# Cesàro summability of Taylor series in higher order weighted Dirichlet type
spaces

Soumitra Ghara  ,  Rajeev Gupta  and  Md. Ramiz Reza  Department of
Mathematics
Indian Institute of Technology Kharagpur
Midnapore-721302  School of Mathematics and Computer Science
Indian Institute of Technology Goa
Goa - 403401  School of Mathematics
Indian Institute of Science Education and Research Thiruvananthapuram
Kerala - 695551<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For a positive integer $m$ and a finite non-negative Borel measure $\mu$ on
the unit circle, we study the Hadamard multipliers of higher order weighted
Dirichlet-type spaces $\mathcal{H}_{\mu,m}$. We show that if
$\alpha>\frac{1}{2},$ then for any $f$ in $\mathcal{H}_{\mu,m},$ the sequence
of generalized Cesàro sums $\\{\sigma_{n}^{\alpha}[f]\\}$ converges to $f$. We
further show that if $\alpha=\frac{1}{2}$ then for the Dirac delta measure
supported at any point on the unit circle, the previous statement breaks down
for every positive integer $m$.

###### Key words and phrases:

Weighted Dirichlet-type integrals, Cesàro mean, Hadamard multiplication

###### 2010 Mathematics Subject Classification:

Primary 41A10, 40G05, 46E20 Secondary 41A17

Dedicated to Professor Jan Stochel on the occasion of his 70th anniversary

###  1\. Introduction

The symbols $\mathbb{T}$ and $\mathbb{D}$ will denote the unit circle and the
open unit disc in the complex plane $\mathbb{C}$ respectively. We use the
symbols $\mathbb{Z},\mathbb{N}$ and $\mathbb{Z}_{\geqslant 0}$ to denote the
set of integers, positive integers and non-negative integers respectively. The
notation $\mathcal{M}_{+}(\mathbb{T})$ stands for the set of all finite non-
negative Borel measures on $\mathbb{T}$. Let $\mathcal{O}(\mathbb{D})$ denote
the space of all complex valued holomorphic functions on $\mathbb{D}.$ For a
holomorphic function $f\in\mathcal{O}(\mathbb{D}),$ which has a power series
representation of the form $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k},$
$z\in\mathbb{D},$ the $n^{\it th}$ Taylor partial sum $s_{n}[f]$ and the
$n^{\it th}$ Cesàro sum $\sigma_{n}[f]$ are defined by

$\displaystyle s_{n}[f](z)$
$\displaystyle:=\sum\limits_{k=0}^{n}a_{k}z^{k},~{}~{}\mbox{and}\,\,$
$\displaystyle\sigma_{n}[f](z):=\frac{1}{n+1}\Big{(}\sum\limits_{k=0}^{n}s_{k}[f](z)\Big{)}$
$\displaystyle=\sum\limits_{k=0}^{n}\Big{(}1-\frac{k}{n+1}\Big{)}a_{k}z^{k},\,\,\,\,~{}~{}n\in\mathbb{Z}_{\geqslant
0}.$

When exploring analytic function spaces, a common concern revolves around
determining the density of the polynomial set within the given function space.
Once the density of polynomials is established, the next natural question to
address is how to construct polynomials that can closely approximate a given
function. In this context, it becomes natural to inquire whether, for a
function $f$ residing in a normed function space, the sequence of Taylor
partial sum $\\{s_{n}[f]\\}$ converge to $f$ in the norm associated with that
space. It is worth noting that in several classical function spaces like Hardy
space, Dirichlet space, and Bergman space defined over the unit disc
$\mathbb{D}$, it is well-established that the sequence $\\{s_{n}[f]\\}$ indeed
converges to $f$ as prescribed by the associated norm. However, it is worth
noting that there are instances where this convergence property does not hold.
For example, it is well known that there exists a function $f$ in the disc
algebra $A(\mathbb{D})$ such that the sequence $\\{s_{n}[f]\\}$ does not
converge to $f,$ see [12, page 57]. Another notable family of examples in the
context of Hilbert function space is the family of weighted Dirichlet type
spaces. Richter introduced the weighted Dirichlet space $D(\mu)$ for each
$\mu\in\mathcal{M}_{+}(\mathbb{T})$ in order to study the structure of closed
$M_{z}$-invariant subspaces of the classical Dirichlet space on $\mathbb{D}$
and to obtain a model for cyclic analytic $2$-isometries, see [16].
Interestingly, when $\mu=\delta_{\lambda}$, the Dirac delta measure at a point
$\lambda\in\mathbb{T}$, it has been established that there exists a function
$f$ in $D(\delta_{\lambda})$ for which the sequence $\\{s_{n}[f]\\}$ does not
converge to $f$ within the space $D(\delta_{\lambda})$, see [7, Exercise 7.3
(2)]. On the contrary, a remarkable discovery surfaced when Mashreghi and
Ransford recently showed that for any arbitrary, but fixed,
$\mu\in\mathcal{M}_{+}(\mathbb{T})$, the sequence of Cesàro sums
$\\{\sigma_{n}[f]\\}$ converges to $f$ for every $f$ in the space $D(\mu)$
(refer to [14, Theorem 1.6]). Furthermore, they refined their result by
establishing that the sequence of generalized Cesàro sums
$\\{\sigma_{n}^{\alpha}[f]\\}$ also converge to $f$ for every $f$ in $D(\mu)$
and for each $\mu\in\mathcal{M}_{+}(\mathbb{T})$ whenever $\alpha>1/2$, see
[15, Theorem 1.1]. For each $n\in\mathbb{Z}_{\geqslant 0},$ the generalized
$n^{\it th}$-Cesàro mean $\sigma_{n}^{\alpha}[f]$ is defined by

$\displaystyle\sigma_{n}^{\alpha}[f](z)=\binom{n+\alpha}{\alpha}^{-1}\sum\limits_{k=0}^{n}\binom{n-k+\alpha}{\alpha}a_{k}z^{k},\quad\alpha\geqslant
0,$

where the binomial coefficient $\binom{m+\alpha}{\alpha}$ is given the
following interpretation:

$\displaystyle\binom{m+\alpha}{\alpha}=\frac{\Gamma(m+\alpha+1)}{\Gamma(\alpha+1)\Gamma(m+1)},$

with $\Gamma$ denoting the usual gamma function. Note that the generalized
$n^{\it th}$-Cesàro mean $\sigma_{n}^{\alpha}[f]$ corresponds to the $n^{\it
th}$-partial sum $s_{n}[f]$ when $\alpha=0$ and corresponds to the the $n^{\it
th}$-Cesàro sum $\sigma_{n}[f]$ when $\alpha=1.$ For $\alpha\geqslant 0,$ the
Taylor series of a function $f$ is said to be $(C,\alpha)$-summable to $f$ if
the sequence $\\{\sigma_{n}^{\alpha}[f]\\}$ converges to $f$ in the associated
normed function space. It is well known that $(C,\alpha)$-summability implies
$(C,\beta)$-summability if $\alpha\leqslant\beta,$ see [11, Theorem 43].

In this article, our primary objective is to investigate whether the sequence
of generalized Cesàro means $\\{\sigma_{n}^{\alpha}[f]\\}$ converges to the
corresponding function $f$ that belongs to the higher order weighted Dirichlet
space $\mathcal{H}_{\mu,m}$ as defined below. These spaces serve as a crucial
framework for modeling a specific sub-class within the broader category of
$m$-isometries. The motivation behind introducing these spaces is rooted in
Agler’s exploration of $m$-isometries see [2], [3], [4]. In order to find a
suitable model for cyclic $m$-isometries, Rydhe delved into the study of
higher order weighted Dirichlet type space, see [17]. Following Rydhe’s
framework, we consider, for a measure $\mu$ in $\mathcal{M}_{+}(\mathbb{T})$,
$f\in\mathcal{O}(\mathbb{D})$ and for a positive integer $m,$ the concept of
weighted Dirichlet-type integral of $f$ of order $m$, $D_{\mu,m}(f),$ defined
by

$\displaystyle
D_{\mu,m}(f):=\frac{1}{m!(m-1)!}\displaystyle\int_{\mathbb{D}}\big{|}f^{(m)}(z)\big{|}^{2}P_{\\!\mu}(z)(1-|z|^{2})^{m-1}dA(z).$

Here $dA(z)$ denotes the normalized Lebesgue measure on the unit disc
$\mathbb{D},$ $f^{(m)}(z)$ represents the $m^{\it th}$-order derivative of $f$
at $z,$ and $P_{\\!\mu}(z)$ is the Poisson integral of the measure $\mu,$ that
is,

$\displaystyle
P_{\\!\mu}(z):=\int_{\mathbb{T}}\frac{1-|z|^{2}}{|z-\zeta|^{2}}d\mu(\zeta),\,\,\,z\in\mathbb{D}.$

When dealing with Dirac delta measure $\delta_{\lambda}$, representing a point
measure at $\lambda\in\mathbb{T}$, we adopt a simpler notation
$D_{\lambda,m}(\cdot)$ in place of $D_{\delta_{\lambda},m}(\cdot)$. For a
measure $\mu\in\mathcal{M}_{+}(\mathbb{T})/\\{0\\}$ and for each
$m\in\mathbb{N},$ we consider the semi-inner product space
$\mathcal{H}_{\mu,m}$ given by

$\mathcal{H}_{\mu,m}:=\big{\\{}f\in\mathcal{O}(\mathbb{D}):D_{\mu,m}(f)<\infty\big{\\}},$

associated to the semi-norm $\sqrt{D_{\mu,m}(\cdot)}.$ In case $\mu$ is
$\delta_{\lambda}$ for some $\lambda\in\mathbb{T},$ we will use a simpler
notation $\mathcal{H}_{\lambda,m}$ in place of
$\mathcal{H}_{\delta_{\lambda},m}$ and we refer it as a local Dirichlet space
of order $m$ at $\lambda$. If $\mu=0,$ we set $\mathcal{H}_{\mu,m}=H^{2},$
Hardy space on $\mathbb{D},$ for every $m\in\mathbb{N}.$ If $\mu=\sigma,$ by a
straightforward computation, it follows that for a holomorphic function
$f=\sum_{k=0}^{\infty}a_{k}z^{k}$ in $\mathcal{O}(\mathbb{D}),$ we have

$D_{\sigma,m}(f)=\sum_{k=m}^{\infty}\binom{k}{m}|a_{k}|^{2},\,\,\,\quad
m\in\mathbb{Z}_{\geqslant 0},$

where $\binom{k}{m}:=\frac{k!}{m!(k-m)!}$ for any $k\geqslant m.$ Using this,
it can be easily verified that $\mathcal{H}_{\sigma,m}$ coincides with the
space $\mathcal{D}_{m}$, studied in [18], where

$\mathcal{D}_{m}=\Big{\\{}\sum_{k=0}^{\infty}a_{k}z^{k}:\sum_{k=0}^{\infty}|a_{k}|^{2}(k+1)^{m}<\infty\Big{\\}}.$

The reader is referred to [8], [9], [13] and [17] for several properties of
the spaces $\mathcal{H}_{\mu,m}$ for an arbitrary non-negative measure $\mu$
and a positive integer $m.$ Note that when $m=1,$ the weighted Dirichlet-type
space $\mathcal{H}_{\mu,1}$ coincides with the weighted Dirichlet type space
$D(\mu)$ as introduced in Richter [16]. Theorem 1.1 and Theorem 1.3 are the
main results of this article. These results in the case of $m=1$ are well
known, see [14, 15].

######  Theorem 1.1.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $m\in\mathbb{N}.$ If
$\alpha>\frac{1}{2}$ then there exists a constant $\kappa>0$ such that

$\displaystyle D_{\mu,m}(\sigma_{n}^{\alpha}[f])\leqslant\kappa
D_{\mu,m}(f),\,\,~{}~{}~{}\mbox{for every~{}}n\in\mathbb{N},\
f\in\mathcal{H}_{\mu,m}.$

Moreover, for every $f\in\mathcal{H}_{\mu,m},$
$D_{\mu,m}\big{(}\sigma_{n}^{\alpha}[f]-f\big{)}\rightarrow 0$ as
$n\to\infty.$

It is known that the space $\mathcal{H}_{\mu,m}$ is contained in the Hardy
space $H^{2}$ (see [8, Corollary 2.5]). This allows us to define a norm
$\|\cdot\|_{\mu,m}$ on $\mathcal{H}_{\mu,m}$ given by

$\|f\|_{\mu,m}^{2}=\|f\|^{2}_{H^{2}}+D_{\mu,m}(f),~{}f\in\mathcal{H}_{\mu,m},$

where $\|f\|_{H^{2}}$ denotes the norm of $f$ in $H^{2}.$ As a consequence of
Theorem 1.1, we obtain the following corollary.

