# Polarized Hardy–Stein identity

Krzysztof Bogdan  Faculty of Pure and Applied Mathematics, Wrocław University
of Science and Technology, Wyb. Wyspiańskiego 27, 50-370 Wrocław, Poland.
<EMAIL_ADDRESS>,  Michał Gutowski  Faculty of Pure and Applied
Mathematics, Wrocław University of Science and Technology, Wyb. Wyspiańskiego
27, 50-370 Wrocław, Poland<EMAIL_ADDRESS>and  Katarzyna
Pietruska-Pałuba  Institute of Mathematics, University of Warsaw, ul. Banacha
2, 02-097 Warsaw, Poland<EMAIL_ADDRESS>
###### Abstract.

We prove the Hardy–Stein identity for vector functions in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$ with $1<p<\infty$ and for the
canonical paring of two real functions in $L^{p}({\mathbb{R}^{d}})$ with
$2\leq p<\infty$. To this end we propose a notion of Bregman co-divergence and
study the corresponding integral forms.

###### Key words and phrases:

Bregman co-divergence, Markovian semigroup, calculus in $L^{p}$

###### 2010 Mathematics Subject Classification:

Primary 46E35; Secondary 31C05

The research was supported by the NCN grant 2018/31/B/ST1/03818.

##  1\. Introduction

Hardy–Stein identity is a disintegration of energy of a function in an
elliptic or parabolic setting. The energy is defined as an integral of
$|f|^{p}$ and the disintegration is expressed in terms of the corresponding
Bregman divergence. This is exemplified by the following result from Bañuelos,
Bogdan, and Luks [1, Theorem 3.2]: If $p>1$ and $f\in L^{p}({\mathbb{R}^{d}})$
then

(1.1) $\displaystyle\int_{\mathbb{R}^{d}}|f(x)|^{p}\,{\rm
d}x=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}F_{p}(P_{t}f(x),P_{t}f(x+y))\,\nu({\rm
d}y){\rm d}x{\rm d}t.$

Here $F_{p}(a,b):=|b|^{p}-|a|^{p}-pa|a|^{p-2}(b-a)\geq 0$, $a,b\in\mathbb{R}$,
is the Bregman divergence and $(P_{t})_{t\geq 0}$ is the convolution semigroup
on ${\mathbb{R}^{d}}$ generated by symmetric Lévy measure $\nu$ satisfying the
Hartman–Wintner condition. We can rewrite (1.1) as

(1.2) $\displaystyle\int_{\mathbb{R}^{d}}|f(x)|^{p}\,{\rm
d}x=p\int_{0}^{\infty}\mathcal{E}_{p}[P_{t}f]\,{\rm d}t,$

where, for $u\\!:{\mathbb{R}^{d}}\to\mathbb{R}$, the Sobolev-Bregman form is
given by

(1.3)
$\displaystyle\mathcal{E}_{p}[u]:=\frac{1}{p}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}F_{p}(u(x),u(x+y))\,\nu({\rm
d}y){\rm d}x.$

Similarly, if $(P_{t})_{t\geq 0}$ is the Gaussian semigroup, then, as we
verify in Appendix E,

(1.4) $\displaystyle\int_{\mathbb{R}^{d}}|f(x)|^{p}\,{\rm
d}x=p(p-1)\int_{0}^{\infty}\int_{\mathbb{R}^{d}}|P_{t}f(x)|^{p-2}|\nabla
P_{t}f(x)|^{2}\,{\rm d}x{\rm d}t,$

In Theorem 3.1 below we prove a variant of (1.1) for vector-valued
functions$F\\!:\mathbb{R}^{d}\to\mathbb{R}^{n}$ with energy
$\int_{\mathbb{R}^{d}}|F|^{p}$ and multidimensional Bregman divergence
$\mathcal{F}_{p}$. We use a novel approach based on differential calculus in
$L^{p}$, but to reduce technicalities, we restrict our attention to Lévy
measures $\nu({\rm d}x)=\nu(x){\rm d}x$, whose density $\nu(x)$ and semigroup
$P_{t}$ satisfy certain relative bounds (P1) and (P2).

Then, in Theorem 4.1, we provide a polarized Hardy–Stein identity for $2\leq
p<\infty$ and the mutual energy of two scalar functions $f,g\in
L^{p}({\mathbb{R}^{d}})$ as follows:

(1.5) $\int_{\mathbb{R}^{d}}f(x)g(x)|g(x)|^{p-2}\,{\rm
d}x=p\int_{0}^{\infty}\mathcal{E}_{p}(P_{t}f,P_{t}g)\,{\rm d}t.$

Here, for suitable functions $u,v\\!:{\mathbb{R}^{d}}\to\mathbb{R}$, we define

(1.6)
$\mathcal{E}_{p}(u,v):=\frac{1}{p}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}\big{(}u(x),v(x);u(y),v(y)\big{)}\nu(x,y)\,{\rm
d}x{\rm d}y,$

which we call the polarized Sobolev-Bregman form; the function
$\mathcal{J}_{p}$ in (1.6) will be called the Bregman co-divergence, see
Section 4. Compared to (1.1), the polarized Hardy–Stein identity is delicate
because its integrands are signed and absolute integrability of the right-hand
side of (1.5) is in question, especially when $2<p<3$.

We see in (1.2) that $p\mathcal{E}_{p}[P_{t}f]$ indeed disintegrates the
energy of $f$. The formula is closely related to the fact that for nice
functions $u$ and Markovian semigroups $(P_{t})$ with generator $L$ we have
$\tfrac{d}{dt}\int|P_{t}f|^{p}=\int
pP_{t}f|P_{t}f|^{p-2}LP_{t}f=-p\mathcal{E}_{p}[P_{t}f]$, see, e.g., the
discussions in Bogdan, Jakubowski, Lenczewska, and Pietruska-Pałuba [8] and
Bogdan, Grzywny, Pietruska-Pałuba, and Rutkowski [5], or Proposition 2.4
below. The connection is known at least since Varopoulos [32, (1.1)]; see also
Davies [17, Chapter 2 and 3], Langer and Maz’ya [24], and Liskevich and
Semenov [26]. Accordingly, as recently highlighted in [8], Sobolev–Bregman
forms are an efficient tool for estimating Markovian semigroups on $L^{p}$
spaces. For related analysis, we refer to Farkas, Jacob and Schilling [18,
(2.4)] and to Jacob [23, (4.294)]. In the same vein, the Bregman divergence
defines the energy in Gagliardo–Nirenberg–Sobolev inequalities, see Carrillo,
Jüngel, Markowich, Toscani, and Unterreiter [15, p. 71], Bonforte, Dolbeault,
Nazaret, and Simonov [11], and Borowski and Chlebicka [12]. For references to
further applications of Bregman divergence in analysis, statistical learning,
and optimization, we refer to Bogdan, Grzywny, Pietruska-Pałuba, and Rutkowski
[6]; see also Bogdan and Wiȩcek [10] for a martingale connection.

Applications of Hardy–Stein identities include characterizations of Hardy
spaces in Bogdan, Dyda, and Luks [3], the Littlewood–Paley theory and
estimates of Fourier multipliers [1], Douglas identities, extension and trace
theorems, Dirichlet-to-Neumann maps, and quasiminimizers; see [5] and Bogdan,
Fafuła, and Rutkowski [4]. On the other hand, to the best of our knowledge,
the Bregman co-divergence $\mathcal{J}_{p}$, the polarized Sobolev-Bregman
form $\mathcal{E}_{p}(\cdot,\cdot)$, and the polarized Hardy–Stein identity
(1.5) are new. Apart from being interesting for their own sake, they should be
useful for studying the variational problems for $\mathcal{E}_{p}[\cdot]$. Let
us also note that the convexity of the function $|a|^{p}$ plays an important
role in (1.1), but (1.5) is achieved by writing $ab|b|^{p-2}$ as a difference
of two convex functions.

The structure of the paper is as follows. The Hardy–Stein identity (1.4) for
the Gaussian semigroup is verified in Appendix E. The result is apparently
new, but [3, Section 4] and [4] discuss elliptic analogues. Theorem 3.1 is
proved in Section 3. Theorem 4.1, including (1.5), is proved in Section 4. In
Section 5 we discuss the polarized Sobolev-Bregman form. In Appendix A we give
auxiliary estimates for Bregman divergence and related quantities. Appendix B
presents elements of calculus in $L^{p}$, in particular Lemma B.5 gives the
derivative of $\int_{\mathbb{R}^{d}}P_{t}fP_{t}g|P_{t}g|^{p-2}\,{\rm d}x$. In
Appendix C we discuss convexity properties related to the Bregman co-
divergence $\mathcal{J}_{p}$. In Appendix D we give a simpler proof of Theorem
4.1, but only for $p\geq 3$. In Appendix F, complementing Appendix E, we
discuss the $L^{p}$ generator of the Gaussian semigroup. For technical
reasons, our main results are restricted to a class of convolution Markovian
semigroups on ${\mathbb{R}^{d}}$, but some arguments are presented in a more
general setting and further extensions are forthcoming. For instance, Gutowski
[20] and Gutowski and Kwaśnicki [21] extend (1.1) to general symmetric
Markovian semigroups with nonlocal Dirichlet forms; see also Bogdan, Kutek,
and Pietruska-Pałuba [9] for a recent probabilistic approach, based on
stochastic integrals.

Acknowledgements. We thank Włodzimierz Ba̧k, Bartłomiej Dyda, Mateusz
Kwaśnicki, Agnieszka Kałamajska, and Bartłomiej Wróbel for helpful
discussions.

##  2\. Preliminaries

###  2.1. Standing assumptions

All the considered sets, functions, and measures are assumed Borel. For
nonnegative functions $f$ and $g$, we write $f(x)\asymp g(x)$ to indicate that
there exist constants, i.e., numbers $0<c\leq C<\infty$ such that $cf(x)\leq
g(x)\leq Cf(x)$ for all the considered arguments $x$. Without warning, the
symbols $c$ or $C$ may denote different constants even within a single line of
text. The symbol $:=$ means definition, e.g., $a\vee b:=\max\\{a,b\\}$,
$a\wedge b:=\min\\{a,b\\}$, $a_{+}:=a\vee 0$, and $a_{-}:=(-a)\vee 0$. We
denote the Euclidean norm of a vector $z\in\mathbb{R}^{n}$ as $|z|$ and the
standard scalar product of vectors $w$ and $z$ in $\mathbb{R}^{n}$ as $(w,z)$
or $w\cdot z$. The unit sphere in ${\mathbb{R}^{n}}$ centered at the origin is
denoted by $\mathbb{S}^{n-1}$. As usual, $\|f\|_{L^{q}({\mathbb{R}^{d}})}$
denotes the $L^{q}({\mathbb{R}^{d}})$ norm of the (extended real-valued)
function $f$, $1\leq q\leq\infty$. More specifically,
$\|f\|_{L^{q}({\mathbb{R}^{d}})}:=\big{(}\int_{{\mathbb{R}^{d}}}|f(x)|^{q}\,{\rm
d}x)^{1/q}$ for $1\leq q<\infty$, where ${\rm d}x$ refers to the Lebesgue
measure on ${\mathbb{R}^{d}}$ and
$\|f\|_{L^{\infty}({\mathbb{R}^{d}})}:=\operatorname*{ess\,sup}|f|$.

Let $d=1,2,\ldots$. Consider a symmetric, absolutely continuous Lévy measure
$\nu$ on the Euclidean space ${\mathbb{R}^{d}}$. Thus, $\nu({\rm
d}z)=\nu(z)\,{\rm d}z$, where
$\nu\\!:\mathbb{R}^{d}\setminus\\{0\\}\to(0,\infty)$, $\nu(-z)=\nu(z)$,
$z\in\mathbb{R}^{d}\setminus\\{0\\}$, and

$\int_{{\mathbb{R}^{d}}}\left(|z|^{2}\wedge 1\right)\nu(z)\,{\rm d}z<\infty.$

The corresponding Lévy–Khinchine exponent is

(2.1) $\psi(\xi):=\int_{\mathbb{R}^{d}}\left(1-\cos(\xi\cdot
x)\right)\nu(x)\,{\rm d}x,\quad\xi\in{\mathbb{R}^{d}}.$

We further assume the following Hartman–Wintner condition on $\psi$ (and
$\nu$):

(2.2) $\lim_{|\xi|\to\infty}\frac{\psi(\xi)}{\log|\xi|}=\infty.$

In particular, $\int_{{\mathbb{R}^{d}}}\nu(z)\,{\rm d}z=\infty$. This gives
rise to a convolution semigroup of probability densities $p_{t}$ by the
Lévy–Khintchine formula, or Fourier inversion, as follows:

(2.3) $p_{t}(x):=(2\pi)^{-d}\int_{\mathbb{R}^{d}}e^{-i\xi\cdot
x}e^{-t\psi(\xi)}\,{\rm d}\xi,\quad t>0,\ x\in{\mathbb{R}^{d}}.$

The function $p_{t}(x)$ is continuous and attains its maximum at $x=0$, which
is

$\displaystyle p_{t}(0)$
$\displaystyle=(2\pi)^{-d}\int_{\mathbb{R}^{d}}e^{-t\psi(\xi)}\,{\rm
d}\xi,\quad t>0.$

By (2.2), $p_{t}(0)$ is finite for every $t>0$ and, by the Dominated
Convergence Theorem, $p_{t}(0)$ converges to zero as $t\to\infty$, so
$\|p_{t}\|_{L^{\infty}({\mathbb{R}^{d}})}\to 0$.

We shall also write $p_{t}(x,y):=p_{t}(y-x)$ and $\nu(x,y):=\nu(y-x)$. Note
that $p_{t}(x,y)$ is a transition density of a pure-jump Lévy stochastic
process $\\{X_{t},t\geq 0\\}$ in ${\mathbb{R}^{d}}$ with Lévy–Khintchine
exponent $\psi$ (see Sato [29]) and $\nu(x,y)$ is the kernel of the
corresponding Dirichlet form; see, e.g., Fukushima, Oshima, and Takeda [19]
and (2.24). (In the following discussion, the process does not play a
significant role.) Encouraged by [8], we also assume the following technical
conditions:

(P1) $\phantom{aaa}p_{t}(x,y)/t\leq c\nu(x,y),\quad t>0,\
x,y\in{\mathbb{R}^{d}},$

with some constant $c$, and

(P2) $\phantom{aaa}p_{t}(x,y)/t\rightarrow\nu(x,y)\mbox{ as }t\rightarrow
0^{+},\ x,y\in{\mathbb{R}^{d}}.$

For instance, the transition density corresponding to the fractional Laplacian
satisfies (P1) and (P2); see examples provided by Bogdan, Grzywny, and Ryznar
[7, Corollary 23] and Cygan, Grzywny, and Trojan [16, Proof of Theorem 6]. The
conditions are convenient in limiting procedures based on the Dominated
Convergence Theorem, but they can certainly be relaxed, as evidenced by
Appendix E, [20], [21], and [9].

###  2.2. Elementary functions and inequalities

Throughout we use the notation

$\displaystyle
a^{\langle\kappa\rangle}:=\left|a\right|^{\kappa}\operatorname*{sgn}a=a|a|^{\kappa-2},\quad
a,\kappa\in\mathbb{R},$

where $0^{\langle\kappa\rangle}:=0$ and, as usual, $\operatorname*{sgn}0=0$,
$0^{0}:=1$, $0^{\kappa}:=\infty$ for $\kappa<0$, and $0\cdot\infty:=0$. Note
that

$\displaystyle\left(|x|^{\kappa}\right)^{\prime}=\kappa
x^{\langle\kappa-1\rangle}\quad\mbox{if }x\in\mathbb{R}\mbox{ and
}\kappa>1\mbox{ or }x\in\mathbb{R}\\!\setminus\\!\\{0\\}\mbox{ and
}\kappa\in\mathbb{R}.$

Furthermore,

$\displaystyle\left(x^{\langle\kappa\rangle}\right)^{\prime}=\kappa|x|^{\kappa-1}\quad\mbox{if
}x\in\mathbb{R}\mbox{ and }\kappa>1\mbox{ or
}x\in\mathbb{R}\\!\setminus\\!\\{0\\}\mbox{ and }\kappa\in\mathbb{R}.$

This has a vector counterpart: for $\kappa>0$, we let

(2.4) $z^{\langle\kappa\rangle}:=|z|^{\kappa-1}z,\quad z\in{\mathbb{R}^{n}},$

again with the convention $0^{\langle\kappa\rangle}:=0$. Note that

(2.5) $\displaystyle\nabla|z|^{\kappa}=\kappa
z^{\langle\kappa-1\rangle}\quad\mbox{if }z\in{\mathbb{R}^{n}}\mbox{ and
}\kappa>1\mbox{ or }z\in{\mathbb{R}^{n}}\\!\setminus\\!\\{0\\}\mbox{ and
}\kappa\in\mathbb{R}.$

Furthermore, the Jacobi matrix $J_{\langle\kappa\rangle}$ for the mapping
$z\mapsto z^{\langle\kappa\rangle}$ equals

(2.6)
$J_{\langle\kappa\rangle}(z)=|z|^{\kappa-1}\left((\kappa-1)\left(\frac{z}{|z|}\otimes\frac{z}{|z|}\right)+\mbox{Id}\right)\in{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}\quad\mbox{
if }z\in{\mathbb{R}^{n}}\\!\setminus\\!\\{0\\}$

and we let $J_{\langle\kappa\rangle}(0):=0$.

