# Regular subspaces of symmetric stable processes

Dongjian Qian  School of Mathematical Sciences, Fudan University, Shanghai
200433, People’s Republic of China<EMAIL_ADDRESS>,  Jiangang Ying
School of Mathematical Sciences, Fudan University, Shanghai 200433, People’s
Republic of China<EMAIL_ADDRESS>and  Yushu Zheng  Shanghai Center for
Mathematical Sciences, Fudan University, Shanghai 200433, People’s Republic of
China<EMAIL_ADDRESS>
###### Abstract.

Roughly speaking, regular subspaces are regular Dirichlet forms that inherit
the original forms with smaller domains. In this paper, regular subspaces of
1-dim symmetric $\alpha$-stable processes are considered. The main result is
that it admits proper regular subspaces if and only if $\alpha\in[1,2]$.
Moreover, for $\alpha\in(1,2)$, the characterization of the regular subspaces
is given. General 1-dim symmetric Lévy processes will also be investigated. It
will be shown that whether it has proper regular subspaces is closely related
to whether its sample paths have finite variation.

###### Key words and phrases:

Dirichlet forms, Regular Dirichlet subspaces, Stable processes, Lévy processes

###### 2010 Mathematics Subject Classification:

Primary 31C25, 60J46; Secondary 60G51, 60G52

The second author was supported by NSFC grant No. 11871162

##  1\. Introduction

The theory of Dirichlet space was formulated by A. Beurling and J. Deny [BD59]
in 1959, which is an axiomatic extension of classical Dirichlet integrals in
the direction of Markovian semigroups, and related to symmetric Markov
processes naturally in terms of the regularity condition presented by M.
Fukushima [Fuk71] in 1971, which paves a new approach to obtain Markov
processes and has attracted a lot of attention since then. Let
$(\mathcal{E},\mathcal{F})$ be a Dirichlet form on $L^{2}(E,m)$, where $E$ is
a nice topological space and $m$ is a Radon measure. These notations will not
be detailed since we mainly consider $L^{2}(\mathbb{R},\mathrm{d}x)$ in this
paper. $(\mathcal{E},\mathcal{F})$ is called regular if $C_{c}\cap\mathcal{F}$
is dense in $\mathcal{F}$ under the norm
$\mathcal{E}_{1}^{1/2}:=(\mathcal{E}+(\cdot,\cdot)_{L_{2}})^{1/2}$ and dense
in $C_{c}$ under the uniform norm, where $C_{c}$ represents the space of
continuous functions on $E$ with compact support. It was proved by Fukushima
[Fuk71] that a regular Dirichlet form is always associated with a Hunt process
and for details, refer to [Fuk80, FOT10].

In this paper, we shall consider the problem regarding regular subspaces of
Dirichlet forms. Precisely, let $(\mathcal{E}^{1},\mathcal{F}^{1})$ and
$(\mathcal{E}^{2},\mathcal{F}^{2})$ be two Dirichlet forms on $L^{2}(E,m)$. We
call $(\mathcal{E}^{2},\mathcal{F}^{2})$ a (Dirichlet) subspace of
$(\mathcal{E}^{1},\mathcal{F}^{1})$, and an (Dirichlet) extension conversely,
if

$\mathcal{F}^{2}\subset\mathcal{F}^{1};\,\mathcal{E}^{2}(u,v)=\mathcal{E}^{1}(u,v),\,\forall\,u,v\in\mathcal{F}^{2}.$

When both $(\mathcal{E}^{1},\mathcal{F}^{1})$ and
$(\mathcal{E}^{2},\mathcal{F}^{2})$ are regular, we call
$(\mathcal{E}^{2},\mathcal{F}^{2})$ a regular subspace of
$(\mathcal{E}^{1},\mathcal{F}^{1})$, and regular extension conversely. When
$\mathcal{F}^{2}\neq\mathcal{F}^{1}$, we call $\mathcal{F}^{2}$ a proper
regular subspace. This concept is closely related to Silverstein subspaces and
the readers may refer to [HY22] for details. There are three basic questions
we are interested in

  1. (1)

Does a regular Dirichlet form have proper regular subspaces/extensions?

  2. (2)

How to characterize all regular subspaces/extensions if it does?

  3. (3)

How is the process associated with a regular subspace/extension related to the
process associated with the original form?

Although the problem is not hard to be understood, there have been not many
results yet. The first result in this direction was presented in 2005 [FFY10],
in which the authors, including M. Fukushima and the second author of the
current paper, investigated the problem of regular subspaces for the Dirichlet
form associated with the 1-dim Brownian motion, namely the Sobolev space
$H^{1}$ with the form

$\frac{1}{2}\mathcal{D}(u,v)=\frac{1}{2}\int_{\mathbb{R}}u^{\prime}(x)v^{\prime}(x)\mathrm{d}x,\quad
u,v\in H^{1}.$

For any strictly increasing and continuous function $s$, called a scale
function, on $\mathbb{R}$, define a symmetric Dirichlet form
$(\mathcal{E}^{(s)},\mathcal{F}^{(s)})$ on $L^{2}(\mathbb{R})$ as

$\displaystyle\left\\{\begin{aligned} &\mathcal{F}^{(s)}=\Big{\\{}u\in
L^{2}(\mathbb{R}):u\ll
s,\int_{\mathbb{R}}\Big{(}\frac{\mathrm{d}u}{\mathrm{d}s}\Big{)}^{2}\mathrm{d}s<\infty\Big{\\}};\\\
&\mathcal{E}^{(s)}(u,v):=\frac{1}{2}\int_{\mathbb{R}}\frac{\mathrm{d}u}{\mathrm{d}s}\frac{\mathrm{d}v}{\mathrm{d}s}\mathrm{d}s.\end{aligned}\right.$

where $u\ll s$ means $u$ is absolutely continuous with respect to $s$. It was
shown in [FFY10] that $(\mathcal{E},\mathcal{F})$ is a regular subspace of
$(\frac{1}{2}\mathcal{D},H^{1}$) if and only if
$(\mathcal{E},\mathcal{F})=(\mathcal{E}^{(s)},\mathcal{F}^{(s)})$ with $s$
satisfying

$\mathrm{d}s\ll\mathrm{d}x,\text{and}\ s^{\prime}=0\text{ or }1\ a.e.$

Hence this gives a complete characterization of regular subspaces for 1-dim
Brownian motion. Actually, the question regarding one-dimensional diffusions
has been completely solved, including the characterization of irreducible
regular extensions of $(\frac{1}{2}\mathcal{D},H^{1})$, in [FHY10]. General
regular extensions of Brownian motion and 1-dim diffusions can be
characterized with the help of effective intervals in [LY19a, LSY20]. It is
seen that the process associated with the regular subspace of Brownian motion
is a diffusion with a scale function of Cantor-type, which can not be
described by an SDE. Hence the problem is not only fundamental in the theory
of Dirichlet forms, but also important probabilistically since every regular
subspace or extension gives birth to a symmetric Markov process, which may not
be known to us before.

However, for other processes, few results have been obtained. Some results
were given in [LY15]. Brownian motion with dimension $d>1$ admits proper
regular subspaces, though a complete characterization is still open. The
Dirichlet form associated with a symmetric step process (for example, a
symmetric pure jump Lévy process with finite Lévy measure) admits no proper
regular subspaces. It was also proved there that the jump measure and killing
measure of a Dirichlet form will be inherited by its regular subspaces.

It is then natural to ask if a pure jump Dirichlet form with infinite jump
measure, including symmetric stable processes, admit proper regular subspaces.
Actually in this case an example admitting proper regular subspace was given
in terms of the trace space of Brownian motion [LY19b].

In this paper, we focus on 1-dim symmetric $\alpha$-stable processes
($0<\alpha<2$). The associated Dirichlet forms are the fractional Sobolev
space

$\displaystyle H^{\alpha/2}=\left\\{f\in
L^{2}(\mathbb{R}):\int_{\mathbb{R}}|\widehat{f}(\xi)|^{2}(1+|\xi|^{\alpha})\mathrm{d}\xi<\infty\right\\},$

where $\widehat{f}(\xi)$ is the Fourier transform of $f$, equipped with the
form

$\displaystyle\mathcal{E}(f,g)=\mathcal{E}^{\alpha/2}(f,g)=\int_{\mathbb{R}}\widehat{f}(\xi)\bar{\widehat{g}}(\xi)|\xi|^{\alpha}\mathrm{d}\xi.$

It’s shown in [FOT10] that it can also be expressed as

(1.1) $\displaystyle\left\\{\begin{aligned} &H^{\alpha/2}=\left\\{f\in
L^{2}(\mathbb{R}):\int_{\mathbb{R}^{2}}\dfrac{(f(x)-f(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y<\infty\right\\};\\\
&\mathcal{E}(f,g)=\frac{C(\alpha)}{2}\int_{\mathbb{R}^{2}}\dfrac{(f(x)-f(y))(g(x)-g(y))}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y,\end{aligned}\right.$

where $C(\alpha)$ is a positive constant. The case $\alpha=2$, corresponding
to Brownian motion, may be included if necessary. The main result of the paper
is stated as follows.

######  Theorem 1.1.

The 1-dim symmetric $\alpha$-stable process has proper regular subspaces if
and only if $\alpha\in[1,2]$.

Moreover, we give a characterization of the regular subspaces for
$1<\alpha\leq 2$. Recall that a scale function refers to a continuous and
strictly increasing function on $\mathbb{R}$. For any scale function $s$, we
define

$\displaystyle\mathbf{L}_{s}$
$\displaystyle:=\big{\\{}f:s(\mathbb{R})\rightarrow\mathbb{R}:f\text{ is
Lipschitz and }\operatorname{supp}(f)\subset s(\mathbb{R})\big{\\}},$
$\displaystyle\mathcal{F}_{s}$ $\displaystyle:=\big{\\{}f\circ
s:f\in\mathbf{L}_{s}\big{\\}}.$

We will mainly focus on the scale function of the following type. Denote by
$H^{\alpha/2}_{\text{loc}}$ the space of functions locally in $H^{\alpha/2}$,
i.e. $f\in H^{\alpha/2}_{\text{loc}}$ if for any compact set $K$, there exists
$g\in H^{\alpha/2}$ with $g=f$ a.e. on $K$. We set

$\mathbf{S}^{\alpha/2}_{\text{loc}}:=\big{\\{}s\in
H^{\alpha/2}_{\text{loc}}:s\text{ is a scale function}\big{\\}}.$

######  Theorem 1.2.

Assume $1<\alpha\leq 2$ and $s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$. Then we
have $\mathcal{F}_{s}\subset H^{\alpha/2}$ and
$\overline{\mathcal{F}_{s}}=\overline{\mathcal{F}_{s}}^{\mathcal{E}^{\alpha/2}_{1}}$
is a regular subspace of $H^{\alpha/2}$. Conversely, any regular subspace of
$H^{\alpha/2}$ equals $\overline{\mathcal{F}_{s}}$ for some
$s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$.