######  Corollary 1.2.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $m\in\mathbb{N}.$ If
$\alpha>\frac{1}{2}$, then $\|\sigma_{n}^{\alpha}[f]-f\|_{\mu,m}\to 0$ as
$n\to\infty$ for any $f\in\mathcal{H}_{\mu,m}$.

Continuing our investigation into the convergence behavior of the sequence of
generalized Cesàro means $\\{\sigma_{n}^{\alpha}[f]\\}$ to the respective
function $f$ when $\alpha\leqslant\frac{1}{2}$ within the higher order
weighted Dirichlet-type space $\mathcal{H}_{\mu,m}$, we bring attention to the
following theorem.

######  Theorem 1.3.

Let $m\in\mathbb{N}$ and $\lambda\in\mathbb{T}$ and $\alpha=\frac{1}{2}.$
There exists a function $f\in\mathcal{H}_{\lambda,m}$ such that
$D_{\lambda,m}\big{(}\sigma_{n}^{\alpha}[f]-f\big{)}\nrightarrow 0$ as
$n\rightarrow\infty$.

We will use the techniques of the Hadamard multiplication operators of
$\mathcal{H}_{\mu,1}$ as developed in [14], in order to prove our results. In
Section 2, first we extend the theory of Hadamard multiplication operators of
$\mathcal{H}_{\mu,1}$ to higher order weighted Dirichlet type spaces
$\mathcal{H}_{\mu,m}$ and then we proceed to establish the main results.

###  2\. Hadamard Multiplication Operators on $\mathcal{H}_{\mu,m}$

For two formal power series $f(z)=\sum_{j=0}^{\infty}a_{j}z^{j}$ and
$g(z)=\sum_{j=0}^{\infty}b_{j}z^{j},$ the Hadamard product $f*g$ of $f$ and
$g$ is defined by a formal power series, given by the formula

$\displaystyle(f*g)(z):=\sum\limits_{j=0}^{\infty}a_{j}b_{j}z^{j}.$

In view of the Cauchy-Hadamard formula for the radius of convergence of a
power series, it is straightforward to verify that $f*g$ is in
$\mathcal{O}(\mathbb{D})$ whenever $f$ and $g$ belong to
$\mathcal{O}(\mathbb{D}).$ For $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and
$m\in\mathbb{N},$ a function $c\in\mathcal{O}(\mathbb{D})$ is said to be a
Hadamard multiplier of $\mathcal{H}_{\mu,m}$ if $(c*f)\in\mathcal{H}_{\mu,m}$
for every $f\in\mathcal{H}_{\mu,m}.$ The following proposition characterizes
the Hadamard multiplier of $\mathcal{H}_{\mu,m}.$ Before delving into the
proposition, for $\mu\in\mathcal{M}_{+}(\mathbb{T})\setminus\\{0\\},$ we
introduce a linear space denoted as $\hat{\mathcal{H}}_{\mu,m}$, associated to
the linear space $\mathcal{H}_{\mu,m},$ for the sake of computational
simplicity.

$\displaystyle\hat{\mathcal{H}}_{\mu,m}:=\Big{\\{}f\in\mathcal{O}(\mathbb{D}):f(z)=\sum\limits_{j=m}^{\infty}a_{j}z^{j},~{}~{}a_{j}\in\mathbb{C},~{}~{}D_{\mu,m}(f)<\infty\Big{\\}}.$
(1)

When $\mu=0,$ we define $\hat{\mathcal{H}}_{\mu,m}$ to be the Hardy space
$H^{2}.$ For a non-zero $\mu$ in $\mathcal{M}_{+}(\mathbb{T}),$ since
$\sqrt{D_{\mu,m}(\cdot)}$ is a semi-norm on the linear space
$\mathcal{H}_{\mu,m}$ and $D_{\mu,m}(f)=0$ if and only if $f$ is a polynomial
of degree at most $(m-1),$ it follows that $\sqrt{D_{\mu,m}(\cdot)}$ induces a
norm on $\hat{\mathcal{H}}_{\mu,m}.$ It is straightforward to verify that
$\hat{\mathcal{H}}_{\mu,m}$ turns out to be a Hilbert space with respect to
this norm. Furthermore, as established in the following lemma,
$\hat{\mathcal{H}}_{\mu,m}$ turns is a reproducing kernel Hilbert space on
$\mathbb{D}.$

######  Lemma 2.1.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})\setminus\\{0\\}$ and $m\in\mathbb{N}.$
The linear space $\hat{\mathcal{H}}_{\mu,m},$ equipped with the norm
$\sqrt{D_{\mu,m}(\cdot)},$ is a reproducing kernel Hilbert space on
$\mathbb{D}.$

###### Proof.

Let $f\in\hat{\mathcal{H}}_{\mu,m}$ and $f(z)=\sum_{j=m}^{\infty}a_{j}z^{j},$
$z\in\mathbb{D}.$ Then we have

$\displaystyle f^{(m)}(z)$
$\displaystyle=\sum\limits_{j=m}^{\infty}\frac{\Gamma(j+1)}{\Gamma(j-m+1)}a_{j}z^{j-m},~{}~{}~{}~{}~{}~{}~{}z\in\mathbb{D}.$

In view of [8, Eq (2.6), p. 454], we obtain that

$\displaystyle D_{\mu,m}(f)$
$\displaystyle\geqslant\frac{\mu(\mathbb{T})}{4m!(m-1)!}\int_{\mathbb{D}}|f^{(m)}|^{2}(z)(1-|z|^{2})^{m}dA(z)$
$\displaystyle=\frac{\mu(\mathbb{T})}{4m!(m-1)!}\sum\limits_{j=m}^{\infty}|a_{j}|^{2}\frac{(\Gamma(j+1))^{2}}{(\Gamma(j-m+1))^{2}}\int_{\mathbb{D}}|z|^{2(j-m)}(1-|z|^{2})^{m}dA(z)$
$\displaystyle=\frac{\mu(\mathbb{T})}{4m!(m-1)!}\sum\limits_{j=m}^{\infty}|a_{j}|^{2}\frac{(\Gamma(j+1))^{2}}{(\Gamma(j-m+1))^{2}}\frac{\Gamma(m+1)\Gamma(j-m+1)}{\Gamma(j+2)}$
$\displaystyle=\frac{\mu(\mathbb{T})}{4(m-1)!}\sum\limits_{j=m}^{\infty}\frac{|a_{j}|^{2}}{(j+1)}\frac{\Gamma(j+1)}{\Gamma(j-m+1)}$

Using Cauchy-Schwarz inequality, we have for each $w\in\mathbb{D},$

$\displaystyle|f(w)|^{2}=|\sum_{j=m}^{\infty}a_{j}w^{j}|^{2}$
$\displaystyle\leqslant\Big{(}\sum\limits_{j=m}^{\infty}\frac{|a_{j}|^{2}}{(j+1)}\frac{\Gamma(j+1)}{\Gamma(j-m+1)}\Big{)}\Big{(}\sum\limits_{j=m}^{\infty}\frac{(j+1)\Gamma(j-m+1)}{\Gamma(j+1)}|w|^{2j}\Big{)}$
$\displaystyle=\frac{4(m-1)!}{\mu(\mathbb{T})}\Big{(}\sum\limits_{j=m}^{\infty}\frac{(j+1)\Gamma(j-m+1)}{\Gamma(j+1)}|w|^{2j}\Big{)}D_{\mu,m}(f).$

This shows that the evaluation at each point $w\in\mathbb{D}$ is a bounded
linear functional on $\hat{\mathcal{H}}_{\mu,m}$ and hence
$\hat{\mathcal{H}}_{\mu,m}$ is a reproducing kernel Hilbert space on
$\mathbb{D}.$ This completes the proof. ∎

Now we establish the result that gives a characterization of the Hadamard
multiplier of $\mathcal{H}_{\mu,m}.$

######  Proposition 2.2.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $m\in\mathbb{N}.$ A function
$c\in\mathcal{O}(\mathbb{D})$ is a Hadamard multiplier of
$\mathcal{H}_{\mu,m}$ if and only if there exists a constant $\kappa>0$ such
that

$\displaystyle D_{\mu,m}(c*f)\leqslant\kappa
D_{\mu,m}(f),~{}~{}~{}~{}~{}~{}~{}\mbox{for
all~{}~{}~{}~{}}f\in\mathcal{H}_{\mu,m}.$

###### Proof.

Suppose that a function $c\in\mathcal{O}(\mathbb{D})$ is a Hadamard multiplier
of $\mathcal{H}_{\mu,m}.$ Thus $(c*f)\in\hat{\mathcal{H}}_{\mu,m}$ for every
$f\in\hat{\mathcal{H}}_{\mu,m}.$ Consider the linear transformation
$H_{c}:\hat{\mathcal{H}}_{\mu,m}\to\hat{\mathcal{H}}_{\mu,m}$ given by
$H_{c}(f)=c*f,$ for $f\in\hat{\mathcal{H}}_{\mu,m}.$ Since, by Lemma 2.1,
$\hat{\mathcal{H}}_{\mu,m}$ is a reproducing kernel Hilbert space consisting
of analytic functions on $\mathbb{D}$, it follows that the linear functional
$L_{j}:f\to\frac{f^{(j)}(0)}{j!}$ on $\hat{\mathcal{H}}_{\mu,m}$ is continuous
for every $j\geqslant m$ (see [5, Lemma 4.1]). This gives us that the graph of
$H_{c}$ is closed, and hence by closed graph theorem, $H_{c}$ is bounded. That
is, $D_{\mu,m}(c*f)\leqslant\|H_{c}\|D_{\mu,m}(f)$ for every
$f\in\hat{\mathcal{H}}_{\mu,m}.$ For an arbitrary $f\in\mathcal{H}_{\mu,m},$
we write

$\displaystyle f=p+g,$

where $p$ is a polynomial of degree at most $m-1$ and
$g\in\hat{\mathcal{H}}_{\mu,m}.$ Hence we obtain

$\displaystyle
D_{\mu,m}(c*f)=D_{\mu,m}(c*g)\leqslant\|H_{c}\|D_{\mu,m}(g)=\|H_{c}\|D_{\mu,m}(f).$

The converse part is trivial. ∎

For any $c(z)=\sum_{j=0}^{\infty}c_{j}z^{j}$ in $\mathcal{O}(\mathbb{D})$ and
for any $m\in\mathbb{N},$ we consider the infinite matrix $T_{c}(m)$ defined
by

$\displaystyle
T_{c}(m):=\begin{pmatrix}c_{m}&c_{m+1}-c_{m}&c_{m+2}-c_{m+1}&c_{m+3}-c_{m+2}&\cdots\\\
0&c_{m+1}&c_{m+2}-c_{m+1}&c_{m+3}-c_{m+2}&\cdots\\\
0&0&c_{m+2}&c_{m+3}-c_{m+2}&\cdots\\\ 0&0&0&c_{m+3}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}.$

Let $W_{m}$ be the linear space in $\mathcal{O}(\mathbb{D})$ given by
$W_{m}=span\\{z^{j}:j\geqslant m-1\\}.$ The matrix $T_{c}(m)$ induces a linear
transformation $A_{c}(m)$ on the linear space $W_{m}$, given by

$\displaystyle
A_{c}(m)\Big{(}\sum\limits_{j=m-1}^{n}b_{j}z^{j}\Big{)}:=\sum\limits_{j=m-1}^{n}\Big{(}c_{j+1}b_{j}+\sum\limits_{k=j+1}^{n}(c_{k+1}-c_{k})b_{k}\Big{)}z^{j}.$

Let $\sigma$ be the normalized Lebesgue measure on $\mathbb{T}.$ Consider the
Hilbert space $\hat{\mathcal{H}}_{\sigma,m-1}$ associated with the norm
$\sqrt{D_{\sigma,m-1}(\cdot)},$ as defined in (1). Suppose $A_{c}(m)$ induces
a bounded operator on the Hilbert space $\hat{\mathcal{H}}_{\sigma,m-1}$.
Since $\\{z^{j}:j\geqslant m-1\\}$ is an orthogonal basis for
$\hat{\mathcal{H}}_{\sigma,m-1},$ it easily follows that

$\displaystyle
D_{\sigma,m-1}\Big{(}A_{c}(m)^{*}(z^{m-1})\Big{)}=|c_{m}|^{2}+\sum_{j=m}^{\infty}|c_{j+1}-c_{j}|^{2}\binom{j}{m-1}^{-1}.$
(2)

Let $b(z)=\sum_{j=0}^{\infty}b_{j}z^{j}\in\mathcal{H}_{\sigma,m-1}.$ Then we
have
$D_{\sigma,m-1}(b)=\sum_{j=m-1}^{\infty}|b_{j}|^{2}\binom{j}{m-1}<\infty$.
This together with (2) shows that the infinite series
$\sum_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}$ converges absolutely for every
$j\geqslant m$. We indeed have that

$\displaystyle
A_{c}(m)\Big{(}\sum\limits_{j=m-1}^{\infty}b_{j}z^{j}\Big{)}=\sum\limits_{j=m-1}^{\infty}\Big{(}c_{j+1}b_{j}+\sum\limits_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}\Big{)}z^{j}.$
(3)

The following theorem tells us that $A_{c}(m)$ being a bounded linear operator
is equivalent to $c$ being a Hadamard multiplier of $\mathcal{H}_{\mu,m}$ for
every $\mu\in\mathcal{M}_{+}(\mathbb{T}).$

######  Theorem 2.3.

Let $m\in\mathbb{N},$ $c\in\mathcal{O}(\mathbb{D}).$ The following statements
are equivalent.