In the following, unless otherwise specified, we consider exponents
$p\in(1,\infty)$.

######  Definition 2.1.

The _Bregman divergence_
$\mathcal{F}_{p}\\!:\mathbb{R}^{n}\times\mathbb{R}^{n}\to(0,\infty)$ is given
by

(2.7) $\displaystyle\mathcal{F}_{p}(w,z):=|z|^{p}-|w|^{p}-pw^{\langle
p-1\rangle}\cdot(z-w)$

and the _symmetrized Bregman divergence_ is

(2.8)
$\displaystyle\mathcal{H}_{p}(w,z):=\frac{1}{2}\left(\mathcal{F}_{p}(w,z)+\mathcal{F}_{p}(z,w)\right)=\frac{p}{2}(z-w)\cdot\left(z^{\langle
p-1\rangle}-w^{\langle p-1\rangle}\right).$

For instance,

(2.9) $\displaystyle\mathcal{F}_{2}(w,z)=\mathcal{H}_{2}(w,z)=|z-w|^{2},\quad
w,z\in\mathbb{R}^{n}.$

Note that $\mathcal{F}_{p}(w,z)=|z|^{p}$ if $w=0$, but
$\mathcal{F}_{p}(w,z)=(p-1)|w|^{p}$ if $z=0$. Of course, the mapping
$\mathbb{R}^{n}\ni z\mapsto|z|^{p}$ is convex, since $p>1$. Its second-order
Taylor remainder is $\mathcal{F}_{p}$, so $\mathcal{F}_{p}\geq 0$ and
$\mathcal{H}_{p}\geq 0$. Also, if $Q$ is an $n\times n$ orthogonal matrix,
then

(2.10) $\displaystyle\mathcal{F}_{p}(Qw,Qz)=\mathcal{F}_{p}(w,z).$

For notational convenience in what follows, we also let

(2.11) $\mathcal{G}_{p}(w,z):=|z-w|^{2}(|w|\vee|z|)^{p-2},\quad
z,w\in{\mathbb{R}^{n}}.$

We further introduce the second-order Taylor remainders of the vector
functions $\\!\mathbb{R}^{n}\\!\ni\\!z\mapsto
z^{\langle\kappa\rangle}\in\mathbb{R}^{n}$ for $\kappa>1$.  More precisely,
we define
$\mathcal{F}_{\langle\kappa\rangle}\\!:{\mathbb{R}^{n}}\times{\mathbb{R}^{n}}\to\mathbb{R}^{n}$
by

$\mathcal{F}_{\langle\kappa\rangle}(w,z):=z^{\langle\kappa\rangle}-w^{\langle\kappa\rangle}-J_{\langle\kappa\rangle}(w)(z-w),\;\quad
w,z\in\mathbb{R}^{n}.$

Of course, the mapping $\mathbb{R}\ni x\mapsto
x^{\langle\kappa\rangle}\in\mathbb{R}$ is in general not convex and
$F_{\langle\kappa\rangle}$ changes sign. In fact,
$F_{\langle\kappa\rangle}(-a,-b)=-F_{\langle\kappa\rangle}(a,b)$.

The scalar versions of the above functions (for $n=1$) are denoted
$F_{\kappa}$ (see Introduction), $H_{\kappa}$, $G_{\kappa}$, and
$F_{\langle\kappa\rangle}$, respectively. In particular,

$F_{\langle\kappa\rangle}(a,b)=b^{\langle\kappa\rangle}-a^{\langle\kappa\rangle}-\kappa|a|^{\kappa-1}(b-a),\quad
a,b\in\mathbb{R}.$

The following estimates are quite important for our analysis.

######  Lemma 2.1.

Let $p\in(1,\infty)$. We have

(2.12) $\displaystyle\mathcal{F}_{p}(w,z)\asymp\mathcal{G}_{p}(w,z),\quad
w,z\in\mathbb{R}^{n},$

and

(2.13) $\displaystyle\mathcal{H}_{p}(w,z)\asymp\mathcal{G}_{p}(w,z),\quad
w,z\in\mathbb{R}^{n}.$

Of, course (2.13) follows from (2.12). It seems that (2.12) was first proved
in Pinchover, Tertikas, and Tintarev [28, (2.19)], but one of the one-sided
bounds was given earlier in Shafrir [30, Lemma 7.4] for $p\geq 2$ and the
other in Barbatis, Filippas, and Tertikas [2, Lemma 3.1]. The one-dimensional
case, $F_{p}(a,b)\asymp(b-a)^{2}(|a|\vee|b|)^{p-2}$, $a,b\in\mathbb{R}$, is
crucial in [3, Lemma 6], with [3, (10), (12)] therein being a part of the
comparison for $n=2$ (the setting of (2.12) is essentially two-dimensional).
Optimal constants are known in some cases: for the lower bound of $F_{p}$ with
$p\in(1,2)$ and for the upper bound with $p\in(2,\infty)$; see [8] and [30,
Lemma 7.4]. The quadratic factor in (2.11) is the reason why Bregman
divergence $\mathcal{F}_{p}$ is integrable against Lévy measures in (1.3),
which is crucial in analysis of nonlocal equations of parabolic and elliptic
type; see the applications of [6, (2.14)] therein. See also [8] for a
martingale setting. In passing we note another important estimate:

(2.14) $\mathcal{F}_{p}(w,z)\asymp\left|z^{\langle p/2\rangle}-w^{\langle
p/2\rangle}\right|^{2},\quad w,z\in\mathbb{R}^{n}.$

We refer to [8, Subsection 1.3] for a discussion of the estimate when $n=1$;
the case of arbitrary $n=1,2,\ldots$ can be found in Huang [22]. Further
estimates concerning functions $\mathcal{F}_{p}$,
$\mathcal{F}_{\langle\kappa\rangle}$, and their cousins are collected and
proved in Appendix A.

###  2.3. Semigroups, generators, and forms

The semigroup is defined by

$\displaystyle P_{t}f(x):=\int_{\mathbb{R}^{d}}f(y)p_{t}(x,y)\,{\rm d}y,\quad
t>0,$

and by $P_{0}f(x):=f(x)$, where $x\in{\mathbb{R}^{d}}$ and
$f\\!:{\mathbb{R}^{d}}\to\mathbb{R}$ is nonnegative or integrable. We briefly
mention a well known probability connection:
$P_{t}f(x)={\mathbb{E}_{x}}f(X_{t})$, where $(X_{t},\mathbb{P}_{x})_{t\geq
0,x\in\mathbb{R}^{d}}$ is our Lévy process, considered as a Markov process
with transition density on ${\mathbb{R}^{d}}$ given by $p_{t}(\cdot,\cdot)$,
and $\mathbb{E}_{x}$ is the expectation with respect to the distribution
$\mathbb{P}_{x}$ of the process starting from $x$.

Since $\int_{\mathbb{R}^{d}}p_{t}(x,y)\,{\rm d}y=1$ for $t>0$,
$x\in\mathbb{R}^{d}$ (conservativeness of the semigroup $P_{t}$), and $p_{t}$
is symmetric in $x,y,$ Fubini-Tonelli theorem yields

(2.15) $\displaystyle\int_{\mathbb{R}^{d}}P_{t}f(x)\,{\rm
d}x=\int_{\mathbb{R}^{d}}f(x)\,{\rm d}x.$

Recall that $1<p<\infty$. It is well known that $(P_{t})_{t\geq 0}$ is a
strongly continuous Markovian semigroup of symmetric operators on
$L^{p}({\mathbb{R}^{d}})$; see for example [29, E 34.10]. For all
$x\in{\mathbb{R}^{d}}$ and $f\in L^{p}({\mathbb{R}^{d}})$, by (2.3) and
Hölder’s inequality with exponents $p$ and $q=p/(p-1)$, we get

(2.16) $\displaystyle\left|P_{t}f(x)\right|$
$\displaystyle=\left|\int_{\mathbb{R}^{d}}f(y)p_{t}(x,y)\,{\rm
d}y\right|\leq\|f\|_{L^{p}({\mathbb{R}^{d}})}\left(\int_{\mathbb{R}^{d}}p_{t}(x,y)^{q}\,{\rm
d}y\right)^{1/q}$
$\displaystyle\leq\|f\|_{L^{p}({\mathbb{R}^{d}})}\left(\sup_{x,y\in{\mathbb{R}^{d}}}p_{t}(x,y)^{q-1}\right)^{1/q}=\|f\|_{L^{p}({\mathbb{R}^{d}})}\,\|p_{t}\|_{L^{\infty}({\mathbb{R}^{d}})}^{1/p}\xrightarrow[t\to\infty]{}0.$

We also need the following maximal inequality of Stein for symmetric Markovian
semigroups; see Stein [31, p. 73] and recall that $1<p<\infty$.

######  Lemma 2.2 (Stein inequality).

If $f\in L^{p}({\mathbb{R}^{d}})$, $f^{*}(x):=\sup_{t\geq 0}|P_{t}f(x)|$,
$x\in{\mathbb{R}^{d}}$, then,

(2.17)
$\|f^{*}\|_{L^{p}({\mathbb{R}^{d}})}\leq\frac{p}{p-1}\|f\|_{L^{p}({\mathbb{R}^{d}})}.$

By (2.17) and (2.16), the semigroup is strongly stable in
$L^{p}({\mathbb{R}^{d}})$: If $f\in L^{p}({\mathbb{R}^{d}})$, then

(2.18) $\|P_{t}f\|_{L^{p}({\mathbb{R}^{d}})}\to 0\mbox{ as }t\to\infty.$

Indeed, since for every $x\in{\mathbb{R}^{d}}$ we have $|P_{t}f(x)|\to 0$ and
$|P_{t}f(x)|\leq f^{*}(x)$ with $f^{*}\in L^{p}({\mathbb{R}^{d}})$, we get
$\|P_{t}f\|_{L^{p}({\mathbb{R}^{d}})}\to 0$ by the Dominated Convergence
Theorem.

Let $L$ be the generator of the semigroup $(P_{t})_{t\geq 0}$, when acting on
$L^{p}({\mathbb{R}^{d}})$. Its natural domain, denoted $\mathcal{D}_{p}(L)$,
consists of those $f\in L^{p}({\mathbb{R}^{d}})$ for which there is a $g\in
L^{p}({\mathbb{R}^{d}})$ such that $(P_{h}f-f)/h\to g$ in
$L^{p}({\mathbb{R}^{d}})$ as $h\to 0^{+}$; we then write $Lf=g$.

We next discuss issues related to the $L^{p}$-differentiability of semigroups.
(To make the exposition self-contained, we include a primer on the $L^{p}$
calculus in Appendix B.) Thus, for $f\in L^{p}({\mathbb{R}^{d}})$ and $t\geq
0$, we write $u(t):=P_{t}f$. Of course, $u(t)\in L^{p}({\mathbb{R}^{d}})$.
Furthermore, if $f\in\mathcal{D}_{p}(L)$ then
$u^{\prime}(t)=LP_{t}f=P_{t}Lf=Lu(t)$, $t\geq 0$. By Lemma B.3 with $n=1$, we
obtain the following result.

######  Corollary 2.3.

Let $f\in\mathcal{D}_{p}(L)$. If $1<\kappa\leq p$ then:

  1. (i)

$|u(t)|^{\kappa}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}})$ and

(2.19) $(|u(t)|^{\kappa})^{\prime}=\kappa
u(t)^{\langle\kappa-1\rangle}u^{\prime}(t)=\kappa
u(t)^{\langle\kappa-1\rangle}P_{t}Lf,\quad t\geq 0,$

  2. (ii)

$u^{\langle\kappa\rangle}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}})$ and

(2.20)
$\displaystyle(u(t)^{\langle\kappa\rangle})^{\prime}=\kappa|u(t)|^{\kappa-1}u^{\prime}(t)=\kappa|u(t)|^{\kappa-1}P_{t}Lf,\quad
t\geq 0.$

Moreover, since $(P_{t})_{t\geq 0}$ is symmetric, it is an analytic semigroup
on $L^{p}({\mathbb{R}^{d}})$ for $p\in(1,\infty)$; see Liskevich and
Perel’muter [25, Corollary 3.2]. Therefore, for all $t>0$ and $f\in
L^{p}({\mathbb{R}^{d}})$, $\frac{\rm d}{{\rm d}t}P_{t}f=u^{\prime}(t)$ exists
in $L^{p}({\mathbb{R}^{d}})$, so $P_{t}f\in\mathcal{D}_{p}(L)$ and
$u^{\prime}(t)=LP_{t}f=Lu(t)$.

As a special case of (1.3), in what follows we consider the integral form

(2.21)
$\displaystyle\mathcal{E}_{p}[u]=\frac{1}{p}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}F_{p}(u(x),u(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

Of course, the form is well-defined (possibly infinite) for every
$u\\!:{\mathbb{R}^{d}}\to\mathbb{R}$ because $F_{p}\geq 0$. By the symmetry of
$\nu$,

(2.22) $\displaystyle\mathcal{E}_{p}[u]$ $\displaystyle=$
$\displaystyle\frac{1}{p}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}H_{p}(u(x),u(y))\nu(x,y)\,{\rm
d}x{\rm d}y$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(u(y)-u(x))\left(u(y)^{\langle
p-1\rangle}-u(x)^{\langle p-1\rangle}\right)\nu(x,y)\,{\rm d}x{\rm d}y.$

The natural domain of $\mathcal{E}_{p}$ is

(2.23) $\displaystyle\mathcal{D}(\mathcal{E}_{p}):=\\{u\in
L^{p}({\mathbb{R}^{d}}):\ \mathcal{E}_{p}[u]<\infty\\}.$

When $p=2$, we get the usual Dirichlet form of the semigroup,

(2.24)
$\displaystyle\mathcal{E}_{2}[u]=\frac{1}{2}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(u(y)-u(x))^{2}\nu(x,y)\,{\rm
d}x{\rm d}y,$

with domain $\mathcal{D}(\mathcal{E}_{2})$. We write
$\mathcal{E}:=\mathcal{E}_{2}$.

For $t>0$, $u\in L^{p}({\mathbb{R}^{d}})$, and $v\in L^{q}({\mathbb{R}^{d}})$,
we define, as usual,

(2.25) $\displaystyle\mathcal{E}^{(t)}(u,v):=\frac{1}{t}\langle
u-P_{t}u,v\rangle=\frac{1}{t}\int_{{\mathbb{R}^{d}}}(u(x)-P_{t}u(x))v(x)\,{\rm
d}x.$

Here and below we use the following notation for the canonical paring:

$\displaystyle\langle u,v\rangle:=\int_{\mathbb{R}^{d}}u(x)v(x)\,{\rm d}x.$

The next result was established in [8, Lemma 7] for the fractional Laplacian,
but since its proof requires only symmetry and the conditions (P1) and (P2),
it applies verbatim in the present setting.

######  Proposition 2.4.

Let $p>1$. For every $u\in L^{p}({\mathbb{R}^{d}})$, we have

(2.26) $\mathcal{E}_{p}[u]=\lim_{t\to 0}\mathcal{E}^{(t)}(u,u^{\langle
p-1\rangle}).$

Furthermore,

(2.27) $\displaystyle\mathcal{D}(\mathcal{E}_{p})$ $\displaystyle=$
$\displaystyle\\{u\in
L^{p}({\mathbb{R}^{d}}):\sup_{t>0}\mathcal{E}^{(t)}(u,u^{\langle
p-1\rangle})<\infty\\}$ (2.28) $\displaystyle=$ $\displaystyle\\{u\in
L^{p}({\mathbb{R}^{d}}):\text{ finite }\lim_{t\to
0}\mathcal{E}^{(t)}(u,u^{\langle p-1\rangle})\text{ exists}\\}.$

For arbitrary $u\\!:{\mathbb{R}^{d}}\to\mathbb{R}$, we have

(2.29) $\frac{4(p-1)}{p^{2}}\mathcal{E}[u^{\langle
p/2\rangle}]\leq\mathcal{E}_{p}[u]\leq 2\mathcal{E}[u^{\langle p/2\rangle}]$

and $\mathcal{D}(\mathcal{E}_{p})=\mathcal{D}(\mathcal{E})^{\langle
2/p\rangle}:=\\{v^{\langle 2/p\rangle}:v\in\mathcal{D}(\mathcal{E})\\}$.
Finally, $\mathcal{D}_{p}(L)\subset\mathcal{D}(\mathcal{E}_{p})$ and

(2.30) $\mathcal{E}_{p}[u]=-\langle Lu,u^{\langle p-1\rangle}\rangle,\quad
u\in\mathcal{D}_{p}(L).$

The disscusion extends to functions with values in $\mathbb{R}^{n}$,
$n=1,2,\ldots$. Namely, let $f_{1},\ldots,f_{n}\in L^{p}({\mathbb{R}^{d}})$,
so $F:=(f_{1},\ldots,f_{n})\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$. We
denote

(2.31) $P_{t}F:=(P_{t}f_{1},\ldots,P_{t}f_{n}),\quad t\geq 0,$

thus $P_{t}F\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$, $t\geq 0$. If,
furthermore, $f_{1},\ldots,f_{n}\in\mathcal{D}_{p}(L)$, then we define

(2.32) $LF:=(Lf_{1},\ldots,Lf_{n}).$

Then, letting $U(t):=P_{t}F(t)$, $t\geq 0$, we get $U^{\prime}(t)=LU(t)$ and
the following multidimensional extension of Corollary 2.3, an easy consequence
of Lemma B.3.