Next, we further give several necessary and sufficient conditions on $s$ for
$\overline{\mathcal{F}_{s}}$ to be a proper subspace. For $\alpha=2$, we can
readily tell whether it’s proper by considering the Lebesgue measure of
$\\{\mathrm{d}s/\mathrm{d}x=0\\}$, as is done by [FHY06]. For
$\alpha\in(0,2)$, set $\alpha^{*}=2-\alpha$. Denote by
$\text{Cap}^{\alpha^{*}/2}$ the capacity relative to
$(H^{\alpha^{*}/2},\mathcal{E}^{\alpha^{*}/2})$. For a Borel set $E$, we say
$\text{Cap}^{\alpha^{*}/2}$ is concentracted on $E$ if for any open set
$U\subset\mathbb{R}$, $\text{Cap}^{\alpha^{*}/2}(E\cap
U)=\text{Cap}^{\alpha^{*}/2}(U)$. We use $\mathrm{d}s$ to represent the Radon
measure induced by a scale function $s$ and write $\langle
f,\mathrm{d}s\rangle=\int_{\mathbb{R}}f\,\mathrm{d}s$ for a Borel function
$f$. For $f\in H^{\alpha^{*}/2}$, $\widetilde{f}$ represents its quasi
continuous modification.

######  Theorem 1.3.

For $1<\alpha<2$ and $s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$, $\mathrm{d}s$
is a smooth measure and the following are equivalent:

  1. (a)

$\overline{\mathcal{F}_{s}}$ is proper;

  2. (b)

$\text{Cap}^{\alpha^{*}/2}$ is not concentrated on the quasi support of
$\mathrm{d}s$;

  3. (c)

there exists an open set $E$ such that $\mathrm{d}s$ is concentrated on $E$
(i.e. $\langle 1_{E^{c}},\mathrm{d}s\rangle=0$) and
$\text{Cap}^{\alpha^{*}/2}$ is not concentrated on $E$;

  4. (d)

there exists a nonzero $f\in H^{\alpha^{*}/2}$ such that
$\langle\widetilde{f},\mathrm{d}s\rangle=0$,

where the quasi notions ‘smooth’, ‘quasi support’, ‘q.e.’, and
‘$\widetilde{f}$’ are all relative to
$(H^{\alpha^{*}/2},\mathcal{E}^{\alpha^{*}/2})$ and a function is nonzero if
it is not identically zero.

We point out that except for the second assertion in Theorem 1.2, all other
conclusions in Theorem 1.2 and 1.3 also hold for $\alpha=1$.

The paper is organized as follows. In Section 2, we prepare a lemma for the
later proof. Section 3$\sim$6 is devoted to the proof of Theorem 1.1$\sim$1.3.
In Section 7, we extend results to general symmetric Lévy processes.

The proofs in the paper use quite different methods from the previous ones
about the Brownian motion. It may shed a light on the existence of regular
subspaces of general pure jump processes, though the methods are highly
analytical.

### Notation

We will use the following notation throughout the paper. For $-\infty\leq
a<b\leq\infty$, we denote by

  * •

$C_{c}(a,b)$ the space of continuous functions on $(a,b)$ with compact
support;

  * •

$C_{\infty}(a,b)$ the space of continuous functions on $(a,b)$ which vanish at
$a$ and $b$;

  * •

$C^{\infty}_{c}(a,b)$ the space of smooth functions on $(a,b)$ with compact
support.

In particular, ‘$(a,b)$’ in the above notation is omitted when
$(a,b)=\mathbb{R}$.

##  2\. A preliminary lemma

In this section, we give a preliminary lemma (Lemma 2.2) which will be
frequently used in the sequel. First, we present a simple generalization of a
result in Sobolev spaces.

######  Proposition 2.1.

Let $\alpha\in(0,1)\cup(1,2]$ and $(a,b)$ be a finite open interval. For $f\in
L^{2}(a,b)$ (if $\alpha\in(0,1)$) or $f\in C_{\infty}(a,b)$ (if
$\alpha\in(1,2)$),

(2.1) $\displaystyle\begin{aligned}
\int_{(a,b)\times\mathbb{R}\backslash(a,b)}\dfrac{f(x)^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y&=\frac{1}{\alpha}\int_{(a,b)}f(x)^{2}\left(\dfrac{1}{|x-a|^{\alpha}}+\dfrac{1}{|x-b|^{\alpha}}\right)\mathrm{d}x\\\
&\leq
c_{\alpha}\left(\int_{(a,b)^{2}}\dfrac{(f(x)-f(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y+\left\lVert
f\right\rVert^{2}_{L^{2}(a,b)}\right),\end{aligned}$

where $c_{\alpha}$ is a constant depending only on $\alpha$.

###### Proof.

It suffices to consider $f$ such that the right-hand side of (2.1) is finite.
In [MM00, Lemma 3.32], the analogous result for $f\in C^{\infty}_{c}(a,b)$ is
given. For $\alpha\in(0,1)$, the generalization is immediate from the fact
that $C^{\infty}_{c}(a,b)$ is dense in

(2.2) $\displaystyle H^{\alpha/2}(a,b)=\left\\{f\in
L^{2}(a,b):\int_{(a,b)^{2}}\dfrac{(f(x)-f(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y<\infty\right\\}.$

For $\alpha\in(1,2)$, it is known (Cf. [MM00, Theorem 3.40]) that

$\overline{C^{\infty}_{c}(a,b)}^{\mathcal{E}^{\alpha/2}_{1}}=H^{\alpha/2}(a,b)\cap
C_{\infty}(a,b),$

which implies the conclusion. ∎

For two non-negative functions $f$ and $g$ in $C_{c}$, we call $f$ an erased
function of $g$ if $f\leq g$ and $f$ is constant on each connected component
of $\\{f<g\\}$. Intuitively, $f$ is obtained by ‘erasing’ some upwards
excursions of $g$. Henceforth we shall simply write
$\mathcal{E}=\mathcal{E}^{\alpha/2}$ whenever the parameter ‘$\alpha$’ is
clear from the context.

######  Lemma 2.2.

Let $\alpha\in(0,1)\cup(1,2]$, $\mathcal{G}$ be a regular subspace of
$H^{\alpha/2}$, and $f$ and $g$ be non-negative functions in $C_{c}$. If
$g\in\mathcal{G}$ and $f$ is an erased function of $g$, then $f\in\mathcal{G}$
and $\left\lVert f\right\rVert_{\mathcal{E}_{1}}\leq
6(1+c_{\alpha})\left\lVert g\right\rVert_{\mathcal{E}_{1}}$.

###### Proof.

The proof is divided into two steps. We first show $f\in H^{\alpha/2}$ and
$\left\lVert f\right\rVert_{\mathcal{E}_{1}}\leq 6(1+c_{\alpha})\left\lVert
g\right\rVert_{\mathcal{E}_{1}}$, and then prove $f\in\mathcal{G}$.

Step 1. Denote $\\{f<g\\}=\bigcup_{i\geq 1}(a_{i},b_{i})$, where
$(a_{i},b_{i})$ are disjoint intervals. Here for simplicity, we also see
$\emptyset$ as an interval, which ensures that the right-hand side can always
be taken as a countable union. For $n\geq 1$, let

$f_{n}(x)=\begin{cases}f(x),\ &x\in\bigcup_{1\leq i\leq n}(a_{i},b_{i});\\\
g(x),\ &x\in\mathbb{R}\backslash\bigcup_{0\leq i\leq
n}(a_{i},b_{i}).\end{cases}$

It suffices to prove $\left\lVert f_{n}\right\rVert_{\mathcal{E}_{1}}\leq
6(1+c_{\alpha})\left\lVert g\right\rVert_{\mathcal{E}_{1}}$. In fact, once it
is proved, we have the $\mathcal{E}_{1}$-norms of $f_{n}$ are uniformly
bounded. It follows that the Cesàro means of a subsequence of $f_{n}$ are
$\mathcal{E}_{1}$-convergent. On the other hand, $f_{n}$ converges uniformly
to $f$. Thus, $f\in H^{\alpha/2}$ and $\left\lVert
f\right\rVert_{\mathcal{E}_{1}}\leq 6(1+c_{\alpha})\left\lVert
g\right\rVert_{\mathcal{E}_{1}}$.

Set $D=\bigcup_{1\leq i\leq n}(a_{i},b_{i})$. We will divide the domain of the
integration in (1.1) and compare the integrations associated with $f_{n}$ and
$g$ term by term.

  1. (1)

Since $f_{n}$ is constant on each $(a_{i},b_{i})$, $1\leq i\leq n$, we have

$\int_{\bigcup_{1\leq i\leq
n}(a_{i},b_{i})^{2}}\dfrac{(f_{n}(x)-f_{n}(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y=0.$

  2. (2)

Since $f_{n}=g$ outside $D$, it is plain that

$\int_{(D^{c})^{2}}\dfrac{(f_{n}(x)-f_{n}(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y=\int_{(D^{c})^{2}}\dfrac{(g(x)-g(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y.$

  3. (3)

For $1\leq i\leq n$,

$\displaystyle\int_{(a_{i},b_{i})\times
D^{c}}\dfrac{(f_{n}(x)-f_{n}(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle 2\int_{(a_{i},b_{i})\times
D^{c}}\left[\dfrac{(g(x)-f_{n}(x))^{2}}{|x-y|^{1+\alpha}}+\dfrac{(g(x)-g(y))^{2}}{|x-y|^{1+\alpha}}\right]\mathrm{d}x\mathrm{d}y.$

  4. (4)

For $1\leq i,j\leq n$ and $i\neq j$,

$\displaystyle\int_{(a_{i},b_{i})\times(a_{j},b_{j})}\dfrac{(f_{n}(x)-f_{n}(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle
3\int_{(a_{i},b_{i})\times(a_{j},b_{j})}\left[\dfrac{(f_{n}(x)-g(x))^{2}}{|x-y|^{1+\alpha}}+\dfrac{(f_{n}(y)-g(y))^{2}}{|x-y|^{1+\alpha}}+\dfrac{(g(x)-g(y))^{2}}{|x-y|^{1+\alpha}}\right]\mathrm{d}x\mathrm{d}y.$

Summing the parts (3) and (4), we have

$\displaystyle\sum_{1\leq i\leq
n}\int_{(a_{i},b_{i})\times(a_{i},b_{i})^{c}}\dfrac{(f_{n}(x)-f_{n}(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle 6\sum_{1\leq i\leq
n}\int_{(a_{i},b_{i})\times(a_{i},b_{i})^{c}}\left[\dfrac{(g(x)-f_{n}(x))^{2}}{|x-y|^{1+\alpha}}+\dfrac{(g(x)-g(y))^{2}}{|x-y|^{1+\alpha}}\right]\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq$ $\displaystyle 6(1+c_{\alpha})\left(\sum_{1\leq i\leq
n}\int_{(a_{i},b_{i})\times(a_{i},b_{i})^{c}}\dfrac{(g(x)-g(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y+\left\lVert
g-f_{n}\right\rVert^{2}_{L^{2}}\right),$

where the last inequality follows from Proposition 2.1. Further adding (1) ,
(2), and $\left\lVert f_{n}\right\rVert^{2}_{L^{2}}$, we get $\left\lVert
f_{n}\right\rVert_{\mathcal{E}_{1}}\leq 6(1+c_{\alpha})\left\lVert
g\right\rVert_{\mathcal{E}_{1}}$.