  * (i)

$c$ is a Hadamard multiplier of $\mathcal{H}_{\mu,m}$ for every
$\mu\in\mathcal{M}_{+}(\mathbb{T}).$

  * (ii)

$c$ is a Hadamard multiplier of $\mathcal{H}_{\lambda,m}$ for some
$\lambda\in\mathbb{T}.$

  * (iii)

The transformation $A_{c}(m)$ defines a bounded operator on
$\hat{\mathcal{H}}_{\sigma,m-1}.$

Moreover, in this case, for every $\mu\in\mathcal{M}_{+}(\mathbb{T}),$ we have

$\displaystyle
D_{\mu,m}(c*f)\leqslant\|A_{c}(m)\|^{2}D_{\mu,m}(f),~{}~{}~{}~{}f\in\mathcal{H}_{\mu,m}.$

###### Proof.

$(iii)\implies(i):$ Assume that the linear transformation $A_{c}(m)$ defines a
bounded operator on the Hilbert space $\hat{\mathcal{H}}_{\sigma,m-1}.$ Fix a
$\lambda\in\mathbb{T}$. Let $f\in\mathcal{\mathcal{missing}}H_{\lambda,m}$ and
the associated power series representation of $f$ be given by
$f(z)=\sum_{j=0}^{\infty}a_{j}z^{j}.$ By [9, Theorem 1.1],
$f(z)=a+(z-\lambda)L_{\lambda}[f](z)$, $z\in\mathbb{D}$, for some
$a\in\mathbb{C}$ and $L_{\lambda}[f]\in\mathcal{H}_{\sigma,m-1}$ with
$D_{\lambda,m}(f)=D_{\sigma,m-1}(L_{\lambda}[f])$. Writing
$L_{\lambda}[f](z)=\sum_{j=0}^{\infty}b_{j}z^{j}$, we obtain the relations

$\displaystyle
a_{0}=a-b_{0}\lambda,~{}~{}~{}a_{k}=b_{k-1}-b_{k}\lambda,~{}~{}~{}k\geqslant
1.$ (4)

Note that

$\displaystyle
D_{\lambda,m}(f)=D_{\sigma,m-1}(L_{\lambda}[f])=\sum_{j=m-1}^{\infty}|b_{j}|^{2}\binom{j}{m-1}=D_{\sigma,m-1}\Big{(}\sum_{j=m-1}^{\infty}b_{j}z^{j}\Big{)}.$

As $A_{c}(m)$ defines a bounded operator on $\hat{\mathcal{H}}_{\sigma,m-1}$
and $|\lambda|=1,$ the series
$\sum_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}\lambda^{k}$ converges absolutely
for every $j\geqslant m$ (see the discussion preceding this theorem) and it
makes sense to consider the formal power series $g$ given by

$\displaystyle g(z)$
$\displaystyle=\sum\limits_{j=0}^{\infty}\Big{(}c_{j+1}b_{j}+\sum\limits_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}\lambda^{k-j}\Big{)}z^{j}$
$\displaystyle=\sum\limits_{j=0}^{\infty}\Big{(}c_{j+1}b_{j}\lambda^{j}+\sum\limits_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}\lambda^{k}\Big{)}(\bar{\lambda}z)^{j}.$

Then by (4), it follows that $(c*f)(z)=A+(z-\lambda)g(z)$ for every
$z\in\mathbb{D}$ for some $A\in\mathbb{C}$. Note that

$D_{\sigma,m-1}\big{(}\sum_{j=m-1}^{\infty}b_{j}\lambda^{j}z^{j}\big{)}=D_{\sigma,m-1}(\sum_{j=m-1}^{\infty}b_{j}z^{j})=D_{\lambda,m}(f),$

and

$\displaystyle D_{\sigma,m-1}(g)$
$\displaystyle=\sum_{j=m-1}^{\infty}\big{|}c_{j+1}b_{j}\lambda^{j}+\sum\limits_{k=j+1}^{\infty}(c_{k+1}-c_{k})b_{k}\lambda^{k}\big{|}^{2}\binom{j}{m-1}$
$\displaystyle=D_{\sigma,m-1}\Big{(}A_{c}(m)\big{(}\sum_{j=m-1}^{\infty}b_{j}\lambda^{j}z^{j}\big{)}\Big{)}<\infty.$

Thus applying [9, Theorem 1.1] a second time, we obtain
$c*f\in\mathcal{H}_{\lambda,m}$, and

$\displaystyle D_{\lambda,m}(c*f)=D_{\sigma,m-1}(g).$ (5)

Hence we obtain that

$\sup\Big{\\{}D_{\lambda,m}(c*f):D_{\lambda,m}(f)=1\Big{\\}}=\|A_{c}(m)\|^{2}.$
(6)

This gives us that
$D_{\lambda,m}(c*f)\leqslant\|A_{c}(m)\|^{2}D_{\lambda,m}(f)$ for every $f\in
H_{\lambda,m}.$ Since $\lambda\in\mathbb{T}$ was arbitrary, for any
$f\in\mathcal{H}_{\mu,m}$, it follows that

$\displaystyle D_{\mu,m}(c*f)$
$\displaystyle=\int_{\mathbb{T}}D_{\lambda,m}(c*f)d\mu(\lambda)$
$\displaystyle\leqslant\|A_{c}(m)\|^{2}\int_{\mathbb{T}}D_{\lambda,m}(f)d\mu(\lambda)$
$\displaystyle=\|A_{c}(m)\|^{2}D_{\mu,m}(f).$

Hence we obtain that $c$ is a Hadamard multiplier of $\mathcal{H}_{\mu,m}$ for
every $\mu\in\mathcal{M}_{+}(\mathbb{T}).$

$(i)\implies(ii)$ is obvious.

$(ii)\implies(iii):$ Assume that $c$ is a Hadamard multiplier of
$\mathcal{H}_{\lambda,m}$ for some $\lambda\in\mathbb{T}.$ Following
Proposition 2.2, we obtain that the linear transformation $H_{c}$ on the
Hilbert space $\hat{\mathcal{H}}_{\lambda,m}$ given by $H_{c}(f)=c*f,$ for
$f\in\hat{\mathcal{H}}_{\lambda,m}$ is bounded. Moreover, there exists a
constant $\kappa=\|H_{c}\|^{2}$ such that

$\displaystyle D_{\lambda,m}(c*f)\leqslant\kappa
D_{\lambda,m}(f),~{}~{}~{}~{}~{}~{}~{}\mbox{for
all~{}~{}~{}~{}}f\in\mathcal{H}_{\lambda,m}.$ (7)

Note that, by [9], every $f\in\mathcal{H}_{\lambda,m}$ can be uniquely written
as $f(z)=a+(z-\lambda)L_{\lambda}[f](z)$, where $a\in\mathbb{C}$ and
$D_{\lambda,m}(f)=D_{\sigma,m-1}(L_{\lambda}[f]).$ As $c$ is a Hadamard
multiplier of $\mathcal{H}_{\lambda,m}$, the inequality in (7) can be
rephrased as

$\displaystyle D_{\sigma,m-1}(L_{\lambda}[c*f])\leqslant\kappa
D_{\sigma,m-1}(L_{\lambda}[f]),~{}~{}~{}~{}~{}~{}~{}\mbox{for
all~{}~{}~{}~{}}f\in\mathcal{H}_{\lambda,m}.$ (8)

In view of the higher order local Douglas formula [9], we have that for any
$g\in\mathcal{H}_{\sigma,m-1},$ the function
$(z-\lambda)g\in\mathcal{H}_{\lambda,m}$ and consequently
$D_{\sigma,m-1}(L_{\lambda}[c*(z-\lambda)g])<\infty$. Since for any
$f\in\mathcal{H}_{\sigma,m-1},$ the function
$f-s_{m-2}[f]\in\hat{\mathcal{H}}_{\sigma,m-1}$ whenever $m\geqslant 2.$ Now
consider the transformation $\hat{H}_{c}(m)$ from the Hilbert space
$\hat{\mathcal{H}}_{\sigma,m-1}$ into itself defined by

$\displaystyle\hat{H}_{c}(m)(g):=\begin{cases}L_{\lambda}[c*(z-\lambda)g]&\mbox{if~{}~{}}m=1,\\\
L_{\lambda}[c*(z-\lambda)g]-s_{m-2}[L_{\lambda}[c*(z-\lambda)g]]&~{}~{}~{}~{}\mbox{if~{}~{}}m\geqslant
2,\end{cases}~{}~{}~{}~{}\,\,\,\,\,\,\,g\in\hat{\mathcal{H}}_{\sigma,m-1}.$

In view of the inequality (8), we obtain that

$\displaystyle D_{\sigma,m-1}\Big{(}\hat{H}_{c}(m)(g)\Big{)}$
$\displaystyle=D_{\sigma,m-1}(L_{\lambda}[c*(z-\lambda)g])$
$\displaystyle\leqslant\kappa D_{\sigma,m-1}(L_{\lambda}[(z-\lambda)g])$
$\displaystyle=\kappa D_{\sigma,m-1}(g),$

for every $g\in\hat{\mathcal{H}}_{\sigma,m-1}.$ Let’s compute the matrix
representation of $\hat{H}_{c}(m)$ with respect to the orthogonal basis
$\\{(\bar{\lambda}z)^{j}:j\geqslant m-1\\}$ of
$\hat{\mathcal{H}}_{\sigma,m-1}.$ Note that for any $j\geqslant(m-1)$

$\displaystyle L_{\lambda}\Big{[}c*(z-\lambda)(\bar{\lambda}z)^{j}\Big{]}$
$\displaystyle=L_{\lambda}\Big{[}c_{j+1}\bar{\lambda}^{j}z^{j+1}-c_{j}\bar{\lambda}^{j-1}z^{j}\Big{]}$
$\displaystyle=c_{j+1}\bar{\lambda}^{j}\frac{z^{j+1}-{\lambda}^{j+1}}{z-\lambda}-c_{j}\bar{\lambda}^{j-1}\frac{z^{j}-{\lambda}^{j}}{z-\lambda}$
$\displaystyle=(c_{j+1}-c_{j})\sum_{k=0}^{j-1}(\bar{\lambda}z)^{k}+c_{j+1}(\bar{\lambda}z)^{j}$

Thus it follows that for any $j\geqslant(m-1),$ we have

$\displaystyle\hat{H}_{c}(m)\Big{(}(\bar{\lambda}z)^{j}\Big{)}=(c_{j+1}-c_{j})\sum_{k=m-1}^{j-1}(\bar{\lambda}z)^{k}+c_{j+1}(\bar{\lambda}z)^{j}.$