######  Corollary 2.5.

Let $n=1,2,\ldots$, $f_{1},\ldots,f_{n}\in\mathcal{D}_{p}(L)$,
$F:=(f_{1},\ldots,f_{n})$, and $U(t)=P_{t}F$, $t\geq 0$. If $1<\kappa\leq p$,
then for $t\geq 0$,

  1. (i)

$|U(t)|^{\kappa}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}})$ with

$\left(|U(t)|^{\kappa}\right)^{\prime}=\kappa
U(t)^{\langle\kappa-1\rangle}\cdot LU(t),$

  2. (ii)

$U(t)^{\langle\kappa\rangle}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}};\mathbb{R}^{n})$ with

$\left(U(t)^{\langle\kappa\rangle}\right)^{\prime}=\left(J_{\langle\kappa\rangle}\circ
U(t)\right)LU(t).$

The following result will be useful in limiting procedures later on.

######  Lemma 2.6.

[8, Lemma 6] If nonnegative functions
$f,f_{k}\\!:{\mathbb{R}^{d}}\to\mathbb{R}$, $k=1,2,\ldots$, satisfy $f_{k}\leq
cf$ and $f=\lim_{k\to\infty}f_{k}$, then $\lim_{k\to\infty}\int f_{k}\,{\rm
d}\mu=\int f\,{\rm d}\mu$ for each measure $\mu$.

##  3\. Hardy–Stein identity

Below we work under the assumptions on $\nu$ formulated in Subsection 2.1. We
will extend (1.1) to arbitrary dimension $n=1,2,\ldots$. We recall that the
proof given in [1, Theorem 3.2] for $n=1$ relies on approximations and
pointwise calculus in ${\mathbb{R}^{d}}$. Here, instead, we use a more
synthetic differential calculus in $L^{p}$.

######  Theorem 3.1.

Let $p>1$, $n=1,2,\ldots$, and $F=(f_{1},\ldots,f_{n})\in
L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$. Then,

(3.1) $\displaystyle\int_{\mathbb{R}^{d}}|F(x)|^{p}\,{\rm
d}x=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

###### Proof.

Let first $F=(f_{1},\ldots,f_{n})\in(\mathcal{D}_{p}(L))^{n}$ and $0\leq t\leq
T<\infty$. Then $U(t):=P_{t}F\in(\mathcal{D}_{p}(L))^{n}$ and
$LP_{t}F=LU(t)=(LP_{t}f_{1},\ldots,LP_{t}f_{n})$. From Corollary 2.5,
$|U(t)|^{p}$ is continuously differentiable in $L^{1}({\mathbb{R}^{d}})$ and
$\left(|U(t)|^{p}\right)^{\prime}=pU(t)^{\langle p-1\rangle}\cdot LF(t)$. As
$f\mapsto\int_{\mathbb{R}^{d}}f\,{\rm d}x$ is a continuous linear functional
on $L^{1}({\mathbb{R}^{d}})$,

(3.2) $\displaystyle\frac{\rm d}{{\rm
d}t}\int_{\mathbb{R}^{d}}|U(t)|^{p}\,{\rm d}x$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\frac{\rm d}{{\rm d}t}|U(t)|^{p}\,{\rm
d}x=\int_{\mathbb{R}^{d}}pU(t)^{\langle p-1\rangle}\cdot LU(t)\,{\rm d}x$
$\displaystyle=$ $\displaystyle\langle LU(t),pU(t)^{\langle
p-1\rangle}\rangle.$

Since $LU(t)=\lim_{h\to 0^{+}}(P_{h}U(t)-U(t))/h$ strongly in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$, $U(t)^{\langle p-1\rangle}$ belongs
to the (dual) space $L^{\frac{p}{p-1}}({\mathbb{R}^{d}};\mathbb{R}^{n})$, and
the semigroup $(P_{t})_{t\geq 0}$ is conservative, we get

(3.3) $\displaystyle\langle LU(t),pU(t)^{\langle p-1\rangle}\rangle$
$\displaystyle=$ $\displaystyle\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}pU(t)(x)^{\langle
p-1\rangle}\cdot(U(t)(y)-U(t)(x))\frac{p_{h}(x,y)}{h}\,{\rm d}x{\rm d}y$
$\displaystyle=$ $\displaystyle\lim_{h\to
0^{+}}\left[\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}pP_{t}F(x)^{\langle
p-1\rangle}\cdot(P_{t}F(y)-P_{t}F(x))\frac{p_{h}(x,y)}{h}\,{\rm d}x{\rm
d}y\right.$
$\displaystyle+\left.\frac{1}{h}\int_{\mathbb{R}^{d}}|P_{t}F(x)|^{p}\,{\rm
d}x-\frac{1}{h}\int_{\mathbb{R}^{d}}P_{h}(|P_{t}F|^{p})(x)\,{\rm d}x\right]$
$\displaystyle=$ $\displaystyle-\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\frac{p_{h}(x,y)}{h}\,{\rm
d}x{\rm d}y$ $\displaystyle=$
$\displaystyle-\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

The last equality (3.3) is justified by Lemma 2.6, the nonnegativity of
$\mathcal{F}_{p}$, and assumptions (P1), (P2). Summarizing,

$\frac{\rm d}{{\rm d}t}\int_{\mathbb{R}^{d}}|U(t)|^{p}{\rm
d}x=-\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

Since $|U(t)|^{p}$ is continuously differentiable in $L^{1}({\mathbb{R}^{d}})$
on ${[0,\infty)}$ and the integration is a continuous linear functional on
$L^{1}({\mathbb{R}^{d}})$, $\int_{\mathbb{R}^{d}}|U(t)|^{p}\,{\rm d}x$ is
continuously differentiable. Integrating from $0$ to $T$ we obtain

$\displaystyle\int_{\mathbb{R}^{d}}|F|^{p}\,{\rm
d}x-\int_{\mathbb{R}^{d}}|U(T)|^{p}\,{\rm d}x$ $\displaystyle=$
$\displaystyle-\int_{0}^{T}\left(\frac{\rm d}{{\rm
d}t}\int_{\mathbb{R}^{d}}|U(t)|^{p}\,{\rm d}x\right)\,{\rm d}t$
$\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

We let $T\to\infty$ and obtain $\int_{\mathbb{R}^{d}}|U(T)|^{p}\,{\rm d}x\to
0$ from the strong stability (2.18).

We now relax the assumption $f_{j}\in\mathcal{D}_{p}(L)$. Let
$F=(f_{1},\ldots,f_{n})\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$ be
arbitrary and let $s>0$. Since $(P_{t})_{t\geq 0}$ is an analytic semigroup on
$L^{p}({\mathbb{R}^{d}})$, $P_{s}f_{j}\in\mathcal{D}_{p}(L)$ for all
$j=1,\ldots,n$, so $U(s)\in(\mathcal{D}_{p}(L))^{n}$. By (3.1) and a change of
variables,

(3.4) $\displaystyle\int_{\mathbb{R}^{d}}|U(s)|^{p}\,{\rm
d}x=\int_{s}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

Let $s$ decrease to $0$. Since $\mathcal{F}_{p}\geq 0$, the right-hand side of
(3.4) increases to
$\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{F}_{p}(P_{t}F(x),P_{t}F(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$. By the strong continuity of $(P_{t})_{t\geq 0}$ in
$L^{p}({\mathbb{R}^{d}})$, $P_{s}f_{j}\to f_{j}$, $j=1,\ldots,n$, in
$L^{p}({\mathbb{R}^{d}})$, so $U(s)=P_{s}F\to F$ in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$, in particular
$\|U(s)\|^{p}_{L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})}\to\|F\|^{p}_{L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})}$.
The proof is complete. ∎

######  Remark 3.2.

Since $\nu$ is symmetric, by (2.8) we get a symmetrized version of the
Hardy–Stein identity for every $F\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$:

$\displaystyle\int\limits_{\mathbb{R}^{d}}|F|^{p}\,{\rm d}x$
$\displaystyle=\frac{p}{2}\int\limits_{0}^{\infty}\\!\int\limits_{\mathbb{R}^{d}}\\!\int\limits_{\mathbb{R}^{d}}(P_{t}F(y)-P_{t}F(x))\\!\cdot\\!\left(P_{t}F(y)^{\langle
p-1\rangle}-P_{t}F(x)^{\langle p-1\rangle}\right)\\!\nu(x,y)\,{\rm d}x{\rm
d}y{\rm d}t.$

##  4\. Polarized Hardy–Stein identity

Having proved the Hardy–Stein identity for a vector of
$L^{p}({\mathbb{R}^{d}})$ functions, we can establish a disintegration of
$\int_{\mathbb{R}^{d}}f(x)g(x)^{\langle p-1\rangle}\,{\rm d}x$ for $f,g\in
L^{p}({\mathbb{R}^{d}})$ with $p\in[2,\infty)$.

To this end we introduce the function
$\mathcal{J}_{p}\\!:\mathbb{R}^{2}\times\mathbb{R}^{2}\to\mathbb{R}$, defined
as follows:

(4.1) $\displaystyle\mathcal{J}_{p}(w,z)=$
$\displaystyle\mathcal{J}_{p}(w_{1},w_{2};z_{1},z_{2}):=z_{1}z_{2}^{\langle
p-1\rangle}-w_{1}w_{2}^{\langle p-1\rangle}$ $\displaystyle-w_{2}^{\langle
p-1\rangle}(z_{1}-w_{1})-(p-1)w_{1}|w_{2}|^{p-2}(z_{2}-w_{2}),$

where $w=(w_{1},w_{2})$, $z=(z_{1},z_{2})$, and
$w_{1},w_{2},z_{1},z_{2}\in\mathbb{R}$. For instance,

(4.2)
$\displaystyle\mathcal{J}_{2}(w,z)=z_{1}z_{2}-w_{1}w_{2}-w_{2}(z_{1}-w_{1})-w_{1}(z_{2}-w_{2})=(z_{1}-w_{1})(z_{2}-w_{2}).$

As complicated as it looks, $\mathcal{J}_{p}$ is just the second-order Taylor
remainder of the mapping $\mathbb{R}^{2}\ni(z_{1},z_{2})\mapsto
z_{1}z_{2}^{\langle p-1\rangle}$, when the argument changes from $w$ to $z$.
Below we mostly apply $\mathcal{J}_{p}$ to $w_{1}=P_{t}f(x)$,
$w_{2}=P_{t}g(x)$, $z_{1}=P_{t}f(y)$, and $z_{2}=P_{t}g(y)$, so $w$
corresponds to the argument $x$ of the vector function $\Phi=(f,g)$, $z$
corresponds to $y$, the subscript $1$ indicates the first function, $f$, and
$2$ indicates the second function, $g$.

Here is the main result of the paper, which we prove below in this section.

######  Theorem 4.1 (Polarized Hardy–Stein identity).

Let $p\geq 2$. For $f,g\in L^{p}({\mathbb{R}^{d}})$, denote

(4.3) $\Phi(x):=(f(x),g(x))\quad{\rm and}\quad
P_{t}\Phi(x):=(P_{t}f(x),P_{t}g(x)),\quad t\geq 0,\;x\in{\mathbb{R}^{d}}.$

Then,

(4.4)
$\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left|\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\right|\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t<\infty$

and

(4.5) $\displaystyle\int_{\mathbb{R}^{d}}fg^{\langle p-1\rangle}\,{\rm
d}x=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

Note that if $w_{1}=w_{2}=:a$ and $z_{1}=z_{2}=:b$, then
$\mathcal{J}_{p}(w,z)=F_{p}(a,b)$, so (4.5) with $f=g$ agrees with (1.1), at
least for $p\geq 2$.

######  Remark 4.2.

If $p=2$ then (4.5) reads

(4.6) $\displaystyle\int_{\mathbb{R}^{d}}fg\,{\rm
d}x=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}[P_{t}f(x)-P_{t}f(y)][P_{t}g(x)-P_{t}g(y)]\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

In this case (4.4) and (4.5) are obtained by polarization from the one-
dimensional Hardy–Stein identity (1.1) and Cauchy-Schwarz inequality, by
considering $f+g$ and $f-g$. Therefore below we let $p>2$.

Had $\mathcal{J}_{p}$ been nonnegative, the proof of (4.5) would follow as
that of (3.1). Unfortunately, this is not the case, so the proof is more
complicated. Indeed, the function $(z_{1},z_{2})\mapsto z_{1}z_{2}^{\langle
p-1\rangle}$ is not convex, even when restricted to $z_{2}>0$. To see this, we
compute its gradient and Hessian matrix for $z_{2}>0$:

$\displaystyle\nabla\left(z_{1}z_{2}^{p-1}\right)=\begin{bmatrix}z_{2}^{p-1}\\\
(p-1)z_{1}z_{2}^{p-2}\end{bmatrix},$ (4.7)
$\displaystyle\nabla^{2}\left(z_{1}z_{2}^{p-1}\right)=\begin{bmatrix}0&(p-1)z_{2}^{p-2}\\\
(p-1)z_{2}^{p-2}&(p-1)(p-2)z_{1}z_{2}^{p-3}\end{bmatrix}.$

Thus, $\det\nabla^{2}\left(z_{1}z_{2}^{p-1}\right)=-(p-1)^{2}z_{2}^{2p-4}<0$,
so the Hessian matrix $\nabla^{2}\left(z_{1}z_{2}^{p-1}\right)$ is not
positive semi-definite and $z_{1}z_{2}^{p-1}$ is not convex. We will rectify
this situation by decomposing the mapping

$[0,\infty)\times\mathbb{R}\ni z=(z_{1},z_{2})\mapsto z_{1}z_{2}^{\langle
p-1\rangle}$

into a difference of two convex mappings. Then (the Taylor remainder)
$\mathcal{J}_{p}$ will be a difference of two nonnegative functions. To this
end, we recall that $a_{+}:=a\vee 0,$ $a_{-}:=(-a)\vee 0$ and introduce the
functions:

$\displaystyle Y^{(+)}(z)$ $\displaystyle:=$ $\displaystyle
z_{1}\left((z_{2})_{+}\right)^{p-1}+|z|^{p},$ $\displaystyle Y^{(-)}(z)$
$\displaystyle:=$ $\displaystyle
z_{1}\left((z_{2})_{-}\right)^{p-1}+|z|^{p},\quad
z=(z_{1},z_{2})\in\mathbb{R}^{2}.$

Lemma C.3 in Appendix C verifies that these functions are convex on
$[0,\infty)\times\mathbb{R}$ indeed. Since $p>2$, they are differentiable
everywhere and their Taylor remainders are nonnegative on
$[0,\infty)\times\mathbb{R}$.

Let $\mathcal{J}_{p}^{(+)}$ and $\mathcal{J}_{p}^{(-)}$ be the second-order
Taylor remainders of the differentiable mappings $\mathbb{R}^{2}\ni z\mapsto
z_{1}\left((z_{2})_{+}\right)^{p-1}$ and $\mathbb{R}^{2}\ni z\mapsto
z_{1}\left((z_{2})_{-}\right)^{p-1}$. Thus, for
$z_{1},z_{2},w_{1},w_{2}\in\mathbb{R}$,

$\displaystyle\mathcal{J}_{p}^{(+)}(w,z)$ $\displaystyle=$ $\displaystyle
z_{1}\left((z_{2})_{+}\right)^{p-1}-w_{1}\left((w_{2})_{+}\right)^{p-1}-\left((w_{2})_{+}\right)^{p-1}(z_{1}-w_{1})$
$\displaystyle-(p-1)w_{1}\left((w_{2})_{+}\right)^{p-2}(z_{2}-w_{2})$

and

$\displaystyle\mathcal{J}_{p}^{(-)}(w,z)$ $\displaystyle=$ $\displaystyle
z_{1}\left((z_{2})_{-}\right)^{p-1}-w_{1}\left((w_{2})_{-}\right)^{p-1}-\left((w_{2})_{-}\right)^{p-1}(z_{1}-w_{1})$
$\displaystyle+(p-1)w_{1}\left((w_{2})_{-}\right)^{p-2}(z_{2}-w_{2}).$

Since $z_{1}((z_{2})_{+})^{p-1}-z_{1}((z_{2})_{-})^{p-1}=z_{1}z_{2}^{\langle
p-1\rangle}$, it follows that

(4.8)
$\displaystyle\mathcal{J}_{p}=\mathcal{J}_{p}^{(+)}-\mathcal{J}_{p}^{(-)}=\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)-\left(\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\right),$

where we consider $\mathcal{F}_{p}$ given by (2.7) with $n=2$ and we have
$\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\geq 0$ and
$\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\geq 0$ on
$\left([0,\infty)\times\mathbb{R}\right)^{2}$. Note also that, if we denote
$\bar{z}:=(z_{1},-z_{2})$, then

(4.9)
$\displaystyle\mathcal{J}_{p}^{(+)}(\bar{w},\bar{z})=\mathcal{J}_{p}^{(-)}(w,z).$

Here is an preliminary version of Theorem 4.1.