Step 2. Denote $h=g-f$. It suffices to prove $h\in\mathcal{G}$. Since $f\in
H^{\alpha/2}$, so is $h$. Then we can take $\varepsilon$-contraction on $h$,

$h_{\varepsilon}:=h-(-\varepsilon)\vee h\wedge\varepsilon.$

It is known that
$h_{\varepsilon}\stackrel{{\scriptstyle\mathcal{E}_{1}}}{{\rightarrow}}h$ as
$\varepsilon\downarrow 0$ and hence it is enough to prove
$h_{\varepsilon}\in\mathcal{G}$ for any $\varepsilon>0$.

Recall that $\\{h>0\\}=\\{f<g\\}=\bigcup_{i\geq 1}(a_{i},b_{i})$. Since $h$ is
uniformly continuous, there exist at most finite $i$ such that

$(a_{i},b_{i})\cap\\{h>\varepsilon\\}\neq\emptyset.$

Without loss of generality, we assume

$h_{\varepsilon}=\left(\sum_{1\leq i\leq
N}h1_{(a_{i},b_{i})}\right)_{\varepsilon}=\sum_{1\leq i\leq
N}\left(h1_{(a_{i},b_{i})}\right)_{\varepsilon}$

for some integer $N$. Then we need only to prove that for each $i$,
$(h1_{(a_{i},b_{i})})_{\varepsilon}\in\mathcal{G}$. Remember that $f$ is
constant on $[a_{i},b_{i}]$, i.e. $f(x)=d_{i}$ for $x\in[a_{i},b_{i}]$ for
some constant $d_{i}$. Hence, $h1_{(a_{i},b_{i})}=(g-d_{i})1_{(a_{i},b_{i})}$.
Since $g(a_{i})=g(b_{i})=d_{i}$, it follows from continuity that there exists
$\delta>0$ such that for $x\in[a_{i},a_{i}+\delta]\cup[b_{i}-\delta,b_{i}]$,

$0\leq g(x)-d_{i}\leq\varepsilon.$

Due to the regularity of $\mathcal{G}$, we can take a continuous function
$g_{i}\in\mathcal{G}$, such that $0\leq g_{i}\leq 1$ and

$g_{i}(x)=\left\\{\begin{array}[]{ll}1,&x\in(a_{i}+\delta,b_{i}-\delta];\\\
0,&x\in(-\infty,a_{i}]\cup[b_{i},\infty).\end{array}\right.$

It follows easily that $(g-d_{i})g_{i}\in\mathcal{G}$. The difference between
$(g-d_{i})1_{(a_{i},b_{i})}$ and $(g-d_{i})g_{i}$ is located on
$(a_{i},a_{i}+\delta]\cup(b_{i}-\delta,b_{i})$ and both will be erased by
$\varepsilon$-contraction. Therefore it holds that

$(h1_{(a_{i},b_{i})})_{\varepsilon}=((g-d_{i})1_{(a_{i},b_{i})})_{\varepsilon}=((g-d_{i})g_{i})_{\varepsilon}\in\mathcal{G}.$

That completes the proof. ∎

##  3\. Proof of the ‘only if’ part of Theorem 1.1

We shall prove that for $0<\alpha<1$, a symmetric $\alpha$-stable process has
no proper regular subspaces. If $H^{\alpha/2}$ has a regular subspace, we
denote it by $\mathcal{G}$ and deduce that in fact $\mathcal{G}=H^{\alpha/2}$.
Before the proof, let us prepare a few lemmas. The first lemma is an easy but
key observation.

######  Lemma 3.1.

For $a<b$, $1_{(a,b]}\in H^{\alpha/2}$ if and only if $\alpha<1$.

###### Proof.

This is a direct computation. When $f=1_{(a,b]}$ and $\alpha\in(0,1)$,

$\displaystyle\int_{\mathbb{R}^{2}}\dfrac{(f(x)-f(y))^{2}}{\left\lvert
x-y\right\rvert^{\alpha+1}}\mathrm{d}x\mathrm{d}y$ $\displaystyle=$
$\displaystyle 2\left(\int_{x\in(a,b],y>b}+\int_{x\in(a,b],y\leq
a}\right)\dfrac{1}{\left\lvert
x-y\right\rvert^{\alpha+1}}\mathrm{d}x\mathrm{d}y$ $\displaystyle=$
$\displaystyle\dfrac{4}{\alpha(1-\alpha)}(b-a)^{1-\alpha}.$

When $\alpha\geq 1$, this integral diverges. ∎

Actually we don’t need to distinguish $1_{(a,b]}$ and $1_{[a,b]}$. Denote by

$S_{0}=\left\\{\sum_{i=1}^{K}c_{i}1_{(a_{i},b_{i}]}:K\ \text{is an
integer},~{}c_{i}\in\mathbb{R},~{}(a_{i},b_{i}]\text{ are disjoint
intervals}\right\\}$

the set of step functions. It is easily seen that $S_{0}\subset H^{\alpha/2}$
when $0<\alpha<1$.

######  Lemma 3.2.

$S_{0}$ is dense in $H^{\alpha/2}$.

###### Proof.

It is well-known that $C^{\infty}_{c}$ is dense in $H^{\alpha/2}$. Hence it
suffices to prove for any $f\in C^{\infty}_{c}$, we can find a sequence
$\\{f_{n}\\}\subset S_{0}$ such that
$f_{n}\stackrel{{\scriptstyle\mathcal{E}_{1}}}{{\longrightarrow}}f$. Assume
that $f$ is supported on $[-N,N]$ and $|f^{\prime}|<c$ for some positive
constants $N$ and $c$.

Take $f_{n}(x)=f(\frac{i}{2^{n}})$, whenever
$\frac{i}{2^{n}}<x\leq\frac{i+1}{2^{n}}$ for any integer $i$. It obvious that
$|f(x)-f_{n}(x)|\leq c/2^{n}$ for any $x\in\mathbb{R}$, $f_{n}\in S_{0}$ and
$f_{n}$ converges to $f$ both pointwise and in $L^{2}$. We now aim to prove
$f_{n}$ converges to $f$ in $\mathcal{E}$-norm. In the following proof,
$c_{i}$ are all positive constants.

$\displaystyle\int_{\mathbb{R}^{2}}\dfrac{((f-f_{n})(x)-(f-f_{n})(y))^{2}}{|x-y|^{1+\alpha}}dxdy$
$\displaystyle=$
$\displaystyle\left(2\int_{[-N,N]\times[-N,N]^{c}}+\int_{[-N,N]\times[-N,N]}\right)\dfrac{((f-f_{n})(x)-(f-f_{n})(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle=$
$\displaystyle\bigg{(}2\int_{[-N,N]\times[-N,N]^{c}}+2\sum_{-N2^{n}\leq
i<j<N2^{n}}\int_{[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]\times[\frac{j}{2^{n}},\frac{j+1}{2^{n}}]}+\sum_{-N2^{n}\leq
i<N2^{n}}\int_{[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]\times[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]}\bigg{)}$
$\displaystyle\dfrac{((f-f_{n})(x)-(f-f_{n})(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y=:I_{1}+I_{2}+I_{3}.$

It can be verified that

$\displaystyle I_{1}$ $\displaystyle\leq
2\int_{[-N,N]\times\mathbb{R}\backslash[-N,N]}\dfrac{(c\frac{1}{2^{n}})^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y\leq
c_{1}2^{-2n};$ $\displaystyle I_{2}$ $\displaystyle\leq 2\sum_{-N2^{n}\leq
i<j<N2^{n}}\int_{[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]\times[\frac{j}{2^{n}},\frac{j+1}{2^{n}}]}\dfrac{(2c\frac{1}{2^{n}})^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq c_{2}\sum_{-N2^{n}\leq i<j\leq N2^{n}}2^{n(\alpha-3)}\leq
4c_{2}N^{2}2^{n(\alpha-1)};$ $\displaystyle I_{3}$
$\displaystyle\leq\sum_{-N2^{n}\leq
i<N2^{n}}\int_{[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]\times[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]}\dfrac{(c(x-y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y$
$\displaystyle\leq c_{3}\sum_{-N\times 2^{n}\leq i\leq N\times
2^{n}}2^{n(\alpha-3)}\leq 2c_{3}N2^{n(\alpha-2)},$

where for $I_{3}$, when $x,y\in[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]$, we use
the estimate $|(f-f_{n})(x)-(f-f_{n})(y)|=|f(x)-f(y)|\leq c|x-y|$. Therefore
$I_{1}+I_{2}+I_{3}\leq c_{4}2^{n(\alpha-1)}$, which converges to 0 as
$n\rightarrow\infty$. ∎

By this lemma, to prove that $\mathcal{G}=H^{\alpha/2}$, we only need to prove
any indicator function $1_{[a,b]}\in\mathcal{G}$. To do this, we introduce a
type of functions close to indicator functions. Fix $a<b$ from now and assume
that a continuous function $h_{\rho}$ satisfies the following condition,

$h_{\rho}(x)=\begin{cases}1,&x\in[a,b];\\\
0,&x\in(-\infty,a-\rho]\cup[b+\rho,\infty);\\\ \text{increasing},\
&x\in(a-\rho,a];\\\ \text{decreasing},\ &x\in(b,b+\rho).\end{cases}$

with some $\rho>0$. Notice that when $x\in(a-\rho,a)\cup(b,b+\rho)$,
$h_{\rho}(x)$ is not specifically defined, as long as it is monotone. Denote
by $M_{\rho}$ the collection of all such functions. Now it suffices to prove
the following lemma

######  Lemma 3.3.

For any $\rho>0$, it holds that

  1. (i)

$M_{\rho}\cap\mathcal{G}\neq\emptyset$;

  2. (ii)

$\big{\\{}\left\lVert f\right\rVert_{\mathcal{E}_{1}}:f\in M_{\rho}\big{\\}}$
are uniformly bounded.