Hence the matrix representation of the operator $\hat{H}_{c}(m)$ with respect
to the orthogonal basis $\\{(\bar{\lambda}z)^{j}:j\geqslant m-1\\}$ of
$\hat{\mathcal{H}}_{\sigma,m-1}$ coincides with the matrix $T_{c}(m).$ Let $V$
be the unitary operator on $\hat{\mathcal{H}}_{\sigma,m-1}$ defined by

$\displaystyle
V\Big{(}\sum_{j=m-1}^{\infty}b_{j}z^{j}\Big{)}=\sum_{j=m-1}^{\infty}b_{j}{(\bar{\lambda}z)}^{j}$

Now it is straightforward to verify that
$V^{-1}\hat{H}_{c}(m)V(z^{j})=A_{c}(m)(z^{j})$ for every $j\geqslant(m-1).$
Hence $A_{c}(m)$ must define a bounded operator on
$\hat{\mathcal{H}}_{\sigma,m-1}$ and $\|A_{c}(m)\|=\|\hat{H}_{c}(m)\|.$ ∎

######  Remark 2.4.

It is straightforward to verify that any function
$c\in\mathcal{O}(\mathbb{D})$ with Taylor coefficients $\\{c_{n}\\}$ is a
Hadamard multiplier for $\mathcal{H}_{\sigma,m}$ if and only if $\\{c_{n}\\}$
is bounded. On the other hand, if we define $c_{n}=1$ if $n$ is odd and
$c_{n}=0$ if $n$ is even, then the linear transformation $A_{c}(m)$ will be
unbounded on $\hat{\mathcal{H}}_{\sigma,m-1}$, as noted in [15, pg 52] in the
case $m=1.$ To see this for general $m$, note that for any even $n\geqslant
m,$

$\displaystyle\big{\|}A_{c}(m)(z^{n})\big{\|}^{2}_{\sigma,m-1}=\sum_{k=m-1}^{n}{k\choose
m-1}={n+1\choose m}.$

Therefore $\|A_{c}(m)\|\geqslant{n\choose m-1}^{-\frac{1}{2}}{n+1\choose
m}^{\frac{1}{2}}\sim\sqrt{\frac{(n+1)^{m}}{n^{m-1}}}$ for all even $n\geqslant
m.$ Hence $A_{c}(m)$ is unbounded on $\hat{\mathcal{H}}_{\sigma,m-1}$.

######  Remark 2.5.

If $c$ is a Hadamard Multiplier of $\mathcal{H}_{\lambda,m}$ for some
$\lambda\in\mathbb{T}$, then by Proposition 2.2 the operator
$H_{c}:\hat{\mathcal{H}}_{\lambda,m}\to\hat{\mathcal{H}}_{\lambda,m}$, defined
by $H_{c}(f)=c*f$, is bounded. In view of (6), it follows that, in this case,

$\|H_{c}:\hat{\mathcal{H}}_{\lambda,m}\to\hat{\mathcal{H}}_{\lambda,m}\|=\|A_{c}(m)\|.$

###  3\. Generalized Cesàro summability in higher order Dirichlet spaces

In this section, we provide proofs of Theorem 1.1 and Theorem 1.3. We shall
need the following property of the Gamma function (see [1, p. 257 (6.1.46)]):

$\lim_{k\to\infty}k^{b-a}\frac{\Gamma(k+a)}{\Gamma(k+b)}=1,~{}a,b\in\mathbb{C}.$
(9)

We start with the following lemma.

######  Lemma 3.1.

Let $\alpha\in\big{(}\frac{1}{2},1\big{)}.$ Then there exists a positive
constant $C$ such that

$\sum_{j=0}^{n}\frac{\Gamma(j+\alpha)^{2}}{\Gamma(j+1)^{2}}\leqslant
C(n+1)^{2\alpha-1}~{}~{}\mbox{for all}~{}n\geqslant 0.$

###### Proof.

By (9), there exists a constant $C_{1}>0$ such that
$\frac{\Gamma(k+\alpha)}{\Gamma(k+1)}\leqslant C_{1}(k+1)^{\alpha-1}$ for all
$k\in\mathbb{Z}_{\geqslant 0}$. Thus we have

$\displaystyle\sum_{j=0}^{n}\frac{\Gamma(j+\alpha)^{2}}{\Gamma(j+1)^{2}}$
$\displaystyle\leqslant C_{1}^{2}\sum_{j=0}^{n}(j+1)^{2\alpha-2}$
$\displaystyle\leqslant
C_{1}^{2}\sum_{j=0}^{n}\int_{j}^{j+1}t^{2\alpha-2}~{}dt$
$\displaystyle=C_{1}^{2}\int_{j=0}^{n+1}t^{2\alpha-2}~{}dt$
$\displaystyle=\frac{C_{1}^{2}}{2\alpha-1}(n+1)^{2\alpha-1}.$

Choosing $C=\frac{C_{1}^{2}}{2\alpha-1}$ completes the proof of the lemma. ∎

######  Theorem 3.2.

Let $\alpha\in\big{(}\frac{1}{2},1\big{)}.$ For each $n\in\mathbb{N},$ let
$h_{n}(z)=\binom{n+\alpha}{\alpha}^{-1}\sum\limits_{k=0}^{n}\binom{n-k+\alpha}{\alpha}z^{k}$.
Then the family of linear transformations

$A_{h_{n}}(m):\hat{\mathcal{H}}_{\sigma,m-1}\to\hat{\mathcal{H}}_{\sigma,m-1},~{}n\in\mathbb{N},$

is uniformly bounded in operator norm.

###### Proof.

Note that $\big{\\{}{j\choose m-1}^{-\frac{1}{2}}z^{j}\big{\\}}_{j\geqslant
m-1}$ forms an orthonormal basis of $\hat{\mathcal{H}}_{\sigma,m-1}.$ Fix
$n\in\mathbb{N}.$ With respect to this orthonormal basis, let
$(\\!(a_{i,j})\\!)_{i,j=m-1}^{\infty}$ denote the matrix of $A_{h_{n}}(m).$
From (3), it is easy to see that

$a_{i,j}=\begin{cases}c_{j+1},&\mbox{if~{}}i=j\\\ \sqrt{\frac{{i\choose
m-1}}{{j\choose m-1}}}(c_{j+1}-c_{j}),&\mbox{if~{}}i+1\leqslant j\\\
0,&\mbox{otherwise},\end{cases}$

where $c_{j}=\binom{n+\alpha}{\alpha}^{-1}\binom{n-j+\alpha}{\alpha}$ for
$j\leqslant n$ and $c_{j}=0$ for $j>n.$ Note that all the entries of the
matrix $(\\!(a_{i,j})\\!)_{i,j=m-1}^{\infty}$ are zero except finitely many.
Thus we have

$\displaystyle\|A_{h_{n}}(m)\|$
$\displaystyle=\|(\\!(a_{i,j})\\!)_{i,j=m-1}^{n}\|$
$\displaystyle\leqslant\left\|\begin{pmatrix}a_{m-1,m-1}&0&\ldots&0\\\
0&a_{m,m}&\ldots&0\\\ \vdots&\vdots&\ddots&0\\\
0&0&\ldots&a_{n,n}\end{pmatrix}\right\|+\left\|\begin{pmatrix}0&a_{m-1,m}&\ldots&a_{m-1,n}\\\
0&0&\ldots&a_{m,n}\\\ \vdots&\vdots&\ddots&a_{n-1,n}\\\
0&0&\ldots&0\end{pmatrix}\right\|$ $\displaystyle\leqslant\max_{m-1\leqslant
i\leqslant n}|a_{i,i}|+\sum_{j=m}^{n}\sum_{i=m-1}^{j}|a_{i,j}|^{2}$
$\displaystyle\leqslant 1+\sum_{j=m}^{n}\sum_{i=m-1}^{j-1}|a_{i,j}|^{2}.$

Here for the second last inequality, we have used the fact that the operator
norm of a matrix is bounded by its Hilbert- Schmidt norm. It is easily
verified that for $0\leqslant j\leqslant n$

$\displaystyle c_{j+1}-c_{j}$
$\displaystyle=\frac{1}{{n+\alpha\choose\alpha}}\left({n-j-1+\alpha\choose\alpha}-{n-j+\alpha\choose\alpha}\right)$
$\displaystyle=-\frac{\alpha~{}\Gamma(n-j+\alpha)}{{n+\alpha\choose\alpha}\Gamma(\alpha+1)\Gamma(n-j+1)}.$

Note that

$\displaystyle\sum_{j=m}^{n}\sum_{i=m-1}^{j-1}|a_{i,j}|^{2}$
$\displaystyle\leqslant\frac{\alpha^{2}}{(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=m}^{n}\sum_{i=m-1}^{j-1}\frac{{i\choose
m-1}}{{j\choose m-1}}\frac{(\Gamma(n-j+\alpha))^{2}}{(\Gamma(n-j+1))^{2}}$
$\displaystyle=\frac{\alpha^{2}}{(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=m}^{n}\frac{1}{{j\choose
m-1}}\frac{(\Gamma(n-j+\alpha))^{2}}{(\Gamma(n-j+1))^{2}}\sum_{i=m-1}^{j-1}{i\choose
m-1}$
$\displaystyle=\frac{\alpha^{2}}{(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=m}^{n}\frac{1}{{j\choose
m-1}}\frac{(\Gamma(n-j+\alpha))^{2}}{(\Gamma(n-j+1))^{2}}{j\choose m}$
$\displaystyle=\frac{\alpha^{2}}{m(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=m}^{n}(j-m+1)\frac{(\Gamma(n-j+\alpha))^{2}}{(\Gamma(n-j+1))^{2}}$
$\displaystyle\leqslant\frac{\alpha^{2}(n-m+1)}{m(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=m}^{n}\frac{(\Gamma(n-j+\alpha))^{2}}{(\Gamma(n-j+1))^{2}}$
$\displaystyle=\frac{\alpha^{2}(n-m+1)}{m(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}\sum_{j=0}^{n-m}\frac{(\Gamma(j+\alpha))^{2}}{(\Gamma(j+1))^{2}}$
$\displaystyle\leqslant
C\frac{\alpha^{2}}{m(\Gamma(\alpha+1))^{2}{n+\alpha\choose\alpha}^{2}}(n-m+1)^{2\alpha}.$

Here, we have used a well-known binomial identity $\sum_{i=m-1}^{j-1}{i\choose
m-1}={j\choose m},$ (see [6, p. 46]), while the last inequality follows from
Lemma 3.1. Also, by (9), we get ${n+\alpha\choose\alpha}\sim(n+1)^{\alpha}$.
This completes the proof. ∎

The following lemma might be well-known to the experts. We provide a proof for
the sake of completeness.

######  Lemma 3.3.

Let $\\{T_{n}\\}_{n\geqslant 1}$ be a sequence of bounded linear operators on
a reproducing kernel Hilbert space $\mathcal{H}$ of holomorphic functions on
$\mathbb{D}$. Suppose that

  1. (i)

$T_{n}$ is finite-rank for each $n\in\mathbb{N}$

  2. (ii)

$T_{n}T_{m}(\mathcal{H})\subseteq T_{m}(\mathcal{H})$ for each
$m,n\in\mathbb{N}$

  3. (iii)

$T_{n}(f)(z)\to f(z)$ as $n\to\infty$ for all $f\in\mathcal{H}$ and for all
$z\in\mathbb{D}.$

Then $T_{n}(f)\to f$ in norm as $n\to\infty$ for all $f\in\mathcal{H}$ if and
only if $\sup_{n}\|T_{n}\|<\infty.$

###### Proof.

Suppose $T_{n}(f)\to f$ in norm as $n\to\infty$ for all $f\in\mathcal{H}$.
Then, by the uniform boundedness principle, it follows that
$\sup_{n}\|T_{n}\|<\infty.$ For the converse, assume that
$\sup_{n}\|T_{n}\|<\infty.$ Let $K(z,w)$ denote the reproducing kernel of
$\mathcal{H}$. Since $T_{n}(f)(z)\to f(z)$ as $n\to\infty$ for all
$z\in\mathbb{D}$, by the reproducing property of $\mathcal{H}$, it follows
that $\langle T_{n}(f),g\rangle\to\langle f,g\rangle$ for all $g$ of the form
$\sum_{i=1}^{\ell}a_{i}K(\cdot,w_{i})$, where
$a_{i}\in\mathbb{C},w_{i}\in\mathbb{D},\ell\in\mathbb{N}.$ Since the set
$\\{\sum_{i=1}^{\ell}a_{i}K(\cdot,z_{i}):a_{i}\in\mathbb{C},z_{i}\in\mathbb{D},\ell\in\mathbb{N}\\}$
is dense in $\mathcal{H}$, and $\sup_{n}\|T_{n}\|<\infty$, it follows that
$T_{n}(f)\to f$ weakly as $n\to\infty$ for all $f\in\mathcal{H}.$ The proof is
now complete by [10, Lemma 2.3]. ∎

We now are ready to prove main theorem of this section.