######  Proposition 4.3.

For $p>2$, $f,g\in L^{p}({\mathbb{R}^{d}})$, $f\geq 0$, and $\Phi(x)$,
$P_{t}\Phi(x)$ as in (4.3),

$\displaystyle\int_{\mathbb{R}^{d}}$
$\displaystyle\left(f(g_{+})^{p-1}+|\Phi|^{p}\right)\,{\rm d}x$ (4.10)
$\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$

and

$\displaystyle\int_{\mathbb{R}^{d}}$
$\displaystyle\left(f(g_{-})^{p-1}+|\Phi|^{p}\right)\,{\rm d}x$ (4.11)
$\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

###### Proof.

We only prove (4.3) since (4.3) follows by substituting $-g$ in place of $g$,
see (2.10) and (4.9). The proof of (4.3) is much alike that of Theorem 3.1. We
use the convexity of $Y^{(+)}$, resulting in the nonnegativity of its Taylor
remainder. the function $\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}$. As before, we
first consider $f,g\in\mathcal{D}_{p}(L)$. Fix some $0\leq t\leq T<\infty$.
Let $u(t):=P_{t}f$, $v(t):=P_{t}g$, and $U(t):=P_{t}\Phi=(u(t),v(t))\in
L^{p}({\mathbb{R}^{d}};\mathbb{R}^{2})$. Actually,
$U(t)\in(\mathcal{D}_{p}(L))^{2}$. As seen in the proof of Theorem 3.1, the
function $t\mapsto|U(t)|^{p}$ is continuously differentiable in
$L^{1}({\mathbb{R}^{d}})$ and $\left(|U(t)|^{p}\right)^{\prime}=pU(t)^{\langle
p-1\rangle}\cdot LU(t)$. Since $(v(t)_{+})^{p-1}=(|v(t)|^{p-1}+v(t)^{\langle
p-1\rangle})/2$, from Corollary 2.3 with $\kappa=p-1>1$, we obtain that
$(v(t)_{+})^{p-1}$ is continuously differentiable in
$L^{\frac{p}{p-1}}({\mathbb{R}^{d}})$ and

$\displaystyle\left((v(t)_{+})^{p-1}\right)^{\prime}$ $\displaystyle=$
$\displaystyle\left(\frac{|v(t)|^{p-1}+v(t)^{\langle
p-1\rangle}}{2}\right)^{\prime}=\frac{p-1}{2}\left(v(t)^{\langle
p-2\rangle}Lv(t)+|v(t)|^{p-2}Lv(t)\right)$ $\displaystyle=$
$\displaystyle(p-1)(v(t)_{+})^{p-2}Lv(t).$

By Lemma B.4, $u(t)(v(t)_{+})^{p-1}$ is continuously differentiable in
$L^{1}({\mathbb{R}^{d}})$ and

(4.12)
$\displaystyle\left(u(t)(v(t)_{+})^{p-1}\right)^{\prime}=(v(t)_{+})^{p-1}Lu(t)+(p-1)u(t)(v(t)_{+})^{p-2}Lv(t).$

In particular, $\left(u(t)(v(t)_{+})^{p-1}\right)^{\prime}$ is well-defined
and continuous in $L^{1}(\mathbb{R}^{d})$. As in (3.2),

$\displaystyle W(t)$ $\displaystyle:=\frac{\rm d}{{\rm
d}t}\int\limits_{\mathbb{R}^{d}}\left(u(t)(v(t)_{+})^{p-1}+|U(t)|^{p}\right)\,{\rm
d}x=\int\limits_{\mathbb{R}^{d}}\frac{\rm d}{{\rm
d}t}\left[u(t)(v(t)_{+})^{p-1}+|U(t)|^{p}\right]\,{\rm d}x$ (4.13)
$\displaystyle=\langle Lu(t),(v(t)_{+})^{p-1}\rangle+\langle
Lv(t),(p-1)u(t)(v(t)_{+})^{p-2}\rangle+\langle LU(t),pU(t)^{\langle
p-1\rangle}\rangle.$

Since the limits defining $Lu$, $Lv$ (respectively, $LU$) exist strongly in
$L^{p}({\mathbb{R}^{d}})$ (respectively, in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{2})$) and $(v(t)_{+})^{p-1}$,
$u(t)(v(t)_{+})^{p-2}$ (respectively, $U(t)^{\langle p-1\rangle}$) belong to
$L^{q}({\mathbb{R}^{d}})$ (respectively, to
$L^{q}({\mathbb{R}^{d}};\mathbb{R}^{2})$), we get

$\displaystyle W(t)=\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}$
$\displaystyle\Big{(}(u(t)(y)-u(t)(x))(v(t){(x)}_{+})^{p-1}$
$\displaystyle+(p-1)(v(t)(y)-v(t)(x))u(t)(x)(v(t)(x)_{+})^{p-1}$
$\displaystyle+p(U(t)(y)-U(t)(x))\cdot U(t)(x)^{\langle
p-1\rangle}\Big{)}\frac{p_{h}(x,y)}{h}\,{\rm d}x{\rm d}y.$

As $(P_{t})_{t\geq 0}$ is conservative, for every $h>0$, we have

$\displaystyle\int_{\mathbb{R}^{d}}|U(t)|^{p}\,{\rm
d}x=\int_{\mathbb{R}^{d}}P_{h}\left(|U(t)|^{p}\right)\,{\rm d}x$

and

$\displaystyle\int_{\mathbb{R}^{d}}u(t)(v(t)_{+})^{p-1}\,{\rm
d}x=\int_{\mathbb{R}^{d}}P_{h}\left(u(t)(v(t)_{+})^{p-1}\right)\,{\rm d}x.$

Taking this into account and rearranging, we get

$W(t)=\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(U(t)(x),U(t)(y))\,\frac{p_{h}(x,y)}{h}\,{\rm
d}x{\rm d}y.$

Because of the assumption $f\geq 0$, we have
$U(t)\in[0,\infty)\times\mathbb{R}$ for all $x\in{\mathbb{R}^{d}}$ and $t\geq
0$, so that
$\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(U(t)(x),U(t)(y))$ is
nonnegative (see the discussion preceding the proposition). Therefore from
(P1), (P2), and Lemma 2.6, we conclude that

(4.14)
$W(t)=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(U(t)(x),U(t)(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

Since $u(t)(v(t)_{+})^{p-1}+|U(t)|^{p}$ is continuously differentiable in
$L^{1}({\mathbb{R}^{d}})$ for $t\in{[0,\infty)}$, $W(t)$ is a continuous
(real) function on $(0,\infty)$. Thus,

$\displaystyle\int_{\mathbb{R}^{d}}\Big{(}u(0)(v(0)_{+})^{p-1}+|U(0)|^{p}\Big{)}\,{\rm
d}x-\int_{\mathbb{R}^{d}}\left(u(T)((v(T))_{+})^{p-1}+|U(T)|^{p}\right)\,{\rm
d}x$ $\displaystyle=$ $\displaystyle-\int_{0}^{T}W(t)\,{\rm
d}t=\int_{0}^{T}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(U(t)(x),U(t)(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

We now let $T\to\infty$. As the integrand in the right-hand side is
nonnegative, $u(0)=f$, $v(0)=g$, and $U(0)=\Phi$, to prove (4.3) it is enough
to show that

$\int_{\mathbb{R}^{d}}\left(u(T)((v(T))_{+})^{p-1}+|U(T)|^{p}\right)\,{\rm
d}x=\int_{\mathbb{R}^{d}}\left(P_{T}f((P_{T}g)_{+})^{p-1}+|P_{T}\Phi|^{p}\right)\,{\rm
d}x\to 0.$

While proving Theorem 3.1 we have already shown that
$\int_{\mathbb{R}^{d}}|U(T)|^{p}\,{\rm d}x\to 0$. Further, since
$|P_{T}f(x)|\leq f^{*}(x)$ and $|P_{T}g(x)|\leq g^{*}(x)$ for every
$x\in{\mathbb{R}^{d}}$ and $T>0$ and $f^{*},g^{*}\in L^{p}({\mathbb{R}^{d}})$
by (2.17), we get $\int_{\mathbb{R}^{d}}P_{T}f((P_{T}g)_{+})^{p-1}\,{\rm
d}x\to 0$ by the Dominated Convergence Theorem. This yields (4.3) for
$f,g\in\mathcal{D}_{p}(L)$.

It remains to get rid of the assumption $f,g\in\mathcal{D}_{p}(L)$. We proceed
as in the proof of Theorem 3.1. Take $f,g\in L^{p}({\mathbb{R}^{d}})$
arbitrary and let $s>0$. Since $(P_{t})_{t\geq 0}$ is an analytic semigroup on
$L^{p}({\mathbb{R}^{d}})$, $P_{s}f,P_{s}g\in\mathcal{D}_{p}(L)$ as well.
Consequently, by (4.3),

$\displaystyle\int_{\mathbb{R}^{d}}$
$\displaystyle\left(P_{s}f((P_{s}g)_{+})^{p-1}+|P_{s}\Phi|^{p}\right)\,{\rm
d}x$
$\displaystyle=\int_{s}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

Let $s\to 0^{+}$. As the integrand of the right-hand side is nonnegative, the
integrals tend to the right-hand side of (4.3).

To get the convergence of the left-hand side we use the strong continuity of
$(P_{t})_{t\geq 0}$ in $L^{p}({\mathbb{R}^{d}})$. The convergence
$|P_{s}\Phi|^{p}\to|\Phi|^{p}$ in $L^{1}({\mathbb{R}^{d}})$ was shown in proof
of Theorem 3.1. Since $P_{s}f\to f$ and $(P_{s}g)_{+}\to g_{+}$ in
$L^{p}({\mathbb{R}^{d}})$, by Lemma B.1,
$((P_{s}g)_{+})^{p-1}\to(g_{+})^{p-1}$ in
$L^{\frac{p}{p-1}}({\mathbb{R}^{d}})$. Moreover, by Lemma B.2,
$P_{s}f((P_{s}g)_{+})^{p-1}\to f(g_{+})^{p-1}$ in $L^{1}({\mathbb{R}^{d}})$.
Thus,
$\int_{\mathbb{R}^{d}}\left(P_{s}f((P_{s}g)_{+})^{p-1}+|P_{s}\Phi|^{p}\right)\,{\rm
d}x\to\int_{\mathbb{R}^{d}}\left(f(g_{+})^{p-1}+|\Phi|^{p}\right)\,{\rm d}x$.
The proof of (4.3) is complete. ∎

###### Proof of Theorem 4.1.

Thanks to Remark 4.2, we only need to consider $p>2$. Let first $f\geq 0$. By
Lemma 4.3 and (4.8),

$\displaystyle\int_{\mathbb{R}^{d}}$ $\displaystyle fg^{\langle
p-1\rangle}\,{\rm
d}x=\int_{\mathbb{R}^{d}}\left(f(g_{+})^{p-1}+|\Phi|^{p}\right)\,{\rm
d}x-\int_{\mathbb{R}^{d}}\left(f(g_{-})^{p-1}+|\Phi|^{p}\right)\,{\rm d}x$
$\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$
$\displaystyle-\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left(\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\right)(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$
$\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t,$

where all the integrals are absolutely convergent. Therefore (4.4) holds in
this case.

To get rid of the assumption $f\geq 0$, we consider an arbitrary $f\in
L^{p}({\mathbb{R}^{d}})$ and write $f=f_{+}-f_{-}$. The result holds for pairs
$\Phi^{(+)}:=(f_{+},g)$ and $\Phi^{(-)}:=(f_{-},g)$. Of course,
$\Phi=\Phi^{(+)}-\Phi^{(-)}$. The operators $P_{t}$ are linear and the
function $\mathcal{J}_{p}(w,z)$ is linear in $w_{1}$ and $z_{1}$, so

$\displaystyle\int_{\mathbb{R}^{d}}fg^{\langle p-1\rangle}\,{\rm d}x$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\left(f_{+}\right)g^{\langle
p-1\rangle}\,{\rm d}x-\int_{\mathbb{R}^{d}}\left(f_{-}\right)g^{\langle
p-1\rangle}\,{\rm d}x$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$
$\displaystyle-\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi^{(-)}(x),P_{t}\Phi^{(-)}(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t.$

The absolute convergence of the integrals is clear from our previous
arguments. ∎

We next present a quantitative version of (4.4).

######  Proposition 4.4.

Under the assumptions of Theorem 4.1,

(4.15)
$\displaystyle\int\limits_{0}^{\infty}\int\limits_{\mathbb{R}^{d}}\int\limits_{\mathbb{R}^{d}}\left|\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\right|\nu(x,y)\,{\rm
d}x{\rm d}y{\rm
d}t\leq(1+2^{p/2})\|f\|_{L^{p}({\mathbb{R}^{d}})}\|g\|_{L^{p}({\mathbb{R}^{d}})}^{p-1}.$

###### Proof.

As in the proof of Theorem 4.1, we let $\Phi^{(+)}=(f_{+},g)$ and
$\Phi^{(-)}=(f_{-},g)$. Then,

$\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))=\mathcal{J}_{p}(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))-\mathcal{J}_{p}(P_{t}\Phi^{(-)}(x),P_{t}\Phi^{(-)}(y)),$

so

$\left|\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\right|\leq\left|\mathcal{J}_{p}(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))\right|+\left|\mathcal{J}_{p}(P_{t}\Phi^{(-)}(x),P_{t}\Phi^{(-)}(y))\right|.$

Because of (4.8),

$\left|\mathcal{J}_{p}\right|\leq\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)+\left(\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\right),$

both terms being nonnegative on $\left([0,\infty)\times\mathbb{R}\right)^{2}$.
As
$P_{t}\Phi^{(+)},P_{t}\Phi^{(-)}\in\left([0,\infty)\times\mathbb{R}\right)^{2}$,

$\displaystyle\left|\mathcal{J}_{p}(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))\right|$
$\displaystyle\leq$
$\displaystyle\left(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p}\right)(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))$
$\displaystyle+\left(\mathcal{J}_{p}^{(-)}+\mathcal{F}_{p}\right)(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y)),$

and a similar inequality holds for $P_{t}\Phi^{(-)}$. From Proposition 4.3,

$\int\limits_{0}^{\infty}\int\limits_{\mathbb{R}^{d}}\int\limits_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(+)}+\mathcal{F}_{p})(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm
d}t=\int\limits_{\mathbb{R}^{d}}\left(f_{+}(g_{+})^{p-1}+|\Phi^{(+)}|^{p}\right)\,{\rm
d}x.$

A similar identity holds for $\mathcal{J}_{p}^{(-)}$. Summing up, we get

$\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left|\mathcal{J}_{p}(P_{t}\Phi^{(+)}(x),P_{t}\Phi^{(+)}(y))\right|\nu(x,y)\,{\rm
d}x{\rm d}y{\rm
d}t\leq\int_{\mathbb{R}^{d}}\left(f_{+}|g|^{p-1}+2|\Phi^{(+)}|^{p}\right)\,{\rm
d}x$

and

$\int_{0}^{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left|\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\right|\nu(x,y)\,{\rm
d}x{\rm d}y{\rm
d}t\leq\int_{\mathbb{R}^{d}}\left(f|g|^{p-1}+2|\Phi|^{p}\right)\,{\rm d}x.$

By Hölder inequality,

$\int_{\mathbb{R}^{d}}f|g|^{p-1}\,{\rm
d}x\leq\|f\|_{L^{p}({\mathbb{R}^{d}})}^{1/p}\|g\|_{L^{p}({\mathbb{R}^{d}})}^{(p-1)/p}.$

On the other hand,

$|\Phi|^{p}=(f^{2}+g^{2})^{p/2}\leq 2^{p/2-1}(|f|^{p}+|g|^{p}).$

Therefore, if
$\|f\|_{L^{p}({\mathbb{R}^{d}})}=\|g\|_{L^{p}({\mathbb{R}^{d}})}=1$, then
(4.15) is true if we replace its right-hand side by $1+2^{p/2}$. If
$\|f\|_{L^{p}({\mathbb{R}^{d}})}=0$ or $\|g\|_{L^{p}({\mathbb{R}^{d}})}=0$,
then (4.15) is obvious. Otherwise, we observe that $\mathcal{J}_{p}$ is
homogeneous in the first coordinates, and $(p-1)$-homogeneous in the second,
to wit,

$\mathcal{J}_{p}((\lambda w_{1},\mu w_{2}),(\lambda z_{1},\mu
z_{2}))=\lambda\mu^{\langle
p-1\rangle}\mathcal{J}_{p}((w_{1},w_{2}),(z_{1},z_{2})),\qquad\lambda,\mu>0.$

Then, by considering $f/\|f\|_{L^{p}({\mathbb{R}^{d}})}$ and
$g/\|g\|_{L^{p}({\mathbb{R}^{d}})}$, we get the result.

∎

##  5\. Polarized Sobolev-Bregman form $\mathcal{E}_{p}(u,v)$

The integral expression appearing in (1.6) and Theorem 4.1, namely

$\mathcal{E}_{p}(u,v):=\frac{1}{p}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y,$

where $\Phi(x)=(u(x),v(x))$, $u,v\\!:\mathbb{R}^{d}\to\mathbb{R}$,
$p\in[2,\infty)$, and $\mathcal{J}_{p}$ is given by (4.1), deserves further
attention. If $u=v$ then
$\mathcal{E}_{p}(u,v)=\mathcal{E}_{p}(u,u)=\mathcal{E}_{p}[u]$. For $p=2$, we
get $\mathcal{E}_{2}(u,v)$, the usual (bilinear) Dirichlet form [19], in
particular, it is symmetric. For $p>2$, in general
$\mathcal{E}_{p}(v,u)\neq\mathcal{E}_{p}(u,v)$ and we are even puzzled by the
question whether the integral in (1.6) is well-defined for general enough
functions $u,v$, for instance for $u,v\in\mathcal{D}(\mathcal{E}_{p})$. The
next theorem asserts that for $p\geq 2$ and $u,v\in\mathcal{D}_{p}(L)$, (1.6)
is well-defined; we also get an extension of the single-function formula
(2.30) from Proposition 2.4.