If the lemma is proved, we take a sequence $\\{\rho_{n}\\}$ decreasing to 0,
say $\rho_{n}=1/n$. Due to (i), for each $n$, there is a function
$h_{\rho_{n}}\in M_{\rho_{n}}\cap\mathcal{G}$. It follows from (ii) that the
norms $\big{\\{}\|h_{\rho_{n}}\|_{\mathcal{E}_{1}}:n\geq 1\big{\\}}$ is
bounded. Also note that $h_{\rho_{n}}$ converges pointwise to $1_{[a,b]}$.
Hence the standard arguments yield that $1_{[a,b]}\in\mathcal{G}$.

###### Proof.

For (i), since $\mathcal{G}$ is regular, there exists $g\in\mathcal{G}\cap
C_{c}$ such that $0\leq g\leq 1$ and

$\displaystyle g(x)=\begin{cases}1,\ &x\in[a,b];\\\ 0,\
&x\in(a-\rho,b+\rho)^{c}.\end{cases}$

Then we define

(3.5) $\displaystyle g^{*}(x):=\left\\{\begin{array}[]{ll}1,&x\in[a,b];\\\
0,&x\in(a-\rho,b+\rho)^{c};\\\ \inf\\{g(t):t\in[x,a]\\},&x\in(a-\rho,a);\\\
\inf\\{g(t):t\in[b,x]\\},&x\in(b,b+\rho).\end{array}\right.$

The construction is just like the Skorokhod decomposition and shares the same
good properties: $g^{*}\in M_{\rho}$ and $g^{*}$ is an erased function of $g$.
By Lemma 2.2, we have $g^{*}\in\mathcal{G}$.

For (ii), we only need to prove that

$\sup_{f\in M_{\rho}}\int_{|\xi|\geq
1}|\hat{f}(\xi)|^{2}|\xi|^{\alpha}d\xi<\infty.$

Using the integration by parts formula and noticing that any $f\in M_{\rho}$
is of finite variation, we have for $\left\lvert\xi\right\rvert\geq 1$,

$\displaystyle\hat{f}(\xi)$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)(\cos(x\xi)+i\sin(x\xi))\mathrm{d}x$
$\displaystyle=-\frac{1}{\sqrt{2\pi}}\left(\int_{\mathbb{R}}\dfrac{\sin(x\xi)}{\xi}\mathrm{d}f(x)-i\int_{\mathbb{R}}\dfrac{\cos(x\xi)}{\xi}\mathrm{d}f(x)\right).$

Observe that the total variation of $f$ is $2$. Then it follows that

(3.6)
$|\hat{f}(\xi)|\leq{\dfrac{4}{\sqrt{2\pi}\left\lvert\xi\right\rvert}}\leq\dfrac{2}{\left\lvert\xi\right\rvert}\,\text{
for }\left\lvert\xi\right\rvert\geq 1.$

Therefore, for any $f\in M_{\rho}$,

$\int_{|\xi|\geq 1}|\hat{f}(\xi)|^{2}|\xi|^{\alpha}d\xi\leq\int_{|\xi|\geq
1}4|\xi|^{\alpha-2}d\xi<\infty.$

That completes the proof of the lemma and also the ‘only if’ part of Theorem
1.1. ∎

##  4\. Proof of Theorem 1.2

In this section, we will prove Theorem 1.2, which gives a characterization of
the regular subspaces of $\alpha$-stable process when $\alpha\in(1,2)$. The
former part of Theorem 1.2 is straightforward. The first conclusion
‘$\mathcal{F}_{s}\subset H^{\alpha/2}$’ comes from $s\in
H^{\alpha/2}_{\text{loc}}$ and the definition of $\mathbf{L}_{s}$. Moreover,
$\mathcal{F}_{s}$ is a lattice separating points. Hence it is dense in $C_{c}$
under the norm $\left\lVert\cdot\right\rVert_{\infty}$. Then it’s standard to
verify that $\overline{\mathcal{F}_{s}}$ is a regular subspace of
$H^{\alpha/2}$.

The latter part of 1.2 is more involved. Assume $\mathcal{G}$ is a regular
subspace of $H^{\alpha/2}$. We shall find
$s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$ such that
$\mathcal{G}=\overline{\mathcal{F}_{s}}$. The key is to decompose a non-
negative function in $C_{c}\cap\mathcal{G}$ into a countable sum of ladder-
like functions, that are functions in $C_{c}$ which first increase to a
maximum, then decrease to $0$ and stay at $0$ afterwards. Based on this, the
main idea of the following proof is to first take a countable set of functions
in $C_{c}\cap\mathcal{G}$ which is dense in $\mathcal{G}$, and then use the
ladder-like functions in the decompositions of these functions to construct
the function $s$ we need.

######  Lemma 4.1.

Let $1<\alpha\leq 2$ and $\mathcal{G}$ be a regular subspace of
$H^{\alpha/2}$. For any non-negative function $f\in C_{c}\cap\mathcal{G}$,
there exists a sequence of ladder-like functions $f^{n}\in\mathcal{G}$, such
that $\sum_{n}f^{n}$ converges in $\mathcal{E}_{1}$-norm to $f$.

###### Proof.

First, we associate every non-negative function $f\in C_{c}$ with a ladder-
like function as follows. Let $T=T^{f}$ be the minimum of the maximum points
of $f$, i.e. $T=\inf\big{\\{}x\in\mathbb{R}:f(x)=\left\lVert
f\right\rVert_{\infty}\big{\\}}$ . We consider the following decomposition of
$f$, which is similar to that in (3.5).

$\displaystyle
f^{*}(x):=\left\\{\begin{array}[]{ll}\inf\\{f(t):t\in[x,T]\\},&x\in(-\infty,T],\\\
\inf\\{f(t):t\in[T,x]\\},&x\in(T,\infty).\end{array}\right.$

We call $f^{*}$ the ladder-like function associated with $f$. Moreover, we
denote $\\{f^{*}<f\\}=\bigcup_{i\geq 1}(a_{i},b_{i})$, where $(a_{i},b_{i})$
are disjoint intervals. Let $f_{i}:=(f-f^{*})1_{(a_{i},b_{i})}$. Then

$f=f^{*}+\sum_{i}f_{i},$

which is called the ladder decomposition of $f$. And $\\{f_{i}:i\geq 1\\}$ are
called the ladder excursions of $f$.

Now we proceed to consider the ladder decomposition of each $f_{i}$. Let
$\\{f_{ij}:j\geq 1\\}$ be the ladder excursions of $f_{i}$. Recursively,
suppose we have defined $f_{n_{1}n_{2}\cdots n_{m}}$ for $n_{1}n_{2}\cdots
n_{m}=(n_{1},n_{2},\cdots,n_{m})\in\mathbb{N}^{m}$, where
$\mathbb{N}:=\\{1,2,\cdots\\}$. Then $\\{f_{n_{1}n_{2}\cdots n_{m}j}:j\geq
1\\}$ is defined to be the ladder excursions of $f_{n_{1}n_{2}\cdots n_{m}}$.
We also denote $f_{\emptyset}=f$. In this way, we have defined
$\big{\\{}f_{u}:u\in\\{\emptyset\\}\cup\bigcup_{n\in\mathbb{N}}\mathbb{N}^{n}\big{\\}}$.

Note that $\\{\emptyset\\}\cup\bigcup_{n\in\mathbb{N}}\mathbb{N}^{n}$ can be
naturally viewed as the vertices set of a tree $\mathbb{T}=(V,E)$ where
$n_{1}n_{2}\cdots n_{m}$ has children $\\{n_{1}n_{2}\cdots n_{m}i:i\geq 1\\}$.
Let $\mathbb{T}_{n}=(V_{n},E_{n})$ be a sequence of finite connected subgraphs
of $\mathbb{T}$, such that $\emptyset\in\mathbb{T}_{n}$ for all $n$ and
$V_{n}\uparrow V$. In the following, we shall prove $\sum_{u\in
V_{n}}f^{*}_{u}$ converges pointwise to $f$. Once it is proved, by further
noting that $\sum_{u\in V_{n}}f^{*}_{u}$ are erased functions of $f$ and using
Lemma 2.2, we have $\sum_{u\in V_{n}}f^{*}_{u}\in\mathcal{G}$ for all $n$ and
the $\mathcal{E}_{1}$-norms of $\sum_{u\in V_{n}}f^{*}_{u}$ are uniformly
bounded. It follows that $\sum_{u\in V_{n}}f^{*}_{u}$ also converges to $f$ in
$\mathcal{E}_{1}$, which concludes the lemma.