Proof of Theorem 1.1 In view of [11, Theorem 43], it is sufficient to consider
$\frac{1}{2}<\alpha<1.$ Suppose $f\in\mathcal{H}_{\mu,m}.$ Note that
$\sigma_{n}^{\alpha}[f](z)=(h_{n}*f)(z),$ where
$h_{n}(z)=\binom{n+\alpha}{\alpha}^{-1}\sum_{k=0}^{n}\binom{n-k+\alpha}{\alpha}z^{k}$,
$n\in\mathbb{N}$. By Theorem 3.2, there exists a constant $\kappa>0$ such that
$\|A_{h_{n}}(m)\|\leqslant\kappa$ for all $n\in\mathbb{N}.$ Using this along
with Theorem 2.3, we get

$\displaystyle
D_{\mu,m}(\sigma_{n}^{\alpha}[f])=D_{\mu,m}(h_{n}*f)\leqslant\|A_{h_{n}}(m)\|^{2}D_{\mu,m}(f)\leqslant\kappa^{2}D_{\mu,m}(f).$

To prove the second part, let $T_{n}$ be the operator on
$(\hat{\mathcal{H}}_{\mu,m},D_{\mu,m}(\cdot))$ defined by
$T_{n}(f)=\sigma_{n}^{\alpha}[f].$ It is easy to see that $T_{n}$ is finite-
rank and $T_{n}T_{\ell}(\hat{\mathcal{H}}_{\mu,m})\subseteq
T_{\ell}(\hat{\mathcal{H}}_{\mu,m})$ for each $n,\ell$. Also, since for any
$f\in\mathcal{O}(\mathbb{D})$, $s_{n}[f](z)\to f(z)$ for each
$z\in\mathbb{D}$, by [11, Theorem 4.3], we obtain that $T_{n}(f)(z)\to f(z)$
for each $z\in\mathbb{D}$. Further, by the first part of this Theorem,
$\sup_{n}\|T_{n}\|<\infty.$ Hence by Lemma 3.3, it follows that
$D_{\mu,m}(\sigma_{n}^{\alpha}[f]-f)\to 0$ as $n\to\infty$ for all
$f\in\hat{\mathcal{H}}_{\mu,m}$. Now let $f=\sum_{i\geqslant
0}a_{i}z^{i}\in{\mathcal{H}}_{\mu,m}.$ Then $f_{1}:=\sum_{i\geqslant
m}a_{i}z^{i}\in\hat{\mathcal{H}}_{\mu,m}$ and
$D_{\mu,m}(\sigma_{n}^{\alpha}[f]-f)=D_{\mu,m}(\sigma_{n}^{\alpha}[f_{1}]-f_{1})\to
0$ as $n\to\infty$. This completes the proof.

Proof of Corollary 1.2 Let $\alpha>\frac{1}{2}.$ It is easy to see that
$\|s_{n}[f]\to f\|_{H^{2}}\to 0$ for all $f\in H^{2}$. Thus by [11, Theorem
4.3], $\|\sigma_{n}^{\alpha}[f]-f\|_{H^{2}}\to 0$ for all $f\in H^{2}$. This
together with Theorem 1.1 completes the proof of this Corollary.

We proceed now to show that the statement of Theorem 3.2 does not hold true if
$\alpha\leqslant 1/2.$ It is enough to disprove the statement for
$\alpha=1/2.$

######  Proposition 3.4.

Suppose for each $n\in\mathbb{N},$
$g_{n}(z)=\sum\limits_{k=0}^{n}(1-\frac{k}{n+1})^{1/2}z^{k}.$ Then for any
$m\in\mathbb{N},$ the family $\\{A_{g_{n}}(m):n\in\mathbb{N}\\}$ is not
uniformly bounded.

###### Proof.

Fix $n\in\mathbb{N}.$ Let $c_{k}=(1-\frac{k}{n+1})^{1/2},\ k=1,\ldots,n.$
Consider the sub-matrix $PAQ$ defined below of $DT_{g_{n}}(m)D^{-1},$ where

$A:=\begin{pmatrix}c_{s+1}-c_{s}&\ldots&c_{n+1}-c_{n}\\\
\vdots&\ddots&\vdots\\\
c_{s+1}-c_{s}&\ldots&c_{n+1}-c_{n}\end{pmatrix}_{(s+1)\times(n-s+1)},$
$P=\mbox{diag}\bigg{(}{{s\choose m-1}^{\frac{1}{2}}},{{s+1\choose
m-1}^{\frac{1}{2}}},\ldots,{{2s\choose m-1}^{\frac{1}{2}}}\bigg{)},$
$Q=\mbox{diag}\bigg{(}{s\choose m-1}^{-\frac{1}{2}},{s+1\choose
m-1}^{-\frac{1}{2}},\ldots,{n\choose m-1}^{-\frac{1}{2}}\bigg{)},$

and

$D=\mbox{diag}\bigg{(}{{m-1\choose m-1}^{\frac{1}{2}}},{{m\choose
m-1}^{\frac{1}{2}}},\ldots\bigg{)}.$

Note that

$\|PAQ\|\geqslant\alpha\beta\|A\|,$

where $\alpha$ and $\beta$ are minimum of the eigenvalues of $P$ and $Q$
respectively, and $\|\cdot\|$ denotes the standard operator norm. Using the
argument as presented in [15, Theorem 2.2], this in turn implies that

$\displaystyle\|PAQ\|\geqslant\sqrt{\frac{{s\choose m-1}}{{n\choose
m-1}}}\sqrt{s+1}\big{(}\sum_{k=s}^{n}|c_{k+1}-c_{k}|^{2})^{1/2}$ (10)

Since ${s\choose m-1}\sim(s+1)^{m-1}$ and ${n\choose m-1}\sim(n+1)^{m-1}$, it
follows from (10) that

$\displaystyle\|PAQ\|\geqslant\sqrt{\frac{(s+1)^{m}}{(n+1)^{m-1}}}\big{(}\sum_{k=s}^{n}|c_{k+1}-c_{k}|^{2})^{1/2}.$

It follows from [15, page 7] that

$\|PAQ\|\geqslant\sqrt{\frac{(s+1)^{m}}{(n+1)^{m-1}}}\frac{1}{2\sqrt{n+1}}\sqrt{log(n+2-s)}=\frac{1}{2}\sqrt{\frac{(s+1)^{m}}{(n+1)^{m}}}\sqrt{log(n+2-s)}.$

Choosing $s=\lfloor{\frac{n}{2}}\rfloor$, we get that

$\displaystyle\|PAQ\|\geqslant\frac{1}{2^{m+1}}\sqrt{log(2+n/2)}.$ (11)

Now as $n$ approaches to $\infty$, the right hand side of (11) tends to
$\infty$ as well. This proves that the family
$\\{T_{g_{n}}(m):n\in\mathbb{N}\\}$ is not uniformly bounded. ∎

Proof of Theorem 1.3: Let $\alpha=\frac{1}{2}$. Note that
$\sigma_{n}^{\alpha}[f](z)=(h_{n}*f)(z),$ where
$h_{n}(z)=\binom{n+\alpha}{\alpha}^{-1}\sum_{k=0}^{n}\binom{n-k+\alpha}{\alpha}z^{k}$,
$n\in\mathbb{N}$. By [15, Theorem 3.2], for any $f\in\mathcal{H}_{\lambda,m}$,
the sequence $\\{h_{n}*f\\}$ does not converge to $f$ in
$\mathcal{H}_{\lambda,m}$ if and only if the sequence $\\{{\phi_{n}*f}\\}$
does not converge to $f$ in $\mathcal{H}_{\lambda,m}$, where for each
$n\in\mathbb{N},$ $\phi_{n}(z)=\sum_{k=1}^{n}c_{k}z^{k}$ with
$c_{k}=(1-\frac{k}{n+1})^{1/2},\ k=1,\ldots,n.$ By Proposition 3.4, the family
$\\{A_{\phi_{n}}(m)\\}_{n\in\mathbb{N}}$ is not uniformly bounded. Thus, by
Remark 2.5, it follows that the family $\\{H_{\phi_{n}}\\}_{n\in\mathbb{N}}$,
where
$H_{\phi_{n}}:\hat{\mathcal{H}}_{\lambda,m}\to\hat{\mathcal{H}}_{\lambda,m}$
is given by $H_{\phi_{n}}(f)=\phi_{n}*f$, is not uniformly bounded. An
application of the uniform boundedness principle now completes the proof.

###  4\. Convergence of Generalized Cesàro sum

In this section, we provide an alternative proof of Theorem 1.1 using the
method of induction. For any $f\in\mathcal{O}(\mathbb{D}),$ let $Lf$ be the
function in $\mathcal{O}(\mathbb{D})$ defined by

$\displaystyle Lf(z)=\frac{f(z)-f(0)}{z},\,\,\,z\in\mathbb{D}.$

It is known that for any $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and
$m\in\mathbb{N},$ the inequality $D_{\mu,m}(Lf)\leqslant D_{\mu,m}(f)$ holds
for every $f\in\mathcal{H}_{\mu,m},$ see [8, Lemma 2.9]. Moreover, from [8,
Lemma 2.10] and [8, Corollary 3.3], we have the following result which will be
crucial for the proof of Theorem 1.1 presented in this section.

######  Lemma 4.1.

Let $m\geqslant 1$ and $\mu\in\mathcal{M}_{+}(\mathbb{T}).$ Then for any
function $f$ in $\mathcal{H}_{\mu,m}$, we have

$\displaystyle\sum\limits_{k=1}^{\infty}D_{\mu,j}(L^{k}f)=D_{\mu,j+1}(f),\quad
0\leqslant j\leqslant m-1.$

Now we start with the following proposition which describes a relationship
between $L^{j}(\sigma^{\alpha}_{n}[f])$ and $\sigma^{\alpha}_{n-j}[L^{j}f]$
for any function $f\in\mathcal{O}(\mathbb{D}).$

######  Proposition 4.2.

For every $f\in\mathcal{O}(\mathbb{D}),$ $n\in\mathbb{Z}_{\geqslant 0},$
$\alpha\geqslant 0$ and $j\in\mathbb{N}$ we have

$\displaystyle L^{j}(\sigma^{\alpha}_{n}[f])$
$\displaystyle=\begin{cases}\frac{\Gamma(n+1)\Gamma(n+\alpha-j+1)}{\Gamma(n-j+1)\Gamma(n+\alpha+1)}\,\sigma^{\alpha}_{n-j}[L^{j}f],&j\leqslant
n,\\\ 0,&j>n.\end{cases}$

###### Proof.

Since for any $f\in\mathcal{O}(\mathbb{D})$ and $n\in\mathbb{Z}_{\geqslant
0},$ $\sigma^{\alpha}_{n}[f]$ is a polynomial of degree at most $n$, it
follows that $L^{j}(\sigma^{\alpha}_{n}[f])=0$ whenever $j>n.$ For the
remaining case, let $f\in\mathcal{O}(\mathbb{D})$. Note that

$\displaystyle\sigma_{n}^{\alpha}[f](z)=\frac{\Gamma(n+1)}{\Gamma(n+\alpha+1)}\sum\limits_{k=0}^{n}\frac{\Gamma(n-k+\alpha+1)}{\Gamma(n-k+1)}a_{k}z^{k},\;\;\;\;\;\;~{}~{}~{}n\in\mathbb{Z}_{\geqslant
0},\;\alpha\geqslant 0.$