######  Theorem 5.1.

Let $p\geq 2$. If $u,v\in\mathcal{D}_{p}(L)$, then $\mathcal{E}_{p}(u,v)$ is
well-defined and

(5.1) $\displaystyle\mathcal{E}_{p}(u,v)=-\frac{1}{p}\langle Lu,v^{\langle
p-1\rangle}\rangle-\frac{1}{p}\langle Lv,(p-1)u|v|^{p-2}\rangle.$

Note that this agrees with (2.30) if $u=v$. Before we prove (5.1), we need to
further decompose $\mathcal{J}_{p}^{(+)}$ (and $\mathcal{J}_{p}$) into a
difference of nonnegative functions.

Let $\mathbf{1}(a):=(1+\operatorname*{sgn}(a))/2$ be the Heaviside step
function. We define

(5.2) $\displaystyle\mathcal{J}_{p}^{(++)}(w,z)$ $\displaystyle:=$
$\displaystyle(z_{1})_{+}\left((z_{2})_{+}\right)^{p-1}-(w_{1})_{+}\left((w_{2})_{+}\right)^{p-1}-\mathbf{1}(w_{1})\left((w_{2})_{+}\right)^{p-1}(z_{1}-w_{1})$
$\displaystyle-(p-1)(w_{1})_{+}\left((w_{2})_{+}\right)^{p-2}(z_{2}-w_{2}),$
(5.3) $\displaystyle\mathcal{J}_{p}^{(-+)}(w,z)$ $\displaystyle:=$
$\displaystyle(z_{1})_{-}\left((z_{2})_{+}\right)^{p-1}-(w_{1})_{-}\left((w_{2})_{+}\right)^{p-1}+\mathbf{1}(-w_{1})\left((w_{2})_{+}\right)^{p-1}(z_{1}-w_{1})$
$\displaystyle-(p-1)(w_{1})_{-}\left((w_{2})_{+}\right)^{p-2}(z_{2}-w_{2}),$

where $w:=(w_{1},w_{2}),z:=(z_{1},z_{2})\in\mathbb{R}^{2}$. We may view these
functions as the second-order Taylor remainders of the mappings
$\mathbb{R}^{2}\ni z\mapsto(z_{1})_{+}\left((z_{2})_{+}\right)^{p-1}$ and
$\mathbb{R}^{2}\ni z\mapsto(z_{1})_{-}\left((z_{2})_{+}\right)^{p-1}$,
respectively, except for nondifferentiability of the mappings on the vertical
positive semi-axis (for more details, see the proof of Lemma C.4 in Appendix
C).

Similarly to (4.8) and (4.9), we get a decomposition of
$\mathcal{J}_{p}^{(+)}$:

(5.4)
$\displaystyle\mathcal{J}_{p}^{(+)}=\mathcal{J}_{p}^{(++)}-\mathcal{J}_{p}^{(-+)}$

and the identity

(5.5)
$\displaystyle\mathcal{J}_{p}^{(++)}(-\bar{w},-\bar{z})=\mathcal{J}_{p}^{(-+)}(w,z).$

In Lemma C.4 in Appendix C we prove that

$\displaystyle\mathcal{J}_{p}^{(++)}(w,z)+\mathcal{F}_{p}(w,z)\geq
0,\quad\mathcal{J}_{p}^{(-+)}(w,z)+\mathcal{F}_{p}(w,z)\geq 0$

for all $z,w\in\mathbb{R}^{2}$. Therefore, by adding and subtracting
$\mathcal{F}_{p}$ in (5.4), we get the desired decomposition of
$\mathcal{J}_{p}^{(+)}$ and we can proceed from there. Let us mention that it
is crucial to define the Heaviside function so that $\mathbf{1}(0)=1/2$. This
is because we use the identity $\mathbf{1}(a)+\mathbf{1}(-a)=1$ for all
$a\in\mathbb{R}$ to derive (5.4).

###### Proof of Theorem 5.1..

Let $u,v\in\mathcal{D}_{p}(L)$. If $p=2$, then, again, the identity is evident
from classical polarization, (2.24) and (2.30).

Thus, we let $p>2$. Denote $\Phi(x):=(u(x),v(x))$. First we prove the
following:

$\displaystyle l_{++}$ $\displaystyle:=$ $\displaystyle-\langle
Lu,\mathbf{1}(u)(v_{+})^{p-1}\rangle-\langle
Lv,(p-1)u_{+}(v_{+})^{p-2}\rangle-\langle L\Phi,p\Phi^{\langle
p-1\rangle}\rangle$ $\displaystyle\ =$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(++)}+\mathcal{F}_{p})(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y$

and

$\displaystyle l_{-+}$ $\displaystyle:=$ $\displaystyle\langle
Lu,\mathbf{1}(-u)(v_{+})^{p-1}\rangle-\langle
Lv,(p-1)u_{-}(v_{+})^{p-2}\rangle-\langle L\Phi,p\Phi^{\langle
p-1\rangle}\rangle$ $\displaystyle\ =$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(-+)}+\mathcal{F}_{p})(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

We start with the proof of (5). By the definition of $L$,

(5.8) $\displaystyle\langle L\Phi,p\Phi^{\langle p-1\rangle}\rangle=\lim_{h\to
0^{+}}\frac{1}{h}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\Phi(x)-\Phi(y))\cdot
p\Phi(x)^{\langle p-1\rangle}\frac{p_{h}(x,y)}{h}\,{\rm d}x{\rm d}y.$

Since the limits defining $Lu$, $Lv$ exist in the strong sense in
$L^{p}({\mathbb{R}^{d}})$, we have

$\displaystyle l_{++}=$ $\displaystyle\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\left[(u(x)-u(y))\mathbf{1}(u(x))(v(x)_{+})^{p-1}\right.$
$\displaystyle\quad\qquad+(v(x)-v(y))(p-1)u(x)_{+}(v(x)_{+})^{p-2}$
$\displaystyle\quad\qquad\left.+(\Phi(x)-\Phi(y))\cdot p\Phi(x)^{\langle
p-1\rangle}\right]\frac{p_{h}(x,y)}{h}\,{\rm d}x{\rm d}y.$

Then, similarly as in the proofs of Theorems 3.1 and 4.1, we take advantage of
the conservativeness of the semigroup $(P_{t})_{t\geq 0}$:

$\displaystyle\int_{\mathbb{R}^{d}}u_{+}(v_{+})^{p-1}\,{\rm d}x$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}P_{h}\left(u_{+}(v_{+})^{p-1}\right)\,{\rm
d}x,$ $\displaystyle\int_{\mathbb{R}^{d}}|\Phi|^{p}\,{\rm d}x$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}P_{h}\left(|\Phi|^{p}\right)\,{\rm
d}x,\quad\mbox{ for }h>0.$

Taking this into account and rearranging, we obtain

$l_{++}=\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(++)}+\mathcal{F}_{p})(\Phi(x),\Phi(y))\,\frac{p_{h}(x,y)}{h}\,{\rm
d}x{\rm d}y.$

From Lemma C.4 in Appendix C, $\mathcal{J}_{p}^{(++)}+\mathcal{F}_{p}\geq 0$,
hence we can pass to the limit as $h\to 0^{+}$ and by Lemma 2.6 we obtain (5).
By substituting $-u$ in place of $u$, we obtain (5), too; see (2.10) and
(5.5).

Further, we claim that for all $u,v\in\mathcal{D}_{p}(L)$,

$\displaystyle l_{+}$ $\displaystyle:=$ $\displaystyle-\langle
Lu,(v_{+})^{p-1}\rangle-\langle Lv,(p-1)u(v_{+})^{p-2}\rangle$ $\displaystyle\
=$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}^{(+)}(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y$

and

$\displaystyle l_{-}$ $\displaystyle:=$ $\displaystyle-\langle
Lu,(v_{-})^{p-1}\rangle+\langle Lv,(p-1)u(v_{-})^{p-2}\rangle$ $\displaystyle\
=$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}^{(-)}(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

Indeed, using (5), (5), and (5.4), we get

$\displaystyle l_{+}=l_{++}-l_{-+}$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(++)}+\mathcal{F}_{p})(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y$
$\displaystyle-\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}(\mathcal{J}_{p}^{(-+)}+\mathcal{F}_{p})(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}^{(+)}(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

Note that the integral on the right-hand side is well-defined as a difference
of finite integrals with nonnegative integrands. This yields (5). Equality (5)
follows from (5) by substituting $-v$ in place of $v$; see (2.10) and (4.9).

To conclude, using (5), (5), and (4.8), we obtain

$\displaystyle-\langle$ $\displaystyle Lu,v^{\langle
p-1\rangle}\rangle-\langle Lv,(p-1)u|v|^{p-2}\rangle=l_{+}-l_{-}=$ (5.11)
$\displaystyle=\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(\Phi(x),\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y=p\mathcal{E}_{p}(u,v).$

Again, the integral defining $\mathcal{E}_{p}(u,v)$ is absolutely convergent
as a difference of two absolutely convergent integrals. The proof is complete.
∎

######  Remark 5.2.

By the above and Lemma B.5,

$\displaystyle p\mathcal{E}_{p}(f,g)$ $\displaystyle:=$
$\displaystyle\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(f(x),g(x);f(y),g(y))\nu(x,y)\,{\rm
d}x{\rm d}y$ $\displaystyle\ =$ $\displaystyle\langle-Lf,g^{\langle
p-1\rangle}\rangle+\langle-Lg,(p-1)f|g|^{p-2}\rangle$ $\displaystyle\ =$
$\displaystyle-\frac{\rm d}{{\rm
d}t}\int_{\mathbb{R}^{d}}P_{t}f(x)(P_{t}g(x))^{\langle p-1\rangle}\,{\rm
d}x\Big{|}_{t=0},$

at least for $f,g\in\mathcal{D}_{p}(L)$ and $p\geq 2$. At this moment, Lemma
B.5 offers a simplifying perspective on (1.5) and Theorem 4.1, but we should
emphasize the importance of absolute integrability asserted in Theorem 4.1 for
arbitrary $f,g\in L^{p}({\mathbb{R}^{d}})$ when $p\geq 2$; see also
Proposition 4.4.

##  Appendix A Estimates for Bregman divergence

The following lemma extends Lemma 5 of [8], where scalar versions of (A.1),
(A.3), (A.4) were given. The inequality (A.2) seems new.

######  Lemma A.1.

There are constants
$C_{\kappa},C_{\kappa}^{\prime},C_{\kappa}^{\prime\prime},C_{\kappa}^{\prime\prime\prime}\in(0,\infty)$
such that for all $w,z\in{\mathbb{R}^{n}}$,

(A.1) $\displaystyle 0\leq\mathcal{F}_{\kappa}(w,z)$ $\displaystyle\leq
C_{\kappa}|z-w|^{\lambda}({|w|\vee|z|})^{\kappa-\lambda},\quad\lambda\in[0,2],\kappa>1,$
(A.2) $\displaystyle|\mathcal{F}_{\langle\kappa\rangle}(w,z)|$
$\displaystyle\leq
C_{\kappa}^{\prime}|z-w|^{\lambda}({|w|\vee|z|})^{\kappa-\lambda},\quad\lambda\in[0,2],\kappa>1,$
(A.3) $\displaystyle||z|^{\kappa}-|w|^{\kappa}|$ $\displaystyle\leq
C_{\kappa}^{\prime\prime}|z-w|^{\lambda}({|w|\vee|z|})^{\kappa-\lambda},\quad\lambda\in[0,1],\kappa>0,$
(A.4) $\displaystyle|z^{\langle\kappa\rangle}-w^{\langle\kappa\rangle}|$
$\displaystyle\leq
C_{\kappa}^{\prime\prime\prime}|z-w|^{\lambda}({|w|\vee|z|})^{\kappa-\lambda},\quad\lambda\in[0,1],\kappa>0.$

###### Proof.

It suffices to prove the inequalities for the maximal value of $\lambda$
(equal to 2 in (A.1), (A.2), and equal to 1 in (A.3), (A.4)). Indeed, for
other values of $\lambda$, it is enough to use the inequality
$|z-w|\leq|z-w|^{\mu}(|z|+|w|)^{1-\mu}$, $\mu\in(0,1),$
$w,z\in\mathbb{R}^{n}$.

Inequality (A.1) follows from (2.12). In particular, for $a,b\in\mathbb{R}$,
$\lambda=2$, we have

(A.5) $0\leq F_{\kappa}(a,b)=|b|^{\kappa}-|a|^{\kappa}-\kappa
a^{\langle\kappa-1\rangle}(b-a)\leq
C_{\kappa}|b-a|^{2}(|b|\vee|a|)^{\kappa-2}.$

To get the other inequalities, observe that they are obvious for $w=0$. For
$w\neq 0$, we divide by $|w|^{\kappa}$ and, denoting
$t:=|z|/|w|\in[0,\infty)$, $v_{1}:=w/|w|\in\mathbb{S}^{n-1}$,
$v_{2}:=z/|z|\in\mathbb{S}^{n-1}$, we arrive at the following equivalent
statements of (A.2), (A.3), (A.4):

(A.6) $\displaystyle|t^{\kappa}v_{2}-v_{1}-\left((\kappa-1)v_{1}\otimes
v_{1}+\mbox{Id}\right)(tv_{2}-v_{1})|$ $\displaystyle\leq
C_{\kappa}^{\prime}|tv_{2}-v_{1}|^{2}(1\vee t)^{\kappa-2},$ (A.7)
$\displaystyle{|t^{\kappa}-1|}$ $\displaystyle\leq
C_{\kappa}^{\prime\prime}{|tv_{2}-v_{1}|}(1\vee t)^{\kappa-1},$ (A.8)
$\displaystyle{|t^{\kappa}v_{2}-v_{1}|}$ $\displaystyle\leq
C_{\kappa}^{\prime\prime\prime}{|tv_{2}-v_{1}|}(1\vee t)^{\kappa-1}.$

We have

(A.9) ${|tv_{2}-v_{1}|}^{2}(1\vee
t)^{\kappa-2}=\left((1-t)^{2}+2t(1-(v_{1},v_{2}))\right)(1\vee t)^{\kappa-2}.$

If we square the right-hand sides of (A.7) and (A.8) then, up to a constant,
we get

(A.10) ${|tv_{2}-v_{1}|}^{2}(1\vee
t)^{2\kappa-2}=\left((1-t)^{2}+2t(1-(v_{1},v_{2}))\right)(1\vee
t)^{2\kappa-2}.$

Denote $\beta=1-(v_{1},v_{2})\in[0,2]$, so that (A.6) becomes

(A.11) $|(t^{\kappa}-t)v_{2}+(\kappa-1)\left((1-t)+\beta t\right)v_{1}|\leq
C_{\kappa}^{\prime}\left((1-t)^{2}+2\beta t\right)({1\vee t})^{\kappa-2}.$

This inequality is evident when $t$ is away from 1, say, $t\in[0,\frac{1}{2}]$
or $t\in[2,\infty)$. Indeed, for $t\leq 1/2$, we estimate the left-hand side
by ${2}\kappa$, while the function on the right-hand side is not smaller than
$\left(\frac{1}{2}\right)^{2}$, and (A.6) follows. When $t\geq 2$, then the
left-hand side is not greater than ${(2\kappa-1)}t^{\kappa}$, and for the
right-hand side, we get

$\left((1-t)^{2}+2\beta t\right)({1\vee
t})^{\kappa-2}\geq\left(\frac{t}{2}\right)^{2}t^{\kappa-2}.$

To deal with the remaining range $t\in(1/2,2)$, we square both sides of
(A.11). The left-hand side yields

(A.12)
$\displaystyle|(t^{\kappa}-t)v_{2}+(\kappa-1)((1-t)+t(1-{(}v_{1},v_{2}{)}))v_{1}|^{2}$
$\displaystyle=$ $\displaystyle|F_{\kappa}(1,t)+(\kappa-1)\beta
t|^{2}-2(t^{\kappa}-t)(\kappa-1)((1-t)+t\beta))\beta.$

In view of (A.5), the first term on the right-hand side of (A.12) is bounded
above by $(C_{\kappa}(1-t)^{2}+(\kappa-1)t\beta)^{2}$. Since the right-hand
side of (A.6) is then not smaller than a constant multiple of
$((1-t)^{2}+2\beta t)$, we get the estimate of this part.

For the other term, we use the estimate $\left|1-t^{\kappa-1}\right|\leq
C(\kappa)\left|1-t\right|$, $t\in(1/2,2]$, so

$2(t-t^{\kappa})(\kappa-1)((1-t)+t\beta)\beta\leq
C[(t-1)^{2}t\beta+4t^{2}\beta^{2}]\leq C((1-t)^{2}+2\beta t)^{2}.$

The estimate (A.6) follows.