For the proof of pointwise convergence, we first present an easy observation.
We can see from the construction of $f_{u}$ that $\sum_{u\in V_{n}}f^{*}_{u}$
is obtained from $f$ by ‘erasing’ the upwards excursions over the intervals
$\operatorname{supp}(f_{u})$, $u\in V\setminus V_{n}$. It follows that

(4.1) $\displaystyle\sum_{u\in V_{n}}f^{*}_{u}=f\text{ on
}A_{n}:=\operatorname{supp}(f)\setminus\Big{(}\bigcup_{u\in V\setminus
V_{n}}\operatorname{supp}(f_{u})\Big{)}.$

Fix $x\in\operatorname{supp}(f)$. Then $x$ is either in finitely many
$\operatorname{supp}(f_{u})$ or in infinitely many of them. In the former
case, it is direct from (4.1) that for $n$ sufficiently large, $\sum_{u\in
V_{n}}f^{*}_{u}(x)=f(x)$. Henceforth we focus on the latter case. We call a
sequence $\\{u_{n}:n\geq 0\\}$ of $V$ a ray in $\mathbb{T}$ if
$u_{0}=\emptyset$ and for any $n\geq 0$, $u_{n}$ is the parent of $u_{n+1}$.
In the latter case, we can always find a ray $\\{u_{n}\\}$ in $\mathbb{T}$,
such that $x\in\operatorname{supp}(f_{u_{n}})$ for all $n$. For convenience,
we denote $T_{n}=T^{f_{u_{n}}}$ and
$[l_{n},r_{n}]=\operatorname{supp}(f_{u_{n}})$ for $n\geq 1$. From the
construction of $f_{u_{n}}$, we see that $l_{n},r_{n},T_{n}\in A_{m}$ whenever
$u_{n}\in V_{m}$. Combining with (4.1), we have $f(y)=\sum_{u\in
V_{m}}f^{*}_{u}(y)$ for $y=l_{n},r_{n},T_{n}$. Hence,

$f(l_{n})=f(r_{n})\leq\sum_{u\in V_{m}}f^{*}_{u}(x)\leq f(T_{n}).$

Note that the inequality still holds if we replace $\sum_{u\in
V_{m}}f^{*}_{u}(x)$ with $f(x)$. So it remains to show
$f(T_{n})-f(l_{n})=f(T_{n})-f(r_{n})$ goes to $0$. Back to the construction of
$f_{u_{n}}$ again, it is seen that $[l_{n},r_{n}]$ is decreasing and either
$[l_{n+1},r_{n+1}]\subset(l_{n},T_{n})$ or
$[l_{n+1},r_{n+1}]\subset(T_{n},r_{n})$. In particular, one of $(l_{n},T_{n})$
and $(T_{n},r_{n})$ is a subset of $[l_{n},r_{n}]\setminus[l_{n+1},r_{n+1}]$.
Thus,

$(T_{n}-l_{n})\wedge(r_{n}-T_{n})\rightarrow 0.$

The conclusion we want then follows from the uniform continuity of $f$. ∎

Now we start the construction of $s$. Since $H^{\alpha/2}$ is separable and
$\mathcal{G}$ is regular, we can find a countable set $\mathcal{B}\subset
C_{c}\cap\mathcal{G}$ which is dense in $\mathcal{G}$. Consider
$\mathcal{B}^{\pm}:=\\{f^{\pm}:f\in\mathcal{B}\\}$ the positive and negative
parts of the functions in $\mathcal{B}$. For any
$f\in\mathcal{B}^{+}\cup\mathcal{B}^{-}$, let $\\{f^{n}:n\geq 1\\}$ be the
ladder-like functions satisfying the conclusion in Lemma 4.1. For simplicity,
we let $\\{h_{n}:n\geq 1\\}$ be an enumeration of $\\{f^{n}:n\geq
1,\,f\in\mathcal{B}^{+}\cup\mathcal{B}^{-}\\}$. Due to Lemma 4.1, the linear
span of $\\{h_{n}:n\geq 1\\}$ is dense in $\mathcal{G}$. We denote
$[l_{n},r_{n}]=\operatorname{supp}(h_{n})$ and $T_{n}=T^{h_{n}}$. Define

$h^{l}_{n}=h_{n}\vee 1_{[T_{n},\infty]}\left\lVert
h_{n}\right\rVert_{\infty},\quad
h^{r}_{n}=h^{l}_{n}-h_{n}=1_{[T_{n},\infty)}(\left\lVert
h_{n}\right\rVert_{\infty}-h_{n}).$

Intuitively, $h^{l}_{n}$ is the ‘left arm’ of the ladder-like function $h_{n}$
and $h^{r}_{n}$ is a reflection of the ‘right arm’ of $h_{n}$.

Recall the definition of $\mathcal{E}$ in (1.1), whose domain can be extended
to all measurable functions. For simplicity, we omit the coefficient
$C(\alpha)/2$ in the definition henceforth. We shall first prove that
$\mathcal{E}\big{(}h^{l}_{n},h^{l}_{n}\big{)}$ and
$\mathcal{E}\big{(}h^{r}_{n},h^{r}_{n}\big{)}$ are both finite. Since
$h^{r}_{n}=h^{l}_{n}+h_{n}$, it suffices to prove
$\mathcal{E}\big{(}h^{l}_{n},h^{l}_{n}\big{)}<\infty$. We have

$\displaystyle\mathcal{E}\big{(}h^{l}_{n},h^{l}_{n}\big{)}$ $\displaystyle=$
$\displaystyle\int_{(-\infty,T_{n})^{2}}\dfrac{(h_{n}(x)-h_{n}(y))^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y+2\int_{(-\infty,T_{n})\times(T_{n},\infty)}\dfrac{(h_{n}(x)-\left\lVert
h_{n}\right\rVert_{\infty})^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y$ $\displaystyle\leq$
$\displaystyle\mathcal{E}(h_{n},h_{n})+\dfrac{2\left\lVert
h_{n}\right\rVert_{\infty}^{2}}{\alpha(\alpha-1)(T_{n}-l_{n})^{\alpha-1}}+2\int_{(l_{n},T_{n})\times(T_{n},\infty)}\dfrac{(h_{n}(x)-\left\lVert
h_{n}\right\rVert_{\infty})^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y.$

It remains to bound the last term. For this, we take $\phi\in C^{\infty}_{c}$
such that $\phi=1$ on $[l_{n},T_{n}]$ and consider
$f=\phi\cdot(h_{n}-\left\lVert h_{n}\right\rVert_{\infty})$. It holds that

$\displaystyle\int_{(-\infty,T_{n})^{2}}\dfrac{(f(x)-f(y))^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y$ $\displaystyle\leq$
$\displaystyle 2\left\lVert
h_{n}\right\rVert_{\infty}^{2}\mathcal{E}(\phi,\phi)+2\left\lVert\phi\right\rVert_{\infty}^{2}\int_{(-\infty,T_{n})^{2}}\dfrac{(h_{n}(x)-h_{n}(y))^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y<\infty.$

Note that $f\in C_{\infty}(-N,T_{n})$ for $N$ large enough. So it follows from
Lemma 2.1 that

$\displaystyle\int_{(l_{n},T_{n})\times(T_{n},\infty)}\dfrac{(h_{n}(x)-\left\lVert
h_{n}\right\rVert_{\infty})^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y=\int_{(l_{n},T_{n})\times(T_{n},\infty)}\dfrac{f(x)^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y$ $\displaystyle\leq
c_{\alpha}\left(\int_{(-\infty,T_{n})^{2}}\dfrac{(f(x)-f(y))^{2}}{\left\lvert
x-y\right\rvert^{1+\alpha}}\mathrm{d}x\mathrm{d}y+\left\lVert
f\right\rVert^{2}_{L^{2}}\right)<\infty.$

Now we define $s$ as follows:

$s=\sum_{n\geq 1}2^{-n}\dfrac{h^{l}_{n}+h^{r}_{n}}{\left\lVert
h^{l}_{n}\right\rVert_{\mathcal{E}}+\left\lVert
h^{r}_{n}\right\rVert_{\mathcal{E}}+\left\lVert h_{n}\right\rVert_{\infty}}.$

We see that $s$ is continuous. From the construction of $\\{h_{n}\\}$, it is
easy to deduce that $s$ is strictly increasing. Moreover, since $\left\lVert
s\right\rVert_{\infty},\left\lVert s\right\rVert_{\mathcal{E}}\leq 2$, we have
for any $f\in C^{\infty}_{c}$, $f\cdot s\in H^{\alpha/2}$. Thus,
$s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$.

The remaining part is devoted to the proof of
$\overline{\mathcal{F}_{s}}=\mathcal{G}$. The ‘$\supset$’ part is simple. In
fact, for each $h_{n}$, it holds that

$\left\lvert h_{n}(x)-h_{n}(y)\right\rvert\leq C\left\lvert
s(x)-s(y)\right\rvert,$

for some constant $C$. So $h_{n}\circ s^{-1}\in\mathbf{L}_{s}$ and
$h_{n}=h_{n}\circ s^{-1}\circ s\in\mathcal{F}_{s}$. Recall that the linear
span of $\\{h_{n}\\}$ is dense in $\mathcal{G}$. Hence we get
$\overline{\mathcal{F}_{s}}\supset\mathcal{G}$.

For the other direction, it suffices to show every function in
$\mathcal{F}_{s}$ can be approximated by functions in $\mathcal{G}$. We only
need to prove this for a certain class of functions. Fix $f\in\mathbf{L}_{s}$.
Let

$s_{k}=\sum_{n=1}^{k}2^{-n}\dfrac{h^{l}_{n}+h^{r}_{n}}{\left\lVert
h^{l}_{n}\right\rVert_{\mathcal{E}}+\left\lVert
h^{r}_{n}\right\rVert_{\mathcal{E}}+\left\lVert h_{n}\right\rVert_{\infty}}.$

Then $s_{k}$ converges pointwise to $s$, which implies $f\circ s_{k}$
converges pointwise to $f\circ s$. Observe that

$\left\lvert s_{k_{1}}(x)-s_{k_{1}}(y)\right\rvert\leq\left\lvert
s_{k_{2}}(x)-s_{k_{2}}(y)\right\rvert\text{ for any $k_{1}<k_{2}$}.$

So for $k$ large enough, $\operatorname{supp}(f\circ s_{k})$ is compact and
decreases in $k$. Further using Proposition 2.1, we have for above $k$, the
$\mathcal{E}_{1}$-norms of $f\circ s_{k}$ are uniformly bounded. Therefore, it
reduces to proving $f\circ s_{k}\in\mathcal{G}$ for large $k$.

We assume $\operatorname{supp}(f\circ s)\subset[-N,N]$ for any above $k$. Fix
$k$ large enough such that $s_{k}(N)<s_{k}(\infty)$ and
$s_{k}(-N)>s_{k}(-\infty)$, i.e. $\mathrm{d}s_{k}$ assigns positive masses on
both $(-\infty,-N)$ and $(N,\infty)$. We denote
$M=\inf\\{x:s_{k}(x)=s_{k}(\infty)\\}$, which is finite and greater than $N$.