Operating $L$ on both sides, we get

$\displaystyle
L(\sigma_{n}^{\alpha}[f](z))=\frac{\Gamma(n+1)}{\Gamma(n+\alpha+1)}\sum\limits_{k=0}^{n-1}\frac{\Gamma(n-k+\alpha)}{\Gamma(n-k)}a_{k+1}z^{k},\;\;\;\;\;\;~{}~{}~{}n\in\mathbb{N},\;\alpha\geqslant
0.$

Hence it follows that

$\displaystyle L(\sigma^{\alpha}_{n}[f])$
$\displaystyle=\begin{cases}\frac{\Gamma(n+1)\Gamma(n+\alpha)}{\Gamma(n+\alpha+1)\Gamma(n)}\sigma^{\alpha}_{n-1}[Lf],&n\geqslant
1,\\\ 0,&n=0.\end{cases}$

This completes the proof of the proposition for the case $j=1$. Fix
$k\in\mathbb{N}$ with $k<n.$ Assume that the statement of the proposition
holds for $j=k.$ Then applying the induction hypothesis for the function $Lf$
we have that for any $m\in\mathbb{Z}_{\geqslant 0}$ satisfying $m\geqslant k,$

$\displaystyle
L^{k}(\sigma^{\alpha}_{m}[Lf])=\frac{\Gamma(m+1)\Gamma(m+\alpha-k+1)}{\Gamma(m-k+1)\Gamma(m+\alpha+1)}\,\sigma^{\alpha}_{m-k}[L^{k+1}f].$

Since
$L(\sigma^{\alpha}_{m+1}[f])=\frac{\Gamma(m+2)\Gamma(m+\alpha+1)}{\Gamma(m+\alpha+2)\Gamma(m+1)}\,\sigma^{\alpha}_{m}[Lf]$
for every $m\in\mathbb{Z}_{\geqslant 0},$ we obtain that

$\displaystyle
L^{k+1}(\sigma^{\alpha}_{m+1}[f])=\frac{\Gamma(m+2)\Gamma(m+\alpha-k+1)}{\Gamma(m+\alpha+2)\Gamma(m-k+1)}\,\sigma^{\alpha}_{m-k}[L^{k+1}f],\,\,m\geqslant
k.$

This gives us that for any $n\in\mathbb{Z}_{\geqslant 0},$

$\displaystyle L^{k+1}(\sigma^{\alpha}_{n}[f])$
$\displaystyle=\begin{cases}\frac{\Gamma(n+1)\Gamma(n+\alpha-k)}{\Gamma(n+\alpha+1)\Gamma(n-k)}\,\sigma^{\alpha}_{n-k-1}[L^{k+1}f],&k+1\leqslant
n,\\\ 0,&k+1>n.\end{cases}$

This finishes the induction step for $j=k+1$ and completes the proof of the
proposition. ∎

Alternative proof of Theorem 1.1: Let $\mu\in\mathcal{M}_{+}(\mathbb{T}),$
$n\in\mathbb{N},$ and $\alpha>\frac{1}{2}.$ We will prove this theorem by
induction on $m.$ From [15, Theorem 1.1], it follows that there exists a
constant $M_{\alpha},$ independent of $n,$ such that
$D_{\mu,1}(\sigma^{\alpha}_{n}[f])\leqslant M_{\alpha}D_{\mu,1}(f)$ for every
$f\in\mathcal{H}_{\mu,1}.$ Let us assume that
$D_{\mu,m}(\sigma^{\alpha}_{n}[f])\leqslant M_{\alpha}D_{\mu,m}(f)$ for every
$f\in\mathcal{H}_{\mu,m},$ and for $m=1,\ldots,k.$ Now take
$f\in\mathcal{H}_{\mu,k+1}.$ Since $\sigma^{\alpha}_{n}[f]$ is a polynomial of
degree at most $n$, from Lemma 4.1, it follows that

$\displaystyle
D_{\mu,k+1}(\sigma^{\alpha}_{n}[f])=\sum\limits_{j=1}^{n}D_{\mu,k}(L^{j}(\sigma^{\alpha}_{n}[f]).$

Now applying the Proposition 4.2 and the induction hypothesis we obtain that

$\displaystyle D_{\mu,k+1}(\sigma^{\alpha}_{n}[f])$
$\displaystyle=\sum\limits_{j=1}^{n}\left(\frac{\Gamma(n+1)\Gamma(n+\alpha-j+1)}{\Gamma(n-j+1)\Gamma(n+\alpha+1)}\right)^{2}D_{\mu,k}(\sigma^{\alpha}_{n-j}[L^{j}f])$
$\displaystyle\leqslant\sum\limits_{j=1}^{n}D_{\mu,k}(\sigma^{\alpha}_{n-j}[L^{j}f])$
$\displaystyle\leqslant M_{\alpha}\sum\limits_{j=1}^{n}D_{\mu,k}(L^{j}f)$
$\displaystyle\leqslant M_{\alpha}\sum\limits_{j=1}^{\infty}D_{\mu,k}(L^{j}f)$
$\displaystyle=M_{\alpha}D_{\mu,k+1}(f).$

This completes the induction step for $m=k+1$ and as well as the proof of the
theorem.

## Declarations

Conflict of interest : The authors declare that there are no conflict of
interest regarding the publication of this paper.

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# Lattice QCD calculation of the pion mass splitting

Xu Feng<EMAIL_ADDRESS>School of Physics and State Key Laboratory of
Nuclear Physics and Technology, Peking University, Beijing 100871, China
Collaborative Innovation Center of Quantum Matter, Beijing 100871, China
Center for High Energy Physics, Peking University, Beijing 100871, China
Luchang Jin<EMAIL_ADDRESS>Department of Physics, University of
Connecticut, Storrs, CT 06269, USA  RIKEN-BNL Research Center, Brookhaven
National Laboratory, Building 510, Upton, NY 11973  Michael Joseph Riberdy
<EMAIL_ADDRESS>Department of Physics, University of
Connecticut, Storrs, CT 06269, USA

(August 27, 2024)

###### Abstract

We use the infinite volume reconstruction method to calculate the
charged/neutral pion mass difference. The hadronic tensor is calculated using
lattice QCD and then combined with an analytic photon propagator, and the mass
splitting is calculated with exponentially-suppressed finite volume errors.
The calculation is performed using six gauge ensembles generated with
$2+1$-flavor domain wall fermions, and five ensembles are at the physical pion
mass. Both Feynman and Coulomb gauges are adopted in the calculation and agree
well when the lattice spacing approaches zero. After performing the continuum
extrapolation and examining the residual finite-volume effects, we obtain the
pion mass splitting $\Delta
m_{\pi}=4.534(42)_{\text{stat}}(43)_{\text{sys}}\leavevmode\nobreak\
\mathrm{MeV}$, which agrees well with experimental measurements.

##  I Introduction

One of the central goals in high energy physics is understanding the nature of
the matter that we observe in the universe. As one of the four known
fundamental interactions, the strong interaction binds together quarks and
gluons into hadrons and most of the hadron mass arises in turn from the
binding energy; the individual quarks provide only a very small portion of the
mass. Precision calculation based on the lattice formulation of quantum
chromodynamics (QCD), the theory of the strong interaction, can manifest its
success by computing the various hadron spectra, which agree well with
experimental measurements [1]. When the precision reaches percent or
subpercent level, another fundamental force, electromagnetic interaction, is
urged to be considered in theoretical calculations, although its effects are
suppressed by a factor of the fine-structure constant
$\alpha_{\mathrm{EM}}\approx 1/137$. Inclusion of the electromagnetic
corrections does not only provide precise hadron spectroscopy [2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12], but also plays an important role in studies of leptonic
and semileptonic decays of hadrons [13, 14, 15, 16, 17, 18, 19], which
dramatically expands the horizon of lattice QCD studies.

Among various hadrons, pions play a unique role in the development of
theoretical particle and nuclear physics. Pions were first proposed by H.
Yukawa in 1935 as the carrier particles of the strong nuclear force [20]. As
Nambu-Goldstone bosons [21, 22], pions result from the spontaneous break down
of chiral symmetries of QCD effected by quark condensation and serve as the
active degrees of freedom sensitive to chiral dynamics [23, 24, 25]. Besides,
the anomalous decay rate of the neutral pion led to the discovery of Adler-
Bell-Jackiw anomaly of quantum electrodynamics (QED) [26, 27], which revealed
for the first time the violation of classical symmetry by quantum corrections.
It can be concluded that the thorough study of the nature of the pions is a
key to our better understanding of QCD and the strong interaction. In this
work, we focus on the study of the mass splitting between the charged and
neutral pions, which represents the interplay between two fundamental
interactions, the strong and the electromagnetic. For two reasons, the pion
mass splitting is ideal for a lattice QCD calculation and for an exploratory
study of new methodology. First, pions are the lightest hadrons and their
correlation functions have very good statistical signals. Second, the isospin
breaking effects of the up and down quark mass difference are suppressed by a
factor of $(m_{u}-m_{d})^{2}/\Lambda_{\mathrm{QCD}}^{2}\sim 10^{-4}$ with
$m_{u/d}$ the up/down quark mass and $\Lambda_{\mathrm{QCD}}\approx 350$ MeV
the nonperturbative QCD scale, leaving the electromagnetic effects the leading
contribution to the pion mass splitting. Thus the ambuguity in separating the
isospin breaking effects from the electromagnetic interaction and the quark
mass difference becomes irrelevant in this study. One can simply perform the
lattice QCD calculation in the isospin symmetric limit and compute the four-
point correlation functions for the QED self-energy diagrams.

In practice we adopt the infinite-volume reconstruction (IVR) method proposed
in Ref. [28], which allows us to calculate electromagnetic corrections to
stable hadron masses with only exponentially suppressed finite-volume effects.
Upon its development, this method has been successfully applied and extended
to various electroweak processes involving photon or massless leptonic
propagators [29, 30, 18, 31, 32]. This is the first time that we apply this
methodology to the calculation of the pion mass splitting. The calculation
includes the complete diagrams with both connected and disconnected quark-
field contractions. We employ two gauge fixings for the photon propagator and
confirm that the lattice results are consistent in the continuum limit. We
utilize the random field sparsening technique proposed in Ref. [33], which
allows us to improve the precision of the correlation functions significantly
with only a modest cost of computational resources. By using five gauge
ensembles generated with $N_{f}=2+1$ domain wall fermions at physical pion
mass and one additional ensemble at $m_{\pi}\approx 340$ MeV, we obtain the
pion mass splitting with a percent-level uncertainty, which is about 5-10
times smaller than previous lattice QCD calculations [34, 35, 5, 7, 9]. (See
Table 1. Refs. [34, 35] present the pioneering quenched calculations and thus
the results are not included in Table 1.) For the first time in the
literature, we have clearly resolved and included the contribution from the
quark disconnected diagram to the pion mass splitting (see Eq. (17) and the
diagram below this equation). This diagram is related to the
$\pi^{0}-\eta-\eta^{\prime}$ mixing and has also been calculated in Ref. [36].

Reference | $m_{\pi^{\pm}}-m_{\pi^{0}}(\text{MeV})$
---|---
RM123 2013 [5] |  $5.33(48)_{\text{stat}}(59)_{\text{sys}}$ 111Converted from $m_{\pi^{\pm}}^{2}-m_{\pi^{0}}^{2}=1.44(13)(16)\times 10^{3}\leavevmode\nobreak\ \text{MeV}^{2}$.
R. Horsley et al. 2016 [7] | $4.60(20)_{\text{stat}}$
RM123 2017 [9] | $4.21(23)_{\text{stat}}(13)_{\text{sys}}$
This work | $4.534(42)_{\text{stat}}(43)_{\text{sys}}$
Table 1:  Previous lattice calculations of $m_{\pi^{\pm}}-m_{\pi^{0}}$ are
compared to this work.