After squaring its sides, the proof of (A.8) is reduced to verifying

$({t^{\kappa}-1})^{2}+2\beta t^{\kappa}\leq C\left((1-t)^{2}+2\beta
t\right)(1\vee t)^{2\kappa-2},$

with a constant $C$, uniformly in $\beta\in[0,2]$. This is done like before.
For $t\geq 1$,

$({t^{\kappa}-1})^{2}\leq C(1-t)^{2}t^{2\kappa-2}.$

For $0\leq t\leq 1/2$, the left-hand side is bounded and the right-hand side
is bounded away from zero (uniformly in $\beta\in[0,2])$, while for
$t\in(1/2,1]$ we use the inequality $t^{\kappa}\leq C(\kappa)t$,
$t\in(1/2,1)$. The square of the left-hand side of (A.7) is smaller than
$(1-t)^{2}+2\beta t^{\kappa}$, i.e., the square of the left-hand side of
(A.8). The proof is complete. ∎

##  Appendix B Calculus in $L^{p}$

Let $p\in[1,\infty)$ be fixed. In the discussion of the multivariate
Hardy–Stein identity above we use the differential calculus in the Banach
space

$L^{p}(\mathbb{R}^{d};\mathbb{R}^{n}):=\left\\{\Upsilon\\!:\mathbb{R}^{d}\to\mathbb{R}^{n}\mbox{
measurable, }\int_{\mathbb{R}^{d}}|\Upsilon(x)|^{p}\,{\rm
d}x<\infty\right\\},\quad n=1,2,\ldots,$

with the norm
$\|\Upsilon\|_{L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})}:=\left(\int_{\mathbb{R}^{d}}|\Upsilon(x)|^{p}\,{\rm
d}x\right)^{1/p}$, or

$\|\Upsilon\|_{\ell_{2}^{n}(L^{p}(\mathbb{R}^{d}))}:=\left(\sum_{i=1}^{n}\|\upsilon_{i}\|_{L^{p}({\mathbb{R}^{d}})}^{2}\right)^{1/2},$

where $\Upsilon=(\upsilon_{1},\ldots,\upsilon_{n})$,
$\upsilon_{1},\ldots,\upsilon_{n}\in L^{p}({\mathbb{R}^{d}})$. The norms are
comparable:

$\displaystyle\left(\int_{\mathbb{R}^{d}}|\Upsilon(x)|^{p}\,{\rm
d}x\right)^{\frac{1}{p}}$ $\displaystyle=$
$\displaystyle\left(\int_{\mathbb{R}^{d}}\left(\sum_{i=1}^{n}|\upsilon_{i}(x)|^{2}\right)^{\frac{p}{2}}\,{\rm
d}x\right)^{\frac{1}{p}}\leq\left(\int_{\mathbb{R}^{d}}\left(\sum_{i=1}^{n}|\upsilon_{i}(x)|\right)^{p}\,{\rm
d}x\right)^{\frac{1}{p}}$ $\displaystyle=$
$\displaystyle\||\upsilon_{1}|+\ldots+|\upsilon_{n}|\|_{L^{p}({\mathbb{R}^{d}})}\leq\sum_{i=1}^{n}\|\upsilon_{i}\|_{L^{p}({\mathbb{R}^{d}})}\leq\sqrt{n}\|\Upsilon\|_{\ell_{2}^{n}(L^{p}(\mathbb{R}^{d}))}.$

Let $\Upsilon\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$ and $\Psi\in
L^{q}({\mathbb{R}^{d}};\mathbb{R}^{n})$, where $p,q\in(1,\infty)$ with
$\frac{1}{p}+\frac{1}{q}=1$. We consider the canonical pairing

(B.1)
$\displaystyle\langle\Upsilon,\Psi\rangle:=\int_{\mathbb{R}^{d}}\Upsilon(x)\cdot\Psi(x)\,{\rm
d}x=\sum_{j=1}^{n}\int_{\mathbb{R}^{d}}\upsilon_{j}(x)\psi_{j}(x)\,{\rm d}x.$

For a mapping $[0,\infty)\ni t\mapsto\Upsilon(t)\in
L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$, we denote

$\Delta_{h}\Upsilon(t)=\Upsilon(t+h)-\Upsilon(t)\quad\textnormal{provided}\,\,t,t+h\geq
0.$

As usual, $\Upsilon$ is called _continuous_ at $t_{0}\geq 0$ if
$\Delta_{h}\Upsilon(t_{0})\to 0$ in $L^{p}(\mathbb{R}^{d};\mathbb{R}^{n})$ as
$h\to 0$. Furthermore, $\Upsilon$ is called _differentiable_ at $t_{0}\geq 0$
if the limit

(B.2) $\lim_{h\to
0}\frac{1}{h}\Delta_{h}\Upsilon(t_{0})=:\Upsilon^{\prime}(t_{0})$

exists in $L^{p}(\mathbb{R}^{d};\mathbb{R}^{n})$. If $\Upsilon^{\prime}(t)$
defined by (B.2) is continuous at $t=t_{0}$, then we say that $\Upsilon$ is
continuously differentiable at $t_{0}$. In other words,
$\Upsilon^{\prime}(t_{0})$ is the Fréchet derivative of the mapping
$[0,\infty)\ni t\mapsto\Upsilon(t)$ at $t_{0}$; $\Upsilon^{\prime}(0)$ denotes
the right-hand side derivative at $0$. Clearly, if $\Upsilon$ is continuously
differentiable on $[0,\infty)$, then $\Upsilon$ is continuous on $[0,\infty)$.

Of course, $\Upsilon=(\upsilon_{1},\ldots,\upsilon_{n})$ is continuous
(respectively, differentiable, continuously differentiable) in
$L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$ if and only if all the functions
$\upsilon_{i},$ $i=1,\ldots,n$, are continuous (respectively, differentiable,
continuously differentiable) in $L^{p}({\mathbb{R}^{d}})$.

We next present a series of auxiliary lemmas.

######  Lemma B.1.

Let $\kappa\in(0,p]$. Then the following mappings are continuous:

(B.3) $\displaystyle L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})\ni\Upsilon$
$\displaystyle\mapsto$ $\displaystyle\Upsilon^{\langle\kappa\rangle}\in
L^{p/\kappa}({\mathbb{R}^{d}};{\mathbb{R}^{n}}),$ (B.4) $\displaystyle
L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})\ni\Upsilon$ $\displaystyle\mapsto$
$\displaystyle|\Upsilon|^{\kappa}\in L^{p/\kappa}({\mathbb{R}^{d}}).$

###### Proof.

First, observe that $|\Upsilon|^{\kappa}$ and
$\Upsilon^{\langle\kappa\rangle}$ are in
$L^{\frac{p}{\kappa}}({\mathbb{R}^{d}})$ and
$L^{p/\kappa}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$, respectively, if
$\Upsilon\in L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$.

To prove (B.3), choose $\lambda\in(0,1)$ such that $\kappa-\lambda>0$ and
suppose $\Upsilon_{k}\to\Upsilon$ in
$L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$ as $k\to\infty$. From (A.4) we get,
for every $x\in{\mathbb{R}^{d}}$,

$\displaystyle|\Upsilon_{k}(x)^{\langle\kappa\rangle}-\Upsilon(x)^{\langle\kappa\rangle}|$
$\displaystyle\leq$ $\displaystyle
C_{\kappa}^{\prime\prime\prime}|\Upsilon_{k}(x)-\Upsilon(x)|^{\lambda}(|\Upsilon_{k}(x)|\vee|\Upsilon(x)|)^{\kappa-\lambda}.$

Using Hölder’s inequality with exponents $\kappa/\lambda$ and
$\kappa/(\kappa-\lambda)$, we get

$\displaystyle\|\Upsilon_{k}^{\langle\kappa\rangle}-\Upsilon^{\langle\kappa\rangle}\|_{L^{p/\kappa}({\mathbb{R}^{d}};\mathbb{R}^{n})}^{\kappa/p}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}|\Upsilon_{k}(x)^{\langle\kappa\rangle}-\Upsilon(x)^{\langle\kappa\rangle}|^{\frac{p}{\kappa}}\,{\rm
d}x$ $\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{d}}\left(C_{\kappa}^{\prime\prime\prime}\right)^{p/\kappa}|\Upsilon_{k}(x)-\Upsilon(x)|^{\frac{\lambda
p}{\kappa}}(|\Upsilon_{k}(x)|\vee|\Upsilon(x)|)^{\frac{(\kappa-\lambda)p}{\kappa}}\,{\rm
d}x$ $\displaystyle\leq$ $\displaystyle
C\|\Upsilon_{k}-\Upsilon\|_{L^{p}(\mathbb{R}^{d};\mathbb{R}^{n})}^{p\lambda/\kappa}\cdot\left(\|\Upsilon_{k}\|_{L^{p}({\mathbb{R}^{d}})}^{p(\kappa-\lambda)/\kappa}+\|\Upsilon\|_{L^{p}({\mathbb{R}^{d}})}^{p(\kappa-\lambda)/\kappa}\right).$

The result follows. The proof of (B.4) is similar. ∎

The following generalization of [8, Lemma 13] follows from Hölder’s
inequality.

######  Lemma B.2.

Let $q\in(1,\infty)$, $r\in\big{[}\frac{q}{q-1},\infty\big{)}$, $\Upsilon\in
L^{q}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$, $\Psi\in
L^{r}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$. Then

$\|\Upsilon\cdot\Psi\|_{L^{\frac{qr}{q+r}}({\mathbb{R}^{d}};{\mathbb{R}^{n}})}\leq\|\Upsilon\|_{L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})}\|\Psi\|_{L^{r}({\mathbb{R}^{d}};{\mathbb{R}^{n}})}.$

Moreover, if $\Upsilon_{n}\to\Upsilon$ in
$L^{q}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$ and $\Psi_{n}\to\Psi$ in
$L^{r}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$, then
$\Upsilon_{n}\cdot\Psi_{n}\to\Upsilon\cdot\Psi$ in
$L^{\frac{qr}{q+r}}({\mathbb{R}^{d}};\mathbb{R}^{n})$, as $n\to\infty$.

The next lemma is an extension of [8, Lemma 15], where the result for
$|\Upsilon|^{\kappa}$ was proved for $n=1$, $\kappa=p$.

######  Lemma B.3.

Let $1<\kappa\leq p<\infty$ be given. If $[0,\infty)\ni t\mapsto\Upsilon(t)$
is continuously differentiable in $L^{p}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$,
then:

  1. (i)

$|\Upsilon|^{\kappa}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}})$ and

(B.5)
$(|\Upsilon|^{\kappa})^{\prime}=\kappa\Upsilon^{\langle\kappa-1\rangle}\cdot\Upsilon^{\prime},$

  2. (ii)

$\Upsilon^{\langle\kappa\rangle}$ is continuously differentiable in
$L^{p/\kappa}({\mathbb{R}^{d}};{\mathbb{R}^{n}})$ and

(B.6)
$\left(\Upsilon^{\langle\kappa\rangle}\right)^{\prime}=\left(J_{\langle\kappa\rangle}\circ\Upsilon\right)\Upsilon^{\prime},$

with $J_{\langle\kappa\rangle}$ defined in (2.6).

###### Proof.

Both statements are proved similarly, therefore we only prove (B.6), as it is
the more complicated of the two.

Observe that for every $a\in\mathbb{S}^{n-1}\subset\mathbb{R}^{n}$ and
$A:=a\otimes a$, the linear mapping $F\mapsto AF$ is a contraction on
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$. Indeed,

$\displaystyle\|AF\|_{L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})}$
$\displaystyle=$ $\displaystyle\left(\int\left|(a,F)\cdot a\right|^{p}\,{\rm
d}x\right)^{\frac{1}{p}}$ $\displaystyle=$
$\displaystyle\left(\int_{\mathbb{R}^{d}}|(a,F)|^{p}\,{\rm
d}x\right)^{\frac{1}{p}}\leq\left(\int_{\mathbb{R}^{d}}\left(|a|\cdot|F|\right)^{p}\,{\rm
d}x\right)^{\frac{1}{p}}=\left(\int_{\mathbb{R}^{d}}|F|^{p}\,{\rm
d}x\right)^{\frac{1}{p}}.$

So, for every $F\in L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$, by Hölder’s
inequality with exponents $\kappa$ and $\kappa/(\kappa-1)$,

(B.7)
$\displaystyle\|(J_{\langle\kappa\rangle}\circ\Upsilon)F\|_{L^{p/\kappa}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}$
$\displaystyle=$
$\displaystyle\left\||\Upsilon|^{\kappa-1}\left((\kappa-1)(\frac{\Upsilon}{|\Upsilon|}\otimes\frac{\Upsilon}{|\Upsilon|})+\mbox{Id}\right)F\right\|_{L^{p/\kappa}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}$
$\displaystyle\leq$
$\displaystyle\|\Upsilon\|_{L^{p}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}^{\kappa-1}\cdot\left\|\left((\kappa-1)(\frac{\Upsilon}{|\Upsilon|}\otimes\frac{\Upsilon}{|\Upsilon|})+\mbox{Id}\right)F\right\|_{L^{p}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}$
$\displaystyle\leq$
$\displaystyle\kappa\|\Upsilon\|_{L^{p}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}^{\kappa-1}\cdot\|F\|_{L^{p}{({\mathbb{R}^{d}};\mathbb{R}^{n})}}.$

For $t\geq 0$, we have the convergence
$\frac{1}{h}\Delta_{h}\Upsilon(t)\to\Upsilon^{\prime}(t)$ as $h\to 0$, in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$, with $\Upsilon^{\prime}$ being
continuous. This and (B.7) yield
$\frac{1}{h}(J_{\langle\kappa\rangle}\circ\Upsilon(t))\Delta_{h}(\Upsilon(t))\to(J_{\langle\kappa\rangle}\circ\Upsilon(t))\Upsilon^{\prime}(t)$
in $L^{p/\kappa}$ (with the limit continuous). Therefore we only need to
verify that for $h\to 0$,

$\displaystyle
W_{h}(t):=\frac{1}{h}\Delta_{h}\Upsilon^{\langle\kappa\rangle}(t)-(J_{\langle\kappa\rangle}\circ\Upsilon(t))\frac{1}{h}\Delta_{h}u(t)\to
0\qquad\mbox{in}\ L^{p/\kappa}({\mathbb{R}^{d}};\mathbb{R}^{n}).$

Since
$W_{h}(t)=\frac{1}{h}\mathcal{F}_{\langle\kappa\rangle}(\Upsilon(t),\Upsilon(t+h))$,
we choose $\lambda\in(1,2]$ such that $\kappa-\lambda>0$, then use the
inequality (A.2) to get:

$\displaystyle|W_{h}(t)|$ $\displaystyle\leq$
$\displaystyle\frac{1}{|h|}C_{\kappa}^{\prime}|\Upsilon(t+h)-\Upsilon(t)|^{\lambda}(|\Upsilon(t+h)|\vee|\Upsilon(t)|)^{\kappa-\lambda}$
$\displaystyle=$
$\displaystyle|h|^{\lambda-1}C_{\kappa}^{\prime}\left|\frac{1}{h}\Delta_{h}\Upsilon(t)\right|^{\lambda}(|\Upsilon(t+h)|\vee|\Upsilon(t)|)^{\kappa-\lambda}.$

Furthermore, by Hölder’s inequality with parameters $\kappa/\lambda$ and
$\kappa/(\kappa-\lambda)$,

(B.8)
$\displaystyle\|W_{h}(t)\|_{{L^{p/\kappa}({\mathbb{R}^{d}};\mathbb{R}^{n})}}$
$\displaystyle=$ $\displaystyle
C_{\kappa}|h|^{\lambda-1}\left\|\left|\frac{1}{h}\Delta_{h}\Upsilon(t)\right|^{\lambda}(|\Upsilon(t+h)|\vee|\Upsilon(t)|)^{\kappa-\lambda}\right\|_{L^{p/\kappa}({\mathbb{R}^{d}})}$
$\displaystyle\leq$ $\displaystyle
C|h|^{\lambda-1}\left\|\frac{1}{h}\Delta_{h}\Upsilon(t)\right\|_{L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})}^{\lambda}\cdot\left\||\Upsilon(t+h)|+|\Upsilon(t)|\right\|_{L^{p}({\mathbb{R}^{d}})}^{\kappa-\lambda}.$

We then conclude as in the proof of Lemma B.1. ∎

Finally we invoke, without proof, an analogue of the Leibniz rule.

######  Lemma B.4 (Product rule).

Let $p>1$ and $r\in\left[\frac{p}{p-1},\infty\right)$ be given. If the
mappings $[0,\infty)\ni t\mapsto\Upsilon(t)\in
L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$ and $[0,\infty)\ni t\mapsto\Psi(t)\in
L^{r}({\mathbb{R}^{d}};\mathbb{R}^{n})$ are continuously differentiable in
$L^{p}({\mathbb{R}^{d}};\mathbb{R}^{n})$ and
$L^{r}({\mathbb{R}^{d}};\mathbb{R}^{n})$, respectively, then
$\Upsilon\cdot\Psi$ is continuously differentiable in
$L^{\frac{pr}{p+r}}({\mathbb{R}^{d}})$ and
$(\Upsilon\cdot\Psi)^{\prime}=\Upsilon^{\prime}\cdot\Psi+\Upsilon\cdot\Psi^{\prime}$.