In the following, we shall show that $\mathcal{G}$ contains a special class of
ladder-like functions. Roughly speaking, the functions have a constant stretch
beyond the maximum and their ‘left arms’ are $h^{l}_{n}$ or $h^{r}_{n}$.
Before the precise statement, we introduce some notation. Let $h$ and $p$ be
two ladder-like functions. We say $h$ and $p$ are compatible if $\left\lVert
h\right\rVert_{\infty}=\left\lVert p\right\rVert_{\infty}$,
$\operatorname{supp}(h)\cap\operatorname{supp}(p)=\emptyset$. For compatible
$h$ and $p$, assuming $T^{h}<T^{p}$ and $\operatorname{supp}(p)=[a,b]$, the
functions $A^{l}(h,p)$ and $A^{r}(h,p)$ are defined as:

$\displaystyle A^{l}(h,p)(x)$ $\displaystyle=\begin{cases}h^{l}(x),\
&x\in(-\infty,T^{p}];\\\ p(x),\ &x\in(T^{p},\infty),\end{cases}$
$\displaystyle A^{r}(h,p)(x)$ $\displaystyle=\begin{cases}h^{r}(x),\
&x\in(-\infty,a];\\\ \left\lVert p\right\rVert_{\infty}-p(x),\
&x\in(a,T^{p}];\\\ 0,\ &(T^{p},\infty).\end{cases}$

Namely, $A^{l}(h,p)$ and $A^{r}(h,p)$ are ladder-like functions. The ‘arms’ of
$A^{l}(h,p)$ consist of the ‘left arm’ of $h$ and the ‘right arm’ of $p$,
while the ‘arms’ of $A^{r}(h,p)$ consist of the reflections of the ‘right arm’
of $h$ and the ‘left arm’ of $p$. Let

$E:=\big{\\{}p\in C_{c}:p\text{ is ladder-like,
}\operatorname{supp}(p)\subset[M+1,M+2]\big{\\}}.$

We claim that for any $n\geq 1$, there exists $p_{n}\in E$ such that $p_{n}$
and $h_{n}$ are compatible and both $A^{l}_{n}:=A^{l}(h_{n},p_{n})$ and
$A^{r}_{n}=A^{r}(h_{n},p_{n})$ are in $\mathcal{G}$. If it is proved, then we
have

$q_{k}:=\sum_{n=1}^{k}2^{-n}\dfrac{A^{l}_{n}+A^{r}_{n}}{\left\lVert
h^{l}_{n}\right\rVert_{\mathcal{E}}+\left\lVert
h^{r}_{n}\right\rVert_{\mathcal{E}}+\left\lVert
h_{n}\right\rVert_{\infty}}\in\mathcal{G}.$

Note that $q_{k}=s_{k}$ on $(-\infty,M+1]$. On $(M+1,\infty)$, $s_{k}$ is
constant while $q_{k}$ decreases to $0$ and stays at $0$ afterwards. This
implies that $f\circ s_{k}\in\mathcal{G}$. Indeed, since $f\circ q_{k}$ is a
contraction of $\left\lVert f^{\prime}\right\rVert_{\infty}q_{k}$, it is also
in $\mathcal{G}$. Moreover, it holds that

$\displaystyle f\circ q_{k}=f\circ s_{k}\text{ on
}(-\infty,M)\supset\operatorname{supp}(f\circ s_{k}),\quad f\circ
q_{k}=0\text{ on }(M,M+1).$

In view of this, we take $h\in\mathcal{G}\cap C_{c}$ such that $h(x)\in(0,1)$,
$h(x)=1$ on $\operatorname{supp}(f\circ s_{k})$, and $h(x)=0$ on
$(M+1,\infty)$. It follows that $f\circ s_{k}=h\cdot(f\circ
q_{k})\in\mathcal{G}$.

Now we prove the claim. Let $p_{n}$ be a function in $E\cap\mathcal{G}$ with
$\left\lVert p_{n}\right\rVert_{\infty}=\left\lVert
h_{n}\right\rVert_{\infty}$. The regularity of $\mathcal{G}$ and the ladder
decomposition ensure the existence of such functions. Since
$A^{l}_{n},A^{r}_{n}\in H^{\alpha/2}$, it suffices to prove for any
$\varepsilon>0$,

  1. (1)

$A^{l}_{n}\wedge(\left\lVert
h_{n}\right\rVert_{\infty}-\varepsilon)\in\mathcal{G}$.

  2. (2)

$(A^{r}_{n})_{\varepsilon}\in\mathcal{G}$, where $(\cdot)_{\varepsilon}$ is
the $\varepsilon$-contraction defined in Section 2.

For (1), it is easy to see that there exists $g\in\mathcal{G}\cap C_{c}$ such
that $\operatorname{supp}(g)\subset(T^{h_{n}},T^{p_{n}})$ and
$g+h_{n}+p_{n}\geq\left\lVert h\right\rVert_{\infty}-\varepsilon$ on
$(T^{h_{n}},T^{p_{n}})$. So

$A^{l}_{n}\wedge(\left\lVert
h_{n}\right\rVert_{\infty}-\varepsilon)=(g+h_{n}+p_{n})\wedge(\left\lVert
h_{n}\right\rVert_{\infty}-\varepsilon)\in\mathcal{G}.$

For (2), we take $g\in\mathcal{G}\cap C_{c}$ such that $g=\left\lVert
h_{n}\right\rVert_{\infty}$ on $[T^{h_{n}},T^{p_{n}}]$ and $g\geq
h_{n}+p_{n}$. Then $g-h_{n}-p_{n}=A^{r}_{n}$ on
$\operatorname{supp}(A^{r}_{n})$. Note that $g-h_{n}-p_{n}=0$ on
$\partial\operatorname{supp}(A^{r}_{n})$. It follows that
$(g-h_{n}-p_{n})_{\varepsilon}=0$ on a neighbourhood
$\partial\operatorname{supp}(A^{r}_{n})$. So we can ‘discard’ the part of
$(g-h_{n}-p_{n})_{\varepsilon}$ in $\operatorname{supp}(A^{r}_{n})^{c}$ by
multiplying a function in $\mathcal{G}\cap C_{c}$. That deduces (2) and
completes the whole proof.

##  5\. Proof of Theorem 1.3

We have seen that for $1<\alpha\leq 2$ and
$s\in\mathbf{S}^{\alpha/2}_{\text{loc}}$, $\overline{\mathcal{F}_{s}}$ is a
regular subspace of $H^{\alpha/2}$. In this section, we shall further explore
the problem: for what kind of $s$, $\overline{\mathcal{F}_{s}}$ is proper? To
begin with, we recall some known results of the extended Dirichlet space of
$H^{\alpha/2}$.

For $0<\alpha\leq 2$, let $H^{\alpha/2}_{e}$ be the extended Dirichlet space
of $H^{\alpha/2}$. When $0<\alpha<1$, $H^{\alpha/2}_{e}$ is Hilbert due to the
transience of $H^{\alpha/2}$. When $1\leq\alpha<2$, it has the following
characterization (Cf. [CF12, §6.5]):

$\displaystyle\left\\{\begin{aligned} H^{\alpha/2}_{e}&=\left\\{f\in
L^{2}_{loc}(\mathbb{R}):\int_{\mathbb{R}^{2}}\dfrac{(f(x)-f(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y<\infty\right\\};\\\
\mathcal{E}(f,f)&=\int_{\mathbb{R}^{2}}\dfrac{(f(x)-f(y))^{2}}{|x-y|^{1+\alpha}}\mathrm{d}x\mathrm{d}y.\end{aligned}\right.$

The quotient space of $H^{\alpha/2}_{e}$ by the space of constant functions
$\mathcal{N}$, denoted by
$\mathring{H}^{\alpha/2}_{e}=H^{\alpha/2}_{e}/\mathcal{N}$, is a Hilbert
space.

In the following, we assume $\alpha\in(1,2)$ and prove Theorem 1.3, where we
exclude the case $\alpha=2$ which has been known. Recall that we denote
$\alpha^{*}=2-\alpha$ and assume all the quasi notions are relative to
$(H^{\alpha^{*}/2},\mathcal{E}^{\alpha^{*}/2})$ henceforth.

For any $g\in H^{\alpha/2}$, we define $Dg$ to be the derivative of $g$ in the
sense of distribution, i.e. for any $\phi\in C^{\infty}_{c}$,
$Dg(\phi)=-\int_{\mathbb{R}}g\phi^{\prime}\mathrm{d}x$. Then we have

(5.1) $\displaystyle\left\lvert
Dg(\phi)\right\rvert=\left\lvert\big{(}g,\phi^{\prime}\big{)}_{L^{2}(\mathbb{R})}\right\rvert=\left\lvert\big{(}(-\Delta)^{\alpha/4}g,(-\Delta)^{\alpha^{*}/4}\phi\big{)}_{L^{2}(\mathbb{R})}\right\rvert\leq\left\lVert
g\right\rVert_{\mathcal{E}^{\alpha/2}}\left\lVert\phi\right\rVert_{\mathcal{E}^{\alpha^{*}/2}},$

where $\Delta$ is the Laplacian and it holds that $\left\lVert
g\right\rVert_{\mathcal{E}^{\alpha/2}}=\left\lVert(-\Delta)^{\alpha/4}g\right\rVert_{L^{2}(\mathbb{R})}$.
It follows from (5.1) that $Dg$ can be generalized to an element in
$\big{(}H^{\alpha^{*}/2}_{e}\big{)}^{*}$ (the dual space of
$H^{\alpha^{*}/2}_{e}$), which we still denote by $Dg$. Since
$(-\Delta)^{\alpha^{*}/4}$ is self-adjoint and injective, we have
$(-\Delta)^{\alpha^{*}/4}(C^{\infty}_{c})$ is dense in $L^{2}(\mathbb{R})$.
Hence the operator norm of $Dg$ equals $\left\lVert
g\right\rVert_{\mathcal{E}^{\alpha/2}}$. Namely, $D$ is an isometry from
$H^{\alpha/2}$ to $\big{(}H^{\alpha^{*}/2}_{e}\big{)}^{*}$, which can be
further extended to an isometry from $\mathring{H}^{\alpha/2}_{e}$ to
$\big{(}H^{\alpha^{*}/2}_{e}\big{)}^{*}$. Now we take $g=f\circ
s\in\mathcal{F}_{s}$. Since $f\circ s$ is of BV, using the formula for
integration by parts, we have

(5.2) $\displaystyle D(f\circ
s)(\phi)=\int_{\mathbb{R}}\phi\,\mathrm{d}(f\circ s),\text{ for all $\phi\in
C^{\infty}_{c}$}.$

Combining with (5.1), we have the measure $\mathrm{d}(f\circ s)$ is of finite
energy integral relative to
$\big{(}H^{\alpha^{*}/2},\mathcal{E}^{\alpha^{*}/2}\big{)}$, which simply
leads to the smoothness of $\mathrm{d}s$. In particular, it doesn’t charge
sets of zero capacity. Based on this, we can generalize (5.2) as follows.
Recall that $\widetilde{\phi}$ represents the quasi continuous version of
$\phi$.