##  II  Infinite Volume Reconstruction Method

The hadron mass extraction relies on the calculation of the hadron QED self-
energy for a stable hadronic state $N$ via the following infinite volume
Euclidean space-time integral

$\Delta M=\mathcal{I}=\frac{1}{2}\int
d^{4}x\,\mathcal{H}_{\mu,\nu}(x)S^{\gamma}_{\mu,\nu}(x),$ (1)

where the hadronic part
$\mathcal{H}_{\mu,\nu}(x)=\mathcal{H}_{\mu,\nu}(t,\vec{x})$ is given by

$\mathcal{H}_{\mu,\nu}(x)=\frac{1}{2M}\langle
N(\vec{0})|T[J_{\mu}(x)J_{\nu}(0)]|N(\vec{0})\rangle,$ (2)

where $|N(\vec{p})\rangle$ indicates a hadronic state $N$ with mass $M$ and
spatial momentum $\vec{p}$,
$J_{\mu}=2e\,\bar{u}\gamma_{\mu}u/3-e\,\bar{d}\gamma_{\mu}d/3-e\,\bar{s}\gamma_{\mu}s/3$
is the electromagnetic current, and $S^{\gamma}_{\mu,\nu}$ is the photon
propagator whose form is analytically known. In Ref. [28], we introduced the
IVR method to relate the infinite volume integration of infinite volume
hadronic matrix elements $\mathcal{H}_{\mu,\nu}$ to a finite lattice volume
integration of finite volume matrix elements $\mathcal{H}^{L}_{\mu,\nu}$ with
only exponentially suppressed finite volume errors. This is accomplished with
the following three steps:

  1. 1.

We pick $t_{s}$ to separate the infinite volume integral into two parts:

$\mathcal{I}=\mathcal{I}^{(s)}+\mathcal{I}^{(l)}\\\ $ (3)

where $\mathcal{I}^{(s)}$ and $\mathcal{I}^{(l)}$ are the “short distance”
($|x_{t}|<t_{s}$) and “long distance” ($|x_{t}|\geq t_{s}$) contributions
respectively.

  2. 2.

For sufficiently large $t_{s}$, $\mathcal{I}^{(l)}$ is dominated by the
lightest single particle intermediate states, and can be calculated using the
hadronic matrix elements at fixed time separation $t_{s}$. The excited-state
effects ignored in this step are exponentially suppressed by large $t_{s}$.

  3. 3.

The next step is to approximate $\mathcal{H}_{\mu,\nu}$ using
$\mathcal{H}^{L}_{\mu,\nu}$ and restrict the integration region to the finite
lattice volume. This step only introduces exponentially suppressed finite
volume errors since we only need to use $\mathcal{H}^{L}_{\mu,\nu}(x)$ for
$|x_{t}|\leq t_{s}\lesssim L$.

The final formula obtained in Ref. [28] is expressed in terms of lattice-
calculable quantaties:

$\displaystyle\mathcal{I}^{(s)}\approx\mathcal{I}^{(s,L)}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{-t_{s}}^{t_{s}}\\!\\!dt\int_{-L/2}^{L/2}\\!\\!d^{3}\vec{x}\,\mathcal{H}^{L}_{\mu,\nu}(\vec{x})S^{\gamma}_{\mu,\nu}(\vec{x})$
(4) $\displaystyle\mathcal{I}^{(l)}\approx\mathcal{I}^{(l,L)}$
$\displaystyle=$
$\displaystyle\int_{-L/2}^{L/2}\\!\\!d^{3}\vec{x}\,\mathcal{H}^{L}_{\mu,\nu}(t_{s},\vec{x})L_{\mu,\nu}(t_{s},\vec{x})$
(5)

where $L_{\mu\nu}(t_{s},\vec{x})$ is an infinite volume QED weighting
function, defined as:

$\displaystyle
L_{\mu,\nu}(t_{s},\vec{x})=\int\\!\\!\frac{d^{3}p}{(2\pi)^{3}}e^{i\vec{p}\cdot\vec{x}}\int_{t_{s}}^{\infty}\\!\\!dt\,e^{-(E_{\vec{p}}-M)(t-t_{s})}$
$\displaystyle\hskip
85.35826pt\times\int\\!d^{3}\vec{x}^{\prime}\,e^{-i\vec{p}\cdot\vec{x}^{\prime}}S^{\gamma}_{\mu,\nu}(t,\vec{x}^{\prime}).$
(6)

In this paper, we use $t_{s}=L/2$ for our final result to ensure both the
excited states and finite volume errors introduced in steps 2 and 3 described
above are exponentially suppressed as we increase the lattice size $L$.

###  II.I Gauge-Specific Expressions

We also present the relevant expressions for the photon propagator
$S^{\gamma}_{\mu,\nu}(x)$ and QED weighting function
$L_{\mu,\nu}(t_{s},\vec{x})$ in Feynman and Coulomb Gauges:

$\displaystyle
S_{\mu,\nu}^{\gamma,\text{Feyn}}(x)=\delta_{\mu,\nu}\int\\!\\!\frac{d^{4}p}{(2\pi)^{4}}\,\frac{e^{ipx}}{p^{2}}=\frac{\delta_{\mu,\nu}}{4\pi^{2}x^{2}}$
(7) $\displaystyle L_{\mu,\nu}^{\text{Feyn}}(t_{s},\vec{x})$ $\displaystyle=$
$\displaystyle\frac{\delta_{\mu,\nu}}{2\pi^{2}}\int_{0}^{\infty}\\!\\!dp\,\frac{\sin(p|\vec{x}|)}{2(p+E_{p}-M)|\vec{x}|}e^{-pt_{s}}$
(8) $\displaystyle
S_{\mu,\nu}^{\gamma,\text{Coul}}(x_{t},\vec{x})=\frac{\delta_{\mu,t}\delta_{\nu,t}}{4\pi|\vec{x}|}\delta(x_{t})$
$\displaystyle+\frac{(1-\delta_{\mu,t})(1-\delta_{\nu,t})}{2(2\pi)^{3}}\int(\delta_{\mu,\nu}-\frac{p_{\mu}p_{\nu}}{\vec{p}^{2}})\frac{e^{-|\vec{p}|x_{t}+i\vec{p}\cdot\vec{x}}}{|\vec{p}|}d^{3}p$
$\displaystyle=$
$\displaystyle\frac{\delta_{\mu,t}\delta_{\nu,t}}{4\pi|\vec{x}|}\delta(x_{t})+\frac{(1-\delta_{\mu,t})(1-\delta_{\nu,t})}{(2\pi)^{2}}$
$\displaystyle\times\int_{0}^{\infty}\frac{e^{-px_{t}}}{|\vec{x}|}\Bigg{[}\delta_{\mu,\nu}\Big{(}\sin(p|\vec{x}|)+\frac{\cos(p|\vec{x}|)}{p|\vec{x}|}-\frac{\sin(p|\vec{x}|)}{p^{2}|\vec{x}|^{2}}\Big{)}$
$\displaystyle\hskip
28.45274pt+\frac{x_{\mu}x_{\nu}}{|\vec{x}|^{2}}\Big{(}-\sin(p|\vec{x}|)-\frac{3\cos(p|\vec{x}|)}{p|\vec{x}|}+\frac{3\sin(p|\vec{x}|)}{p^{2}|\vec{x}|^{2}}\Big{)}\Bigg{]}dp$
$\displaystyle L_{\mu,\nu}^{\text{Coul}}(t_{s},\vec{x})$ $\displaystyle=$
$\displaystyle 0$ (11)

At the origin ($x=0$), the continuum photon propagator is divergent. In our
lattice calculation, we regularize this divergence with the following choice:

$\displaystyle S_{\mu,\nu}^{\gamma,\text{Feyn}}(0)$ $\displaystyle=$
$\displaystyle\frac{\delta_{\mu,\nu}}{a^{8}}\int_{-a}^{a}dx\int_{-a}^{a}dy\int_{-a}^{a}dz\int_{-a}^{a}dt$
$\displaystyle\hskip
28.45274pt\times\frac{(a-|x|)(a-|y|)(a-|z|)(a-|t|)}{4\pi^{2}(x^{2}+y^{2}+z^{2}+t^{2})}$
$\displaystyle=$ $\displaystyle 2.76963\leavevmode\nobreak\
\delta_{\mu,\nu}/(4\pi^{2}a^{2})$ (13) $\displaystyle
S_{t,t}^{\gamma,\text{Coul}}(0)$ $\displaystyle=$
$\displaystyle\frac{1}{a^{7}}\int_{-a}^{a}dx\int_{-a}^{a}dy\int_{-a}^{a}dz$
$\displaystyle\hskip
28.45274pt\times\frac{(a-|x|)(a-|y|)(a-|z|)}{4\pi\sqrt{x^{2}+y^{2}+z^{2}}}$
$\displaystyle=$ $\displaystyle 5.63371/(4\pi a^{2})$ (15)

##  III Relevant Diagrams

Contributions to the charged/neutral QED pion mass splitting are derived from
the following hadronic matrix element:

$\mathcal{H}_{\mu,\nu}^{\pm,0}(x)=\frac{1}{2M_{\pi}}\langle\pi^{\pm,0}(\vec{0})|T[J_{\mu}(x)J_{\nu}(0)]|\pi^{\pm,0}(\vec{0})\rangle$
(16)

where $\pi(\vec{0})$ represents a pion with zero momentum. Only two
contractions contribute to this matrix element [5]. One yields (where $S$
represents the light quark propagator)

$\displaystyle
C^{1}_{\mu,\nu}(x-y)=\big{\langle}\text{Tr}\big{(}S(x;t_{\text{src}})\gamma_{5}S(t_{\text{src}};x)\gamma_{{\mu}}\big{)}$
(17) $\displaystyle\hskip
71.13188pt\times\text{Tr}\big{(}S(y;t_{\text{snk}})\gamma_{5}S(t_{\text{snk}};y)\gamma_{\nu}\big{)}\big{\rangle}_{\text{QCD}}$

which is related to the following quark disconnected diagram.

The other possible contraction yields (up to the insertion of a photon
propagator)

$\displaystyle
C^{2}_{\mu,\nu}(x-y)=\big{\langle}\text{Tr}\big{(}\gamma_{{\mu}}S(x;t_{\text{src}})\gamma_{5}S(t_{\text{src}};y)$
(18) $\displaystyle\hskip
71.13188pt\times\gamma_{\nu}S(y;t_{\text{snk}})\gamma_{5}S(t_{\text{snk}};x)\big{)}\big{\rangle}_{\text{QCD}}$

which is represented by the following quark connected diagram.

To ensure projection onto the pion state, we fix the time separation between
the pion interpolating operators and the closest electromagnetic current
operators to be $t_{\text{sep}}=t_{\text{snk}}-x_{t}=y_{t}-t_{\text{src}}$ for
both diagrams (assuming $x_{t}\geq y_{t}$). The values of $t_{\text{sep}}$ for
each ensemble are listed in Table 2. Coulomb gauge fixed wall source is used
for the pion interpolating operators. Combining these diagrams yields the
hadronic contribution to the mass shift, which is represented by

$\displaystyle\mathcal{H}_{\mu,\nu}^{\pm}(x)-\mathcal{H}_{\mu,\nu}^{0}(x)$
$\displaystyle=$ $\displaystyle
e^{2}Z_{V}^{2}L^{3}{C^{1}_{\mu,\nu}(x)+C^{1}_{\nu,\mu}(-x)-C^{2}_{\mu,\nu}(x)-C^{2}_{\nu,\mu}(-x)\over
2\,C^{\text{AW}}_{\pi}(|x_{t}|+2t_{\text{sep}})},$

where

$\displaystyle C^{\text{AW}}_{\pi}(t+2t_{\text{sep}})$ $\displaystyle=$
$\displaystyle C_{\pi}(2t_{\text{sep}})e^{-M_{\pi}t}$ (20)
$\displaystyle\hskip
14.22636pt\times\big{(}1-e^{-M_{\pi}(T-4t_{\text{sep}})}\big{)},$
$\displaystyle C_{\pi}(t_{2}-t_{1})$ $\displaystyle=$
$\displaystyle\big{\langle}\text{Tr}\big{(}S(t_{2};t_{1}))\gamma_{5}S(t_{1};t_{2})\gamma_{{5}}\big{)}\big{\rangle}_{\text{QCD}}.$

Local lattice vector current operators are used in the contraction, and the
corresponding renormalization factor $Z_{V}$ is included in the above formula.
We have taken the around the world effects into account when calculating the
pion correlation function in Eq. (20). In principle, both the isospin breaking
due to the up / down quark mass difference and the QED effects described above
contribute to the pion mass difference. For many hadronic observables, the
corrections due to the up / down quark mass difference are at the same order
as the QED corrections. Therefore, we treat
$\mathcal{O}((m_{d}-m_{u})/\Lambda_{\text{QCD}})$ corrections to be of the
same perturbative order as corrections of order
$\mathcal{O}(\alpha_{\text{QED}})$. For pion mass splitting, the
$\mathcal{O}((m_{d}-m_{u})/\Lambda_{\text{QCD}})$ effect is zero. [5]
Therefore, to order
$\mathcal{O}(\alpha_{\text{QED}},(m_{d}-m_{u})/\Lambda_{\text{QCD}})$, the two
diagrams discussed above represent the only contributions to the pion mass
splitting.