######  Lemma B.5.

Let $p>1$. If $f,g\in\mathcal{D}_{p}(L)$, then for $t\in[0,\infty)$,

$\frac{\rm d}{{\rm d}t}\int_{\mathbb{R}^{d}}P_{t}f(P_{t}g)^{\langle
p-1\rangle}{\rm d}x=\int_{\mathbb{R}^{d}}\left((P_{t}g)^{\langle
p-1\rangle}LP_{t}f+(p-1)P_{t}f|P_{t}g|^{p-2}LP_{t}g\right){\rm d}x.$

If $f,g\in L^{p}({\mathbb{R}^{d}})$, then the formula holds for
$t\in(0,\infty)$.

###### Proof.

Of course, $P_{t}f$ and $P_{t}g$ are continuously differentiable at $t\geq 0$
in $L^{p}({\mathbb{R}^{d}})$ and $\frac{\rm d}{{\rm d}t}P_{t}f=LP_{t}f$,
$\frac{\rm d}{{\rm d}t}P_{t}g=LP_{t}g$. Hence, by Lemma B.3 with $n=1$,
$(P_{t}g)^{\langle p-1\rangle}$ is continuously differentiable at $t\geq 0$ in
$L^{\frac{p}{p-1}}({\mathbb{R}^{d}})$ and $\frac{\rm d}{{\rm
d}t}(P_{t}g)^{\langle p-1\rangle}=(p-1)|P_{t}g|^{p-2}LP_{t}g$. By Lemma B.4
with $r=p/(p-1)$, $P_{t}f(P_{t}g)^{\langle p-1\rangle}$ is continuously
differentiable at $t\geq 0$ in $L^{1}({\mathbb{R}^{d}})$ and

(B.9) $\frac{\rm d}{{\rm d}t}\left(P_{t}f(P_{t}g)^{\langle
p-1\rangle}\right)=(P_{t}g)^{\langle
p-1\rangle}LP_{t}f+(p-1)P_{t}f|P_{t}g|^{p-2}LP_{t}g.$

Since $u\mapsto\int_{\mathbb{R}^{d}}u(x)\,{\rm d}x$ is a continuous linear
functional on $L^{1}({\mathbb{R}^{d}})$, we get the result (the case of
arbitrary $f,g\in L^{p}({\mathbb{R}^{d}})$ follows since the semigroup $P_{t}$
is analytic). ∎

##  Appendix C Convexity properties

We provide here precise statements and proofs of convexity properties needed
in Sections 4 and 5. First, we recall some facts from the theory of convex
functions.

Let $T\\!:A\to\mathbb{R}$, where the set $A\subset\mathbb{R}^{n}$ is convex.
By definition, $d(w)\in\mathbb{R}^{n}$ is a _subgradient_ of $T$ at $w\in A$
if

(C.1) $\displaystyle T(z)\geq T(w)+d(w)\cdot(z-w)\quad\mbox{for all }z\in A.$

The function $T$ is convex in $A$ if and only if for every $w\in A$, a
subgradient $d(w)$ exists. If $T$ is convex and the first-order partial
derivatives of $T$ exist at some $w\in A$, then $T$ has exactly one
subgradient at the point $w$, which is equal to its gradient $\nabla T(w)$. In
such a case, denoting by $\frac{\partial T}{\partial v}$ the directional
derivative of $T$ along a given vector $v\in\mathbb{R}^{n}$, we have that
$d(w)$ is a subgradient of the function $T$ at the point $w\in A$ if and only
if

$\displaystyle\frac{\partial T}{\partial v}(w)\geq d(w)\cdot v,\quad
v\in\mathbb{R}^{n}.$

For more details see Borwein and Lewis [13, Chapter 3].

We need the following lemma.

######  Lemma C.1.

Let $p\geq 2$. The function

$\displaystyle Y(z):=z_{1}\left(z_{2}\right)^{p-1}+|z|^{p},\quad
z=(z_{1},z_{2})\in[0,\infty)^{2},$

is convex on $[0,\infty)^{2}$.

###### Proof.

As $Y$ is continuous on $[0,\infty)^{2}$, it is enough to prove the convexity
on $(0,\infty)^{2}$. Recall (2.5) and

$\displaystyle\nabla^{2}|z|^{p}=p(p-2)|z|^{p-4}\begin{bmatrix}z_{1}^{2}&z_{1}z_{2}\\\
z_{1}z_{2}&z_{2}^{2}\end{bmatrix}+p|z|^{p-2}\mbox{Id},\quad
z\in\mathbb{R}^{2}\setminus\\{0\\}.$

The Hessian $\nabla^{2}\left(z_{1}z_{2}^{p-1}\right)$ is calculated in (4.7).
The Hessian $\nabla^{2}Y(z)$ of $Y$ is:

$\displaystyle\begin{bmatrix}{p|z|^{p-2}+p(p-2)z_{1}^{2}|z|^{p-4}}&(p-1)z_{2}^{p-2}+p(p-2)z_{1}z_{2}|z|^{p-4}\\\
(p-1)z_{2}^{p-2}+p(p-2)z_{1}z_{2}|z|^{p-4}&(p-1)(p-2)z_{1}z_{2}^{p-3}+{p|z|^{p-2}+p(p-2)z_{2}^{2}|z|^{p-4}}\end{bmatrix}$

We will verify that for $z\in(0,\infty)^{2}$, the matrix is positive semi-
definite. Clearly,

$\left[p|z|^{p-2}+p(p-2)z_{1}^{2}|z|^{p-4}\right]>0.$

Moreover, after long, but elementary, calculations we get:

$\displaystyle\det\nabla^{2}Y(z)$ $\displaystyle=$
$\displaystyle\left[p|z|^{p-2}+p(p-2)z_{1}^{2}|z|^{p-4}\right](p-1)(p-2)z_{1}z_{2}^{p-3}$
$\displaystyle+p^{2}|z|^{2p-4}+p^{2}(p-2)|z|^{2p-4}$
$\displaystyle-(p-1)^{2}z_{2}^{2p-4}-p(p-1)(p-2)z_{1}z_{2}^{p-1}|z|^{p-4}$
$\displaystyle=$ $\displaystyle p^{2}(p-1)|z|^{2p-4}-(p-1)^{2}z_{2}^{2p-4}$
$\displaystyle+p(p-1)(p-2)|z|^{p-4}\left((p-1)z_{1}^{3}z_{2}^{p-3}-z_{1}z_{2}^{p-1}\right).$

We have $z_{2}\leq|z|$, so applying Young’s inequality with exponents $p$ and
$q=p/(p-1)$ to the product $z_{1}z_{2}^{p-1}$ we obtain

$\displaystyle z_{1}z_{2}^{p-1}$
$\displaystyle\leq\frac{z_{1}^{p}}{p}+\frac{(p-1)z_{2}^{p}}{p}=\frac{1}{p}(z_{1}^{2})^{\frac{p}{2}}+\frac{p-1}{p}(z_{2}^{2})^{\frac{p}{2}}\leq|z|^{p}.$

Summarizing,

$\displaystyle\det\nabla^{2}Y(z)$ $\displaystyle\geq$ $\displaystyle
p(p-1)^{2}(p-2)|z|^{p-4}z_{1}^{3}z_{2}^{p-3}$
$\displaystyle+|z|^{2p-4}(p-1)\left(p^{2}-p(p-2)-(p-1)\right)$
$\displaystyle=$ $\displaystyle
p(p-1)^{2}(p-2)|z|^{p-4}z_{1}^{3}z_{2}^{p-3}+|z|^{2p-4}(p-1)(p+1)>0.$

∎

If $w_{1}\leq z_{1},w_{2}\leq z_{2},\ldots,w_{k}\leq z_{k}$ implies
$T(w_{1},\ldots,w_{k})\leq T(z_{1},\ldots,z_{k})$ in the domain of a real-
valued function $T$, then we say $T$ is _coordinate-wise nondecreasing_. The
following fact is self-explanatory, see also Boyd and Vandenberghe [14,
Section 3.2.4].

######  Lemma C.2.

Let $S\\!:A\to\mathbb{R}^{k}$, $S(A)\subset B$, and $T\\!:B\to\mathbb{R}$,
where $A\subset\mathbb{R}^{n}$ and $B\subset\mathbb{R}^{k}$ are convex. If
each coordinate of $S$ is convex and $T$ is coordinate-wise nondecreasing and
convex, then the composition $T\circ S\\!:A\to\mathbb{R}$ is convex.

The following two lemmas are critical for our development.

######  Lemma C.3.

Let $p\geq 2$ and define, for $z=(z_{1},z_{2})\in\mathbb{R}^{2}$,

$Y^{(+)}(z):=z_{1}\left((z_{2})_{+}\right)^{p-1}+|z|^{p},\qquad
Y^{(-)}(z):=z_{1}\left((z_{2})_{-}\right)^{p-1}+|z|^{p}.$

The functions are convex on $[0,\infty)\times\mathbb{R}$.

###### Proof.

Define $T\\!:[0,\infty)\times\mathbb{R}\to[0,\infty)^{2}$ as

$\displaystyle T(z):=(z_{1},(z_{2})_{+}),\quad
z=(z_{1},z_{2})\in[0,\infty)\times\mathbb{R},$

and let $Y\\!:[0,\infty)^{2}\to\mathbb{R}$ be as in Lemma C.1. Since each
coordinate of $T$ is convex and the function $Y$ is convex and coordinate-wise
nondecreasing, the composition

$\displaystyle(Y\circ
T)(z)=z_{1}((z_{2})_{+})^{p-1}+\left((z_{1})^{2}+((z_{2})_{+})^{2}\right)^{p/2}$

is convex on $[0,\infty)\times\mathbb{R}$ (from Lemma C.2). Therefore,

$\displaystyle Y^{(+)}(z)=\max\\{(Y\circ T)(z),|z|^{p}\\}$

is convex on $[0,\infty)\times\mathbb{R}$ as the maximum of convex functions
[14, Section 3.2.3].

To prove the convexity of $Y^{(-)}$ we just notice that
$Y^{(-)}(z_{1},z_{2})=Y^{(+)}(z_{1},-z_{2})$. ∎

######  Lemma C.4.

If $p>2$ then for all $z,w\in\mathbb{R}^{2}$,

$\displaystyle\mathcal{J}_{p}^{(++)}(w,z)+\mathcal{F}_{p}(w,z)\geq
0,\quad\mathcal{J}_{p}^{(-+)}(w,z)+\mathcal{F}_{p}(w,z)\geq 0,$

where $\mathcal{J}_{p}^{(++)}(w,z)$, $\mathcal{J}_{p}^{(-+)}(w,z)$ are given
by (5.2) and (5.3).

###### Proof.

Because of (2.10) and (5.5), we only need to show that
$\mathcal{J}_{p}^{(++)}(w,z)+\mathcal{F}_{p}(w,z)\geq 0$. We rewrite this
inequality as

(C.2) $\displaystyle Y^{(++)}(z)\geq Y^{(++)}(w)+d(w)\cdot(z-w),$

where $Y^{(++)}(z):=(z_{1})_{+}\left((z_{2})_{+}\right)^{p-1}+|z|^{p}$ and

(C.3) $\displaystyle
d(w):=\left(\mathbf{1}(w_{1})\left((w_{2})_{+}\right)^{p-1},(p-1)(w_{1})_{+}\left((w_{2})_{+}\right)^{p-2}\right)+pw^{\langle
p-1\rangle}.$

Therefore the proof of (C.2) amounts to checking that $d(w)$ is a subgradient
of the function $Y^{(++)}$ at the point $w\in\mathbb{R}^{2}$.

To show (C.2), we first establish the convexity of $Y^{(++)}$. Define
$T\\!:\mathbb{R}^{2}\to[0,\infty)^{2}$ as

$\displaystyle T(z):=((z_{1})_{+},(z_{2})_{+}),\quad
z=(z_{1},z_{2})\in\mathbb{R}^{2}.$

Let $Y\\!:[0,\infty)^{2}\to\mathbb{R}$ as in Lemma C.1. Since each coordinate
of $T$ is convex and the function $Y$ is convex and coordinate-wise
nondecreasing, the convexity on $\mathbb{R}^{2}$ of the composition

$\displaystyle(Y\circ
T)(z)=(z_{1})_{+}((z_{2})_{+})^{p-1}+\left(((z_{1})_{+})^{2}+((z_{2})_{+})^{2}\right)^{p/2}$

follows from Lemma C.2. Since

$\displaystyle Y^{(++)}(z)=\max\\{(Y\circ T)(z),|z|^{p}\\},$

it is convex on $\mathbb{R}^{2}$ as a maximum of two convex functions.

If $w=0$ then $Y^{(++)}(w)=0$ and $d(w)=0$, hence (C.2) is true for every $z$.

If $w\neq 0$, to show that $d(w)$ is a subgradient of $Y^{(++)}$ at $w$, we
need to prove that

$\displaystyle\frac{\partial Y^{(++)}}{\partial v}(w)\geq d(w)\cdot v,\quad
w\in\mathbb{R}^{2}\setminus\\{{0}\\},\quad\mbox{for every
}v=(v_{1},v_{2})\in\mathbb{R}^{2}.$

Denote $B:=\\{(w_{1},w_{2})\in\mathbb{R}^{2}\\!:w_{1}=0,w_{2}>0\\}$ as
vertical positive semi-axis. The function $Y^{(++)}$ is differentiable
everywhere but on $B$. Thus when $w\notin B$, the gradient of $Y^{(++)}$
exists, is given by (C.3) and

$\frac{\partial Y^{(++)}}{\partial v}(w)=\nabla Y^{(++)}(w)\cdot v=d(w)\cdot
v.$

In the remaining case $w\in B$, we have two possibilities. If $v_{1}\geq 0$,
then

$\displaystyle\frac{\partial Y^{(++)}}{\partial v}(w)$ $\displaystyle=$
$\displaystyle\left((w_{2})_{+}\right)^{p-1}v_{1}+pw^{\langle p-1\rangle}\cdot
v$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\left((w_{2})_{+}\right)^{p-1}v_{1}+pw^{\langle
p-1\rangle}\cdot v=d(w)\cdot v.$

Otherwise, when $v_{1}<0$, then

$\displaystyle\frac{\partial Y^{(++)}}{\partial v}(w)=pw^{\langle
p-1\rangle}\cdot
v\geq\frac{1}{2}\left((w_{2})_{+}\right)^{p-1}v_{1}+pw^{\langle
p-1\rangle}\cdot v=d(w)\cdot v.$

The proof is complete. ∎

##  Appendix D Alternative proof of polarization for $p\geq 3$

The main difficulty in the proof of Theorem 4.1 above is to justify the
limiting procedure in the absence of nonnegativity in the integrands. For
$p\geq 3$, we can proceed differently: the absolute value of the function
$\mathcal{J}_{p}$ is dominated by the function $\mathcal{G}_{p}$, which helps
with the integrability issues in the proof of the polarized Hardy–Stein
formula.

