\overline{z_{1}}\frac{z_{2}d\overline{z_{2}}+\overline{z_{2}}dz_{2}}{(1+|z_{2}|^{2})^{2}}\wedge
d\overline{z_{2}})$

Hence $\omega(I,I)=\omega$ , similarly $\rho(I,I)=\rho$. It follows that
$\omega,\rho$ are $I,J$ invariant. Consequently
$\frac{1}{|\rho|}\rho=\omega_{I}$ where $I$ is the complex structure of
$\mathbb{CP}^{2}$.

Note that $\rho\wedge\rho=-2(\frac{m}{(|z|^{2}+|z|^{2}|u|^{2}+m)^{2}})^{2}vol$
which implies

$\omega_{I}=(1+|u|^{2})dz\wedge
d\overline{z}+\frac{(|z|^{2}+|z|^{2}|u|^{2}+m)(|u|^{2}-1)-m|u|^{2}}{(1+|u|^{2})^{2}}du\wedge
d\overline{u}+\overline{u}zdu\wedge d\overline{z}+u\overline{z}dz\wedge
d\overline{u}$.

We also have
$\frac{1}{2}d(\omega_{J}-\omega_{I})=d\ln(|z|^{2}+|z|^{2}|u|^{2}+m)\wedge\frac{1}{2}(\omega_{J}-\omega_{I})=d\ln(||z||^{2}+m)\wedge\frac{1}{2}(\omega_{J}-\omega_{I})$.
Hence the Burn’s metric is of Calabi type.

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