##  IV Numerical Results

| Volume |  $a^{-1}$ (GeV) |  $L$ (fm) |  $M_{\pi}$ (MeV) |  $t_{\text{sep}}$ (a)
---|---|---|---|---|---
48I | $48^{3}\times 96$ | 1.730(4) | 5.5 | 135 | 12
64I | $64^{3}\times 128$ | 2.359(7) | 5.4 | 135 | 18
24D | $24^{3}\times 64$ | 1.0158(40) | 4.7 | 142 | 8
32D | $32^{3}\times 64$ | 1.0158(40) | 6.2 | 142 | 8
32Dfine | $32^{3}\times 64$ | 1.378(7) | 4.6 | 144 | 10
24DH | $24^{3}\times 64$ | 1.0158(40) | 4.7 | 341 | 8
Table 2:  List of the ensembles used in the calculations and their properties. They are generated by the RBC and UKQCD collaborations. [37] Note we use a partially quenched quark mass for 48I and 64I ensembles. The unitary pion mass for both 48I and 64I ensembles is $139\leavevmode\nobreak\ \mathrm{MeV}$. We use unitary quark masses for all the other ensembles.  | Feyn (MeV) | Coul (MeV) | Coul-t (MeV)
---|---|---|---
48I | $4.283(21)\ $ | $4.375(25)\ $ | $1.884(13)\ $
64I | $4.415(14)\ $ | $4.459(15)\ $ | $1.875(6)\ $
24D | $3.632(10)\ $ | $3.823(12)\ $ | $1.872(6)\ $
32D | $3.598(12)\ $ | $3.825(13)\ $ | $1.866(7)\ $
32Dfine | $4.002(18)\ $ | $4.109(21)\ $ | $1.863(12)\ $
24DH | $2.406(37)\ $ | $2.509(12)\ $ | $1.498(8)\ $
Table 3:  Contributions to the pion mass shift are shown by ensemble. The
fourth column displays the Coulomb potential contribution. The statistical
uncertainty is shown in parenthesis.

Calculation of the hadronic function $\mathcal{H}_{\mu\nu}^{L}(x)$ was
performed on six ensembles generated by the RBC and UKQCD collaborations. The
names and attributes of the ensembles are shown in Table 2. The properties of
these ensembles are studied in detail in Ref. [37]. In particular, the lattice
spacing and other basic parameters of the ensemble are determined by matching
the lattice calculation of the masses of pion, kaon, and the $\Omega$ baryon
to their experimental values. We use All-modes-averaging (AMA) [38, 39],
zMöbius [40], and compressed eigenvector deflation [41] methods to accelerate
the calculation of the propagators. Figure LABEL:fig:ts_dependence shows the
mass shift $\Delta m_{\pi}\equiv m_{\pi^{+}}-m_{\pi^{0}}$ as a function of
$t_{s}$ in the Feynman and Coulomb gauges for the 24D and 32D ensembles. It
can be seen from the plots that the finite volume effects are very small as
the differences between 24D and 32D are barely visible. Also, for
$t_{s}\gtrsim 1.5\leavevmode\nobreak\ \mathrm{fm}$, the final results have a
very mild dependence on $t_{s}$, suggesting the excited states contribution
beyond $t_{s}$, which is exponentially suppressed and ignored in the IVR
method, is indeed quite small. In the following analysis, we stick to:

$\displaystyle t_{s}=L/2.$ (21)

With this choice, the finite volume effects at fixed $t_{s}$ and the excited
intermediate states’ contribution beyond $t_{s}$ are both exponentially
suppressed by the spatial lattice size $L$, and we will refer to the sum of
these two effects as the finite volume effects in the following analysis. In
Feynman gauge, the difference between 32D and 24D is
$-0.035(16)\leavevmode\nobreak\ \mathrm{MeV}$. This is consistent with a
scalar QED calculation, which yields $-0.022\leavevmode\nobreak\ \mathrm{MeV}$
[28]. In Coulomb gauge, the difference between 32D and 24D is
$0.002(17)\leavevmode\nobreak\ \mathrm{MeV}$.

The results for each ensemble are presented in Table 3. In the table, we also
show the Coulomb potential contribution to the pion mass difference. This
contribution comes from the time component of the Coulomb gauge photon
propagator, $S^{\gamma,\text{Coul}}_{t,t}$ in Eq. (II.I).

##  V Extrapolation to the physical point

We perform the continuum extrapolation for the Iwasaki ensembles (48I and 64I)
and the I-DSDR ensembles 24D, 32Dfine, and 24DH separately using the following
formula ($m_{\pi,\text{phys}}=135\leavevmode\nobreak\ \mathrm{MeV}$).

$\displaystyle m_{\pi}\Delta m_{\pi}(a^{2},m_{\pi})$ $\displaystyle=$
$\displaystyle m_{\pi,\text{phys}}\Delta m_{\pi}(0,m_{\pi,\text{phys}})$
$\displaystyle\times\Big{(}1+c_{1}a^{2}+c_{2}({m_{\pi}}^{2}-m_{\pi,\text{phys}}^{2})\Big{)}$

When using Domain wall fermions in lattice calculations, lattice artifacts,
which scale as $\mathcal{O}(a)$ or $\mathcal{O}(a^{2n+1})$, are absent [42,
37]. We choose to include an $\mathcal{O}(a^{2})$ term to account for the
lattice artifact. We also include a naive pion mass dependence term in this
extrapolation formula to accommodate the small pion mass difference in the
I-DSDR ensembles. This pion mass dependence term does not effect our main
result, which is taken from the continuum extrapolation of the Iwasaki
ensembles exactly at the physical pion mass. The extrapolation results of the
Iwasaki 48I and 64I ensembles are shown in Figure 2, where separate fits were
conducted for Feynman and Coulomb gauges.

Figure 2: Feynman and Coulomb gauge mass shifts are shown as a function of
$a^{2}$ for the Iwasaki ensembles 48I and 64I.

To estimate the discretization systematic error for the fitted value of
$\Delta m_{\pi}(0,m_{\pi,\text{phys}})$, we perform two slightly different
fits by:

  1. 1.

Using the same fitting formula Eq. (V), but excluding the contribution from
$x=0$ for all ensembles.

  2. 2.

Using a different fitting formula which includes an additional
$\mathcal{O}(a^{4})$ term:

$\displaystyle m_{\pi}\Delta m_{\pi}(a^{2},m_{\pi})$ $\displaystyle=$
$\displaystyle m_{\pi,\text{phys}}\Delta m_{\pi}(0,m_{\pi,\text{phys}})$
$\displaystyle\times\Big{(}1+c_{1}a^{2}+c_{1}|c_{1}|a^{4}+c_{2}({m_{\pi}}^{2}-m_{\pi,\text{phys}}^{2})\Big{)},$

where the magnitude of the $\mathcal{O}(a^{4})$ term is assumed to equal to
the square of the $\mathcal{O}(a^{2})$ term as an estimate of the remaining
systematic effects.

We obtain the differences between the results of the above fits and the
original fits. The maximum of the two differences is used as the estimation of
the remaining discretization systematic error.

After continuum extrapolation, we use the differences between 32D and 24D to
correct the finite volume effects. The absolute size of the correction is used
as the estimation of the remaining finite volume systematic error. The
continuum extrapolated, finite volume corrected results are shown in Table 4
and Table 5. The discretization and finite volume systematic errors are
combined in quadrature. As expected, the continuum extrapolations from the
I-DSDR ensembles (Table 5) have larger discretization systematic errors due to
larger lattice spacings. Therefore, we use the continuum extrapolation from
the Iwasaki ensembles (Table 4) as our main results.

| Disc (MeV) | Conn (MeV) | Total (MeV)
---|---|---|---
Feyn | $0.051(9)(22)$ | $4.483(40)(28)$ | $4.534(42)(43)$
Coul | $0.052(2)(13)$ | $4.508(46)(42)$ | $4.560(46)(41)$
Coul-t | $0.018(1)(4)$ | $1.840(22)(39)$ | $1.858(22)(41)$
Table 4:  The continuum results from the two Iwasaki ensembles 48I and 64I. Finite volume corrections calculated with the difference of the 32D and 24D ensembles are already included. The second column and third column show the quark disconnected and connected diagrams’ contributions respectively. The bottom row shows the Coulomb potential contribution. The statistical and systematic errors are shown respectively in the first and second set of parentheses.  | Disc (MeV) | Conn (MeV) | Total (MeV)
---|---|---|---
Feyn | $0.035(12)(21)$ | $4.671(49)(99)$ | $4.706(50)(106)$
Coul | $0.050(3)(13)$ | $4.703(57)(158)$ | $4.753(58)(160)$
Coul-t | $0.016(2)(4)$ | $1.931(32)(157)$ | $1.947(32)(160)$
Table 5:  Similar to Table 4 but extrapolated to the continuum limit with the
coarser I-DSDR ensembles 24D, 32Dfine, and 24DH.

As a byproduct of the calculation, we plot the Coulomb potential contribution
as a function of spatial separation at $t=0$ in Figure 3. This plot provides
some indication of the size and the shape of the pion.

Figure 3:  The Coulomb potential contribution to the pion mass difference. The
curve is the partial sum, $I_{C}(x)=\frac{1}{2}\int\\!\\!dt\int_{|\vec{y}|\leq
x}\\!\\!d^{3}\vec{y}\,\mathcal{H}^{L}_{t,t}(t,\vec{y})S^{\gamma}_{t,t}(t,\vec{y})$.
Error bars are statistical only. We use the results from the 48I and 64I
ensembles to obtain the continuum limit, and include the finite volume
corrections from the 32D and 24D difference.

##  VI Conclusion

We have used the IVR method to calculate the pion mass splitting to order
$\mathcal{O}(\alpha_{\text{QED}},(m_{u}-m_{d})/\Lambda_{\text{QCD}})$ on the
lattice. The calculation is directly performed at the physical pion mass
($m_{\pi}=135\leavevmode\nobreak\ \mathrm{MeV}$), with different lattice
spacings and lattice sizes. Both the connected and the disconnected diagrams
are included. We have obtained the continuum and infinite volume limit results
with Feynman gauge: $\Delta m_{\pi}=4.534(42)(43)\leavevmode\nobreak\
\mathrm{MeV}$, where the quantities in the first and second sets of
parenthesis correspond respectively to the statistical and systematic
uncertainties. We have also performed the same calculation using Coulomb
gauge. Results in both gauges are consistent with each other and with the
experimental value $4.5936(5)\leavevmode\nobreak\ \mathrm{MeV}$ [43]. Such
high precision agreement demonstrates the success of both lattice QCD and the
IVR methods. We plan to use the IVR method in future lattice QCD + QED
spectroscopy calculations.

###### Acknowledgements.

##  VII Acknowledgements

We would like to thank the RBC/UKQCD for the ensembles they have supplied and
useful discussion many of the members therein have provided. We also thank
Prof. F.-K. Guo for the useful communications. L.C.J. acknowledges support by
DOE Office of Science Early Career Award DE-SC0021147 and DOE grant DE-
SC0010339. X.F. is supported in part by NSFC of China under Grants No.
11775002, No. 12125501 and No. 12070131001 and National Key Research and
Development Program of China under Contracts No. 2020YFA0406400. We developed
the computational code based on the Columbia Physics System
(https://github.com/RBC-UKQCD/CPS) and Grid (https://github.com/paboyle/Grid).
The computation is performed under the ALCC Program of the US DOE on the Blue
Gene/Q (BG/Q) Mira computer at the Argonne Leadership Class Facility, a DOE
Office of Science Facility supported under Contract DE-AC02-06CH11357.
Computations for this work were carried out in part on facilities of the USQCD
Collaboration, which are funded by the Office of Science of the U.S.
Department of Energy.

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