######  Lemma 5.1.

For any $f\circ s\in\mathcal{F}_{s}$,

(5.3) $\displaystyle D(f\circ
s)(\phi)=\int_{\mathbb{R}}\widetilde{\phi}\,\mathrm{d}(f\circ s),\text{ for
all $\phi\in H^{\alpha^{*}/2}_{e}$}.$

###### Proof.

For any bounded $\phi\in H^{\alpha^{*}/2}_{e}$, we take a sequence of
uniformly bounded functions $\\{\phi_{n}\\}\subset C^{\infty}_{c}$ such that
$\phi_{n}$ converges to $\phi$ in $\mathcal{E}^{\alpha^{*}/2}$ and $\phi_{n}$
converges q.e. to $\widetilde{\phi}$. Then by substituting $\phi_{n}$ in place
of $\phi$ in (5.2) and letting $n\uparrow\infty$, we reach (5.3) for bounded
$\phi$. The further genralization to any $\phi\in H^{\alpha^{*}/2}_{e}$ is
obtained by truncation. ∎

Now let us prove (d)$\Rightarrow$(a). Suppose $f\in H^{\alpha^{*}/2}$ is
nonzero and $\langle\widetilde{f},\mathrm{d}s\rangle=0$. Then by slightly
extending (5.1), we have for any $\phi\circ s\in\mathcal{F}_{s}$,

$\displaystyle\left\lvert\langle\widetilde{f},\mathrm{d}(\phi\circ
s)\rangle\right\rvert=\left\lvert\big{(}(-\Delta)^{\alpha/4}\phi\circ
s,(-\Delta)^{\alpha^{*}/4}f\big{)}_{L^{2}(\mathbb{R})}\right\rvert=\left\lvert\big{(}\phi\circ
s,(-\Delta)^{1/2-\alpha/2}f\big{)}_{\mathcal{E}^{\alpha/2}}\right\rvert=0,$

which deduces $\mathcal{F}_{s}$ is not dense in $H^{\alpha/2}$, since
otherwise we can approximate $(-\Delta)^{1/2-\alpha/2}f\in H^{\alpha/2}$ by
elements in $\mathcal{F}_{s}$, which implies

$\sup_{\phi\circ s\in\mathcal{F}_{s}}\left\lvert\big{(}\phi\circ
s,(-\Delta)^{1/2-\alpha/2}f\big{)}_{\mathcal{E}^{\alpha/2}}\right\rvert>0.$

For the proof of (a)$\Rightarrow$(d), we need another lemma. Recall that
$\mathcal{N}$ is the space of constant functions.

######  Lemma 5.2.

Suppose $\overline{\mathcal{F}_{s}}$ is a proper regular subspace of
$H^{\alpha/2}$. Let $\mathcal{G}$ be the extended Dirichlet space of
$\overline{\mathcal{F}_{s}}$ and
$\mathring{\mathcal{G}}:=(\mathcal{G}+\mathcal{N})/\mathcal{N}$. Then
$\mathring{\mathcal{G}}$ is a proper closed subspace of
$\mathring{H}^{\alpha/2}_{e}$.

###### Proof.

It is easy to see that $\overline{\mathcal{F}_{s}}$ is irreducible. So it is
either recurrent or transient. The proof proceeds separately in these two
cases.

If $\overline{\mathcal{F}_{s}}$ is recurrent, then
$\mathcal{N}\subset\mathcal{G}$, which implies $\mathring{\mathcal{G}}$ is
proper, since otherwise we have $\mathcal{G}=H^{\alpha/2}_{e}$ and

$\overline{\mathcal{F}_{s}}=\mathcal{G}\cap
L^{2}(\mathbb{R})=H^{\alpha/2}_{e}\cap L^{2}(\mathbb{R})=H^{\alpha/2}.$

Now let us show $\mathring{\mathcal{G}}$ is closed. Recall that
$\mathring{H}^{\alpha/2}_{e}$ is Hilbert. It suffices to prove that if
$\\{f_{n}\\}\subset\mathcal{G}$ converges in $\mathcal{E}$-norm to $f\in
H^{\alpha/2}_{e}$, then $f\in\mathcal{G}$. Note that the convergence is
equivalent to that $g_{n}(x,y):=f_{n}(x)-f_{n}(y)$ converges to
$g(x,y):=f(x)-f(y)$ in $L^{2}(\mu)$, where $\mu:=\left\lvert
x-y\right\rvert^{-(1+\alpha)}\mathrm{d}x\mathrm{d}y$. It follows that
$g_{n}(x,y)\rightarrow g_{n}(x,y)$ for a.e. $(x,y)$ (or a subsequence at
least). We can find some $x_{0}$ such that $g_{n}(x_{0},y)\rightarrow
g(x_{0},y)$ for a.e. $y$. Since $\mathcal{N}\subset\mathcal{G}$, without loss
of generality, we assume $f_{n}(x_{0})=f(x_{0})$ for all $n$. Then
$f_{n}(x)\rightarrow f(x)$ a.e.. Finally, we consider any approximating
sequence $\big{\\{}f^{(m)}_{n}:m\geq
1\big{\\}}\subset\overline{\mathcal{F}_{s}}$ of $f_{n}$. We can find a
sequence $\big{\\{}f^{(m_{n})}_{n}:n\geq 1\big{\\}}$ and $x_{1}\in\mathbb{R}$
such that $f^{(m_{n})}_{n}\rightarrow f$ in $\mathcal{E}$-norm and
$f^{(m_{n})}_{n}(x_{1})\rightarrow f(x_{1})$. Using the same arguments as
before, we have $f^{(m_{n})}_{n}(x)\rightarrow f(x)$ a.e.. Hence
$f\in\mathcal{G}$.

If $\overline{\mathcal{F}_{s}}$ is transient, then $\mathcal{G}$ contains no
nonzero constant functions and $\mathcal{G}$, $\mathring{\mathcal{G}}$ are
both Hilbert. It remains to show $\mathring{\mathcal{G}}$ is proper. The proof
goes by contradiction. Suppose
$\mathring{\mathcal{G}}=\mathring{H}^{\alpha/2}_{e}$. Then for each non-
negative $f\in C^{\infty}_{c}$, there exists a unique constant $c=c(f)$ such
that $f+c\in\mathcal{G}$. We shall verify that any possible case leads to a
contradiction.

  * •

If $c=0$ for all $f$, then $\overline{\mathcal{F}_{s}}=\mathcal{G}\cap
L^{2}(\mathbb{R})$ contains all non-negative functions in $C^{\infty}_{c}$. So
$\overline{\mathcal{F}_{s}}=H^{\alpha/2}$, which contradicts.

  * •

If $c>0$ for some $f$, then $c=c\wedge(c+f)\in\mathcal{G}$, which contradicts
the transience.

  * •

If $c<0$ for some $f$, we take $g\in\overline{\mathcal{F}_{s}}\cap C_{c}$ such
that $g+f\leq 0$. Then $c=c\vee(g+f+c)\in\mathcal{G}$. Again, it contradicts
the transience.

Therefore, it follows that $\mathring{\mathcal{G}}$ is proper. ∎

From the Lemma 5.2, there exists a non-zero $g\in\mathring{H}^{\alpha/2}_{e}$
such that $g\perp\mathcal{F}_{s}$ in $\mathcal{E}$-norm. Since $D$ is
isometric, it also holds that

$Dg\perp\\{\mathrm{d}(f\circ s):f\circ s\in\mathcal{F}_{s}\\}\text{ in
}\big{(}H^{\alpha^{*}/2}_{e}\big{)}^{*}.$

In particular, since $g$ is non-constant, $Dg$ is also nonzero. Thus, by Hahn-
Banach theorem and Lemma 5.1, there exists a nonzero $h\in
H^{\alpha^{*}/2}_{e}$ such that $\int_{\mathbb{R}}\widetilde{h}\,d(f\circ
s)=0$ for all $f\circ s\in\mathcal{F}_{s}$. Or equivalently,
$\langle\widetilde{h},\mathrm{d}s\rangle=0$. We further take any $\phi\in
C^{\infty}_{c}$ and set $q:=(h\wedge 1)\cdot\phi$. Then $q\in
H^{\alpha^{*}/2}_{e}\cap L^{2}(\mathbb{R})=H^{\alpha^{*}/2}$. It is nonzero
and satisfies $\langle\widetilde{q},\mathrm{d}s\rangle=0$.

Next, we shall further prove the equivalence of (b)$\sim$(d). For this, we
first present an equivalent condition for $\text{Cap}^{\alpha^{*}/2}$ to be
concentrated on a Borel set $E$.

######  Theorem 5.3 ([AH99, Theorem 11.3.2]).

Let $0<\alpha^{*}\leq 1$ and $E$ be a Borel set. The following are equivalent:

  1. (1)

$\text{Cap}^{\alpha^{*}/2}$ is not concentrated on $E$;

  2. (2)

there exists a nonzero $f\in H^{\alpha^{*}/2}$ such that $\widetilde{f}(x)=0$
q.e. on $E$.

###### Proof.

This theorem is proved in [AH99, Theorem 11.3.2], where it is stated in terms
of the Bessel $(\alpha^{*},2)$-capacity which is equivalent to
$\text{Cap}^{\alpha^{*}/2}$. For the sake of completeness, we supply here an
alternative proof with a probabilistic method. We write
$\text{Cap}=\text{Cap}^{\alpha^{*}/2}$ in this proof.

(1)$\Rightarrow$(2): Suppose there exists an open set $U$ such that
$\text{Cap}(E\cap U)<\text{Cap}(U)$. Let $u\in H^{\alpha^{*}/2}$ be the
capacitary function for $E\cap U$, that is characterized by:

$\widetilde{u}(x)=1\text{ q.e. on }E\cap U,\text{ and
}\mathcal{E}_{1}(u,u)=\text{Cap}(E\cap U).$

It holds that $u(x)<1$ on a subset of $U\setminus E$ of positive Lebesgue
measure, since otherwise $\text{Cap}(U\setminus E)=\text{Cap}(U)$. Then there
exists a compact set $K\subset U$ such that $K\cap\\{u<1\\}$ has positive
Lebesgue measure. We take any $\varphi\in C_{c}^{\infty}(\mathbb{R})$ with
$\varphi(x)=1$ on $K$ and $\operatorname{supp}(\phi)\in U$. Then
$f:=\varphi\cdot(1-u)=\varphi-\varphi\cdot u\in H^{\alpha^{*}/2}$ and it is
nonzero. Moreover, $\widetilde{f}=\varphi\cdot(1-\widetilde{u})$ vanishes q.e.
on $E$.