######  Lemma D.1.

For every $p\geq 3$, there is a constant $c_{p}>0$ such that

(D.1) $\displaystyle|\mathcal{J}_{p}(w,z)|\leq
c_{p}\mathcal{G}_{p}(w,z)\asymp\mathcal{H}_{p}(w,z),\quad
w,z\in\mathbb{R}^{2}.$

###### Proof.

The formula (4.1), defining $\mathcal{J}_{p}$, can be rewritten as

$\displaystyle\mathcal{J}_{p}(w,z)$ $\displaystyle=$ $\displaystyle
z_{1}z_{2}^{\langle p-1\rangle}-w_{1}w_{2}^{\langle p-1\rangle}-w_{2}^{\langle
p-1\rangle}(z_{1}-w_{1})-(p-1)w_{1}|w_{2}|^{p-2}(z_{2}-w_{2})$
$\displaystyle=$ $\displaystyle(z_{1}-w_{1})(z_{2}^{\langle
p-1\rangle}-w_{2}^{\langle p-1\rangle})+w_{1}[z_{2}^{\langle
p-1\rangle}-w_{2}^{\langle p-1\rangle}-(p-1)|w_{2}|^{p-2}(z_{2}-w_{2})].$

Using (A.3) we can estimate the first summand above in the following manner

$\displaystyle|z_{1}-w_{1}||z_{2}^{\langle p-1\rangle}-w_{2}^{\langle
p-1\rangle}|$ $\displaystyle\leq$ $\displaystyle
C_{p}^{\prime}|w_{1}-z_{1}|\cdot|w_{2}-z_{2}|\cdot\left(|w_{2}|\vee|z_{2}|\right)^{p-2}$
$\displaystyle\leq$
$\displaystyle\frac{C_{p}^{\prime}}{2}\left(|w_{1}-z_{1}|^{2}+|w_{2}-z_{2}|^{2}\right)\left(|w_{2}|\vee|z_{2}|\right)^{p-2}$
$\displaystyle\leq$
$\displaystyle\frac{C_{p}^{\prime}}{2}|w-z|^{2}\left(|w|\vee|z|\right)^{p-2}=\frac{C_{p}^{\prime}}{2}\mathcal{G}_{p}(w,z).$

For the second summand we use (A.2),

$\displaystyle|w_{1}[z_{2}^{\langle p-1\rangle}-w_{2}^{\langle p-1\rangle}-$
$\displaystyle(p-1)|w_{2}|^{p-2}(z_{2}-w_{2})]|\leq
C_{p-1}^{\prime\prime}|w_{1}||z_{2}-w_{2}|^{2}(|w_{2}|\vee|z_{2}|)^{p-3}$
$\displaystyle\leq
C_{p-1}^{\prime\prime}|w-z|^{2}(|w_{1}|\vee|z_{1}|)(|w_{2}|\vee|z_{2}|)^{p-3}$
$\displaystyle\leq
C_{p-1}^{\prime\prime}|w-z|^{2}(|w|\vee|z|)^{p-2}=C_{p-1}^{\prime\prime}\mathcal{G}_{p}(w,z).$

Thus,

$\displaystyle|\mathcal{J}_{p}(w,z)|\leq c_{p}\mathcal{G}_{p}(w,z)$

and $\mathcal{G}_{p}(w,z)\asymp\mathcal{H}_{p}(w,z)$ by (2.13). ∎

######  Remark D.2.

The statement (D.1) stays true also for $p=2$. Indeed, by (4.2),

$\displaystyle|\mathcal{J}_{2}(w,z)|=|(z_{1}-w_{1})(z_{2}-w_{2})|\leq|z-w|^{2}=\mathcal{G}_{2}(w,z).$

On the other hand, it fails in general for $p\in(1,3)\setminus\\{2\\}$.
Indeed, for $k=1,2,\ldots$, let $w^{(k)}:=\left(1,\frac{1}{k}\right)$,
$z^{(k)}:=\left(1,\frac{2}{k}\right)$. Then, by (4.1),

$\displaystyle|\mathcal{J}_{p}(w^{(k)},z^{(k)})|$ $\displaystyle=$
$\displaystyle\left|\frac{2^{p-1}}{k^{p-1}}-\frac{1}{k^{p-1}}-\frac{p-1}{k^{p-1}}\right|=\left|2^{p-1}-p\right|\frac{1}{k^{p-1}},$
$\displaystyle\mathcal{G}_{p}(w^{(k)},z^{(k)})$ $\displaystyle=$
$\displaystyle\frac{1}{k^{2}}\left(1+\frac{4}{k^{2}}\right)^{\frac{p-2}{2}}.$

Our claim is verified by notifying that

$\displaystyle\frac{|\mathcal{J}_{p}(w^{(k)},z^{(k)})|}{\mathcal{G}_{p}(w^{(k)},z^{(k)})}=\frac{k^{3-p}\left|2^{p-1}-p\right|}{\left(1+\frac{4}{k^{2}}\right)^{\frac{p-2}{2}}}\to\infty\quad\mbox{as
}k\to\infty.$

Estimate (D.1) permits to substantially simplify the proof of the polarized
Hardy–Stein identity (4.5). Indeed, for $f,g\in L^{p}({\mathbb{R}^{d}})$,
$u(t)=P_{t}f$, $v(t)=P_{t}g$, $\Phi=(u,v)$, and $p\geq 3$, from Theorem 3.1 we
have that

(D.2)
$\displaystyle\int_{0}^{\infty}\int_{{\mathbb{R}^{d}}}\int_{{\mathbb{R}^{d}}}\mathcal{H}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y{\rm d}t<\infty,$

so in view of Lemma D.1, an analogous integral of
$\left|\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\right|$ is convergent as
well. We next review the proof of Theorem 4.1: for $f,g\in\mathcal{D}_{p}(L)$,
we differentiate $u(t)v(t)^{\langle p-1\rangle}$ in $L^{1}({\mathbb{R}^{d}})$
as in (4.12) and we have:

$\frac{\rm d}{{\rm d}t}\int_{\mathbb{R}^{d}}u(t)v(t)^{\langle
p-1\rangle}\,{\rm d}x=-\lim_{h\to
0^{+}}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\frac{p_{h}(x,y)}{h}\,{\rm
d}x{\rm d}y.$

Since $|\mathcal{J}_{p}|\leq\mathcal{H}_{p}$ and the integral in (D.2) is
convergent, we can pass to the limit when $h\to 0^{+}$ (we use the Dominated
Convergence Theorem, (P1), and (P2)) to obtain

(D.3) $\frac{\rm d}{{\rm d}t}\int_{\mathbb{R}^{d}}u(t)v(t)^{\langle
p-1\rangle}\,{\rm
d}x=-\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\mathcal{J}_{p}(P_{t}\Phi(x),P_{t}\Phi(y))\nu(x,y)\,{\rm
d}x{\rm d}y.$

The rest of the proof remains unchanged: we integrate from 0 to $T$ with $T>0$
fixed, then we pass to the limit $T\to\infty$. Then, we relax the assumption
that $f,g\in\mathcal{D}_{p}(L)$ by using the analyticity of the semigroup.

##  Appendix E Proof of (1.4)

Let $\\{B_{t},t\geq 0\\}$ be the Brownian motion on the Euclidean space
${\mathbb{R}^{d}}$ running at twice the usual speed, and let $(P_{t})_{t\geq
0}$ be its semigroup:

$\displaystyle
P_{t}f(x):={\mathbb{E}_{x}}f(B_{t})=\int_{\mathbb{R}^{d}}f(y)p_{t}(x,y)\,{\rm
d}y=(p_{t}\ast f)(x),\quad t>0,x\in{\mathbb{R}^{d}},$

where

$\displaystyle p_{t}(x)=(4\pi t)^{-d/2}e^{-\frac{|x|^{2}}{4t}},\quad
t>0,x\in{\mathbb{R}^{d}}$

and $p_{t}(x,y):=p_{t}(x-y)$, as before. Let $1<p<\infty$. It is well known
that $(P_{t})_{t\geq 0}$ is a strongly continuous, analytic, Markovian
semigroup of symmetric operators in $L^{p}({\mathbb{R}^{d}})$. In particular,
for every $t>0$ and $f\in L^{p}({\mathbb{R}^{d}})$, $P_{t}f$ belongs to the
domain of the generator of this semigroup. Estimates (2.16) and (2.17) hold
true as well, therefore the key ingredients needed to prove Hardy–Stein
identity remain satisfied for the Brownian motion. Thus, for every $u\in
L^{p}({\mathbb{R}^{d}})$, we define, as before,

$\displaystyle\mathcal{E}_{p}[u]:=\lim_{t\to 0}\mathcal{E}^{(t)}(u,u^{\langle
p-1\rangle})$

and

$\displaystyle\mathcal{D}(\mathcal{E}_{p})$ $\displaystyle=\\{u\in
L^{p}({\mathbb{R}^{d}}):\text{ finite }\lim_{t\to
0}\mathcal{E}^{(t)}(u,u^{\langle p-1\rangle})\text{ exists}\\}.$

Similarly as in the proof of Theorem 3.1, we obtain

(E.1) $\displaystyle\int_{\mathbb{R}^{d}}|f|^{p}\,{\rm
d}x=p\int_{0}^{\infty}\mathcal{E}_{p}[P_{t}f]\,{\rm d}t,\quad f\in
L^{p}({\mathbb{R}^{d}}).$

The generator of the Gaussian semigroup $(P_{t})_{t\geq 0}$ acting on $u\in
L^{p}({\mathbb{R}^{d}})$ is

$\displaystyle Lu:=\lim_{h\to 0^{+}}\frac{1}{h}(P_{h}u-u),\quad\textit{if the
limit exists in }L^{p}({\mathbb{R}^{d}}).$

We can also write

$\displaystyle Lu=\sum_{j=1}^{d}\frac{\partial^{2}u}{\partial x_{j}^{2}},\quad
u\in L^{p}({\mathbb{R}^{d}}),$

where the partial derivatives of $u$ are understood in the distributional
sense. We kept the letter $L$ here, to be in accordance with previous
devlopment. The domain of the generator is

$\displaystyle\mathcal{D}_{p}(L)$ $\displaystyle:=$ $\displaystyle\\{u\in
L^{p}({\mathbb{R}^{d}}):\lim_{h\to 0^{+}}(P_{h}u-u)/h\text{ exists in
}L^{p}({\mathbb{R}^{d}})\\}$ $\displaystyle\ =$ $\displaystyle\left\\{u\in
L^{p}({\mathbb{R}^{d}}):\sum_{j=1}^{d}\frac{\partial^{2}u}{\partial
x_{j}^{2}}\in L^{p}({\mathbb{R}^{d}})\right\\}.$

In Appendix F we explain and justify the above statements.

As earlier, for $u\in\mathcal{D}_{p}(L)\subset\mathcal{D}(\mathcal{E}_{p})$,

(E.2) $\displaystyle\mathcal{E}_{p}[u]=-\langle Lu,u^{\langle
p-1\rangle}\rangle.$

To express the Hardy–Stein identity in a more explicit form, we need the
following identity, which was proved by Metafune and Spina [27].

######  Lemma E.1.

Let $1<p<\infty$. For $u\in W^{2,p}({\mathbb{R}^{d}})$,

(E.3) $\displaystyle\int_{\mathbb{R}^{d}}u^{\langle p-1\rangle}Lu\,{\rm
d}x=-(p-1)\int_{\mathbb{R}^{d}}|u|^{p-2}|\nabla u|^{2}\,{\rm d}x,$

where $W^{k,p}({\mathbb{R}^{d}})$ is the Sobolev space of order $k$.

It is not hard to see that for $t>0$ and $f\in L^{p}({\mathbb{R}^{d}})$, we
have $P_{t}f\in W^{2,p}({\mathbb{R}^{d}})$. Indeed, for every multi-index
$\alpha=(\alpha_{1},\ldots,\alpha_{d})$, we denote, as usual,
$|\alpha|:=\alpha_{1}+\ldots+\alpha_{d}$ and
$\partial^{\alpha}:=\frac{\partial^{|\alpha|}}{\partial
x_{1}^{\alpha_{1}}\ldots\partial x_{d}^{\alpha_{d}}}$. Then,

$\displaystyle\left\|\partial^{\alpha}P_{t}f\right\|_{L^{p}({\mathbb{R}^{d}})}=\left\|(\partial^{\alpha}p_{t})\ast
f\right\|_{L^{p}({\mathbb{R}^{d}})}\leq\left\|\partial^{\alpha}p_{t}\right\|_{L^{1}({\mathbb{R}^{d}})}\cdot\left\|f\right\|_{L^{p}({\mathbb{R}^{d}})}<\infty.$

By (E.2) and (E.3), for $f\in L^{p}({\mathbb{R}^{d}})$ and $t>0$,

(E.4) $\displaystyle\mathcal{E}_{p}[P_{t}f]=-\langle\Delta
P_{t}f,(P_{t}f)^{\langle
p-1\rangle}\rangle=(p-1)\int_{\mathbb{R}^{d}}|P_{t}f|^{p-2}|\nabla
P_{t}f|^{2}\,{\rm d}x.$

Since $P_{t}f\in C^{\infty}({\mathbb{R}^{d}})$, the above derivatives are
taken in the classical sense. Here $\Delta$ is the classical Laplacian. Using
(E.4) we can express Hardy–Stein identity for the Gaussian semigroup (E.1) in
the desired form. This finishes the proof of (1.4).

##  Appendix F The generator of the Gaussian semigroup in $L^{p}$

For completeness we prove the equivalence of two definitions of the Laplacian
on $L^{p}({\mathbb{R}^{d}})$ used in Appendix E. Let $1<p<\infty$. Let
$(P_{t})_{t\geq 0}$ be the Gaussian semigroup. It is well known that for
$\varphi\in C_{c}^{\infty}({\mathbb{R}^{d}}),$ the infinitely differentiable
functions on ${\mathbb{R}^{d}}$ with compact support.

$\displaystyle\lim_{h\to
0^{+}}\frac{1}{h}(P_{h}\varphi-\varphi)=\sum_{j=1}^{d}\frac{\partial^{2}\varphi}{\partial
x_{j}^{2}}=:\Delta\varphi,\quad\text{ (the limit taken in
}L^{p}({\mathbb{R}^{d}})).$

We show that the equality is satisfied also for those functions from
$L^{p}({\mathbb{R}^{d}})$, for which the right-hand side limit exists, without
further regularity assumptions.

The semigroup Laplacian is defined as

(F.1) $\displaystyle Lf:=\lim_{h\to 0^{+}}\frac{1}{h}(P_{h}f-f),\quad
f\in\mathcal{D}_{p}(L)\subset L^{p}({\mathbb{R}^{d}}),$

where the limit above is taken in $L^{p}({\mathbb{R}^{d}})$, for $f$ in the
natural domain:

$\displaystyle\mathcal{D}_{p}(L):=\\{u\in L^{p}({\mathbb{R}^{d}}):\lim_{h\to
0^{+}}(P_{h}u-u)/h\text{ exists in }L^{p}({\mathbb{R}^{d}})\\}.$

Since $L$ is the generator of a strongly continuous semigroup in
$L^{p}({\mathbb{R}^{d}})$, the operator $\lambda I-L\\!:\mathcal{D}_{p}(L)\to
L^{p}({\mathbb{R}^{d}})$ is a bijection for every $\lambda>0$.

We then recall the notion of the distributional Laplacian $\tilde{L}$ of $f\in
L^{p}({\mathbb{R}^{d}})$. If there exists $g\in L^{p}({\mathbb{R}^{d}})$ such
that

$\displaystyle\langle g,\varphi\rangle=\langle f,L\varphi\rangle=\left\langle
f,\sum_{j=1}^{d}\frac{\partial^{2}\varphi}{\partial
x_{j}^{2}}\right\rangle=\langle f,\Delta\varphi\rangle$

for all test functions $\varphi\in C_{c}^{\infty}({\mathbb{R}^{d}})$, then we
let $\tilde{L}f:=g$. The class of all such functions $f$ is denoted
$\mathcal{D}(\tilde{L})$. In other words,

$\displaystyle\tilde{L}f=\sum_{j=1}^{d}\frac{\partial^{2}f}{\partial
x_{j}^{2}},$

where the partial derivatives are taken in distributional sense. The operators
$L$ and $\tilde{L}$ coincide, which we prove below.

######  Lemma F.1.

The operator $\tilde{L}$ is an extension of $L$.

###### Proof.

Let $f\in\mathcal{D}_{p}(L)$. For every $\varphi\in
C_{c}^{\infty}({\mathbb{R}^{d}})$, by symmetry of the operators $P_{h}$,

$\displaystyle\langle\tilde{L}f,\varphi\rangle=\langle
f,L\varphi\rangle=\lim_{h\to 0^{+}}\frac{1}{h}\left(\langle
f,P_{h}\varphi\rangle-\langle f,\varphi\rangle\right)=\lim_{h\to
0^{+}}\frac{1}{h}\left(\langle P_{h}f,\varphi\rangle-\langle
f,\varphi\rangle\right)=\langle Lf,\varphi\rangle.$

Thus, $f\in\mathcal{D}(\tilde{L})$ and $\tilde{L}f=Lf$. ∎

######  Lemma F.2.

For every $\lambda>0$, the operator $\lambda I-\tilde{L}$ defined on
$\mathcal{D}(\tilde{L})$ is one-to-one.

###### Proof.

Assume that $f\in\mathcal{D}(\tilde{L})$ and $\lambda f-\tilde{L}f=0$. We
prove that $f=0.$ Take $\varphi\in C_{c}^{\infty}({\mathbb{R}^{d}})$, then
using properties of convolutions and distributional derivatives, we can write

(F.2) $\displaystyle\lambda f\ast\varphi-\tilde{L}(f\ast\varphi)=\lambda
f\ast\varphi-(\tilde{L}f)\ast\varphi=(\lambda
f-\tilde{L}f)\ast\varphi=0\ast\varphi=0.$

This yields $f\ast\varphi=0.$ Indeed, assuming the contrary, since
$f\ast\varphi\in C_{0}^{\infty}({\mathbb{R}^{d}})$, there is a point
$x_{0}\in{\mathbb{R}^{d}}$ which is the positive maximum or the negative
minimum of $f\ast\varphi$. If $x_{0}$ is the positive maximum of
$f\ast\varphi$, then
$\tilde{L}(f\ast\varphi)(x_{0})=\Delta(f\ast\varphi)(x_{0})\leq 0$ and

$\displaystyle 0=(\lambda
f\ast\varphi)(x_{0})-\tilde{L}(f\ast\varphi)(x_{0})\geq(\lambda
f\ast\varphi)(x_{0})>0,$

a contradiction. The case of $s_{0}$ being the negative minimum is handled
similarly. Therefore $f\ast\varphi=0$ for every $\varphi\in
C_{c}^{\infty}({\mathbb{R}^{d}})$, meaning $f=0$. ∎

######  Proposition F.3.

We have $\mathcal{D}_{p}(L)=\mathcal{D}(\tilde{L})$ and $L=\tilde{L}$.

###### Proof.

Take any $\lambda>0$. In view of Lemma F.1 and F.2, the bijection$\lambda
I-L\\!:\mathcal{D}_{p}(L)\to L^{p}({\mathbb{R}^{d}})$ and its injective
extension $\lambda I-\tilde{L}\\!:\mathcal{D}(\tilde{L})\to
L^{p}({\mathbb{R}^{d}})$ are equal,
$\mathcal{D}_{p}(L)=\mathcal{D}(\tilde{L})$, $\lambda I-L=\lambda
I-\tilde{L}$. Thus, $L=\tilde{L}$. ∎

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