(2)$\Rightarrow$(1): Suppose $\text{Cap}(E\cap U)=\text{Cap}(U)$ for an open
set $U$. Denote by $\tau_{E\cap U}$ and $\tau_{U}$ the hitting times of $E\cap
U$ and $U$ by a symmetric stable process with parameter $\alpha^{*}$
respectively. Then $\mathbb{P}_{x}(e^{-\tau_{E\cap U}})$ (resp.
$\mathbb{P}_{x}(e^{-\tau_{U}})$) is a quasi-continuous version of the
capacitary function for $E\cap U$ (resp. $U$). We have
$\mathbb{P}_{x}(e^{-\tau_{E\cap U}})=\mathbb{P}_{x}(e^{-\tau_{U}})$ for q.e.
$x$, since otherwise $\text{Cap}(E\cap U)<\text{Cap}(U)$. It follows that for
q.e. $x\in U$ (indeed, for every $x$ considering the transition probability
has a density), $x$ is a regular point of $E\cap U$. Therefore, if $f\in
H^{\alpha^{*}/2}$ with $\widetilde{f}(x)=0$ q.e. on $E\cap U$, it also
vanishes q.e. at the regular points of $E\cap U$ due to the quasi continuity.
So $\widetilde{f}(x)=0$ q.e. on $U$. That immediately leads to the conclusion.
∎

Back to Theorem 1.3, (b)$\Leftrightarrow$(d) is immediate from Theorem 5.3 and
the fact that for any non-negative quasi continuous function $\widetilde{f}\in
H^{\alpha/2}$, $\langle\widetilde{f},\mathrm{d}s\rangle=0$ if and only if
$\widetilde{f}=0$ q.e. on the quasi support of $\mathrm{d}s$.
(b)$\Rightarrow$(c) is obtained by noting that the quasi support of
$\mathrm{d}s$ is a quasi closed set, whose capacity can be approximated by a
descending sequence of open sets. For (c)$\Rightarrow$(d), suppose $E$
satisfies the conditions in (c). By Theorem 5.3, there exists a nonzero $f\in
H^{\alpha^{*}/2}$ such that $\widetilde{f}=0$ q.e. on $E$. Then

$\langle\widetilde{f},\mathrm{d}s\rangle=\langle\widetilde{f}1_{E},\mathrm{d}s\rangle+\langle\widetilde{f}1_{E^{c}},\mathrm{d}s\rangle=0.$

Namely, $f$ satisfies the conditions in (d). That concludes the theorem.

######  Remark 5.4.

Almost all the conclusions of Theorem 1.2 and 1.3 can actually be generalized
to the case $\alpha\in(0,1]$. The only exception is the second part of Theorem
1.2 may not hold in the case $\alpha=1$. The proofs are basically the same as
that in Section 4 and Section 5. We mention that for the proof of
‘(a)$\Rightarrow$(d)’ in Theorem 1.3, one needs to occaasionally consider
$\mathring{H}^{\alpha^{*}/2}_{e}$ instead of $H^{\alpha^{*}/2}_{e}$. The
following are some consequences of this generalization.

  * •

For $\alpha\in(0,1)$, since all the functions in $H^{\alpha^{*}/2}$ is
continuous, using (a)$\Leftrightarrow$(d) in Theorem 1.3, we get
$H^{\alpha/2}$ has no proper subspace, which is another proof the ‘only if’
part of Theorem 1.1. However, this proof relies on the duality of
$H^{\alpha/2}$ and $H^{\alpha^{*}/2}$; while the proof in Section 3 can
actually be generalized to the Lévy processes with finite variation (see
Section 7 for details);

  * •

For $\alpha=1$, the corresponding versions of the first part of Theorem 1.2
and Theorem 1.3 allow us to construct the special proper regular subspaces of
$H^{1/2}$, which will be presented in the next section.

##  6\. Proof of the ‘if’ part of Theorem 1.1

In this section, we will construct special proper regular subspaces to
conclude the ‘if’ part of Theorem 1.1.

Due to Theorem 1.3 and Remark 5.4, it suffices to find
$s\in\mathbf{S}^{1}_{\text{loc}}$ satisfying (c) in Theorem 1.3. We focus on
the following type of $s$. Let $G$ be a dense open subset of $\mathbb{R}$.
Define

(6.1) $\displaystyle s(x):=\int_{0}^{x}1_{G}(y)\mathrm{d}y,\ x\in\mathbb{R}.$

Then $s\in\mathbf{S}^{1}_{\text{loc}}$ and $\mathrm{d}s$ is concentrated on
$G$. It is known (Cf. [AH99, Chapter 5]) that for any interval
$I_{r}:=(-r,r)$, there exists $C,A,a>1$ such that

$\displaystyle\begin{cases}C^{-1}r^{\alpha-1}\leq\text{Cap}^{\alpha^{*}/2}(I_{r})\leq
Cr^{\alpha-1},&\text{ for }1<\alpha<2;\\\
A^{-1}(\log(a/r))^{-1}\leq\text{Cap}^{\alpha^{*}/2}(I_{r})\leq
A(\log(a/r))^{-1},&\text{ for }\alpha=1.\end{cases}$

Hence for any $\alpha\in(1,2)$ (resp. $\alpha=1$), we can take $r_{i}$, $i\geq
1$, small enough, such that

$\sum_{i}Cr_{i}^{\alpha-1}<\text{Cap}^{\alpha^{*}/2}(I_{1})\,(resp.\,\sum_{i}A(\log(a/r_{i}))^{-1}<\text{Cap}^{1/2}(I_{1})).$

Define $G=G(\alpha)=[-1,1]^{c}\cup\bigcup_{i\geq 1}(x_{i}+I_{r_{i}})$, where
$\\{x_{i}\\}_{i\geq 1}$ is a dense subset of $I_{1}$. Then $G$ is dense and
$\text{Cap}^{\alpha^{*}/2}(G\cap I_{1})<\text{Cap}^{\alpha^{*}/2}(I_{1})$. It
implies that the corresponding $s$ satisfies (c) in Theorem 1.3. We have thus
proved the theorem.

##  7\. Generalizations to Lévy processes

The characteristic exponent of a symmetric Lévy process on $\mathbb{R}$ is

$\psi(\xi)=\frac{1}{2}\sigma\xi^{2}+\int_{\mathbb{R}}(1-\cos(\xi
x))\nu(\mathrm{d}x),$

where $\sigma\geq 0$, $\nu$ is a symmetric Radon measure on
$\mathbb{R}\setminus\\{0\\}$ with integrability

$\int_{\mathbb{R}}(1\wedge x^{2})\nu(\mathrm{d}x)<\infty.$

It is known that a Lévy process has finite variation, i.e. its sample path has
bounded variation over any finite interval a.s., if and only if $\sigma=0$ and
$\int_{\mathbb{R}}(1\wedge|x|)\nu(\mathrm{d}x)<\infty$. The latter is
equivalent to for any $\lambda>0$,

$\int_{\mathbb{R}}(\lambda\wedge|x|)\nu(\mathrm{d}x)<\infty.$

Assume $\sigma=0$ in the first part of this section and then the Dirichlet
form of the process can be written as

$\displaystyle\left\\{\begin{aligned} &\mathcal{F}=\left\\{f\in
L^{2}(\mathbb{R}):\int_{\mathbb{R}}\big{|}\widehat{f}(\xi)\big{|}^{2}\psi(\xi)d\xi<\infty\right\\};\\\
&\mathcal{E}(f,g)=\frac{1}{2}\int_{\mathbb{R}^{2}}(f(x)-f(y))(g(x)-g(y))\nu(\mathrm{d}x-y)\mathrm{d}y=\int_{\mathbb{R}}\widehat{f}(\xi)\bar{\widehat{g}}(\xi)\psi(\xi)\mathrm{d}\xi.\end{aligned}\right.$

In particular, $C_{c}^{\infty}\cap\mathcal{F}$ is a core of
$(\mathcal{E},\mathcal{F})$.

The following lemma, obtained by direct computation, generalizes Lemma 3.1.

######  Lemma 7.1.

For the indicator function $f=1_{[a,b]}$ with $a<b$, it holds that

$\int_{\mathbb{R}^{2}}(f(x)-f(y))^{2}\nu(\mathrm{d}x-y)\mathrm{d}y=2\int_{\mathbb{R}}((b-a)\wedge|x|)\nu(\mathrm{d}x).$

Hence $1_{[a,b]}\in\mathcal{F}$ if and only if
$\displaystyle{\int_{\mathbb{R}}(1\wedge|x|)\nu(dx)<\infty}$.

Then the following theorem is quite anticipated.

######  Theorem 7.2.

Any 1-dim symmetric Lévy process with finite variation has no proper regular
subspaces.

###### Proof.

The proof of the ‘only if’ part of Theorem 1.1 also works in this case, if we
can check the analogs of Lemma 3.2 and 3.3 for the Lévy process with finite
variation. Therefore, we only need to verify the analogs of Lemma 3.2 and 3.3,
i.e.

  1. (i)

$S_{0}$ is dense in $\mathcal{F}$;

  2. (ii)

with the same notations as in Section 3, $h_{\rho}\in\mathcal{F}$ for
$\rho>0$, and $\\{\mathcal{E}(h_{\rho},h_{\rho}):\rho>0\\}$ is bounded.

For the proof of (i), because it is known that
$C_{c}^{\infty}(\mathbb{R})\cap\mathcal{F}$ is dense in $\mathcal{F}$ in
$\mathcal{E}_{1}$-norm, we need only to prove that $S_{0}$ is dense in
$C^{\infty}_{c}\cap\mathcal{F}$ in $\mathcal{E}_{1}$-norm. Take $f\in
C_{c}^{\infty}\cap\mathcal{F}$ with supp$(f)\subset(-N,N)$ and
$|f^{\prime}|\leq c$ and $f_{n}$ is defined as in Lemma 3.2. Write
$g_{n}=f-f_{n}$, then $|g_{n}(x)|\leq c/2^{n}$ for all $x\in\mathbb{R}$. It
suffices to show that $\mathcal{E}(g_{n},g_{n})\longrightarrow 0$.

The idea is similar to the proof of Lemma 3.2 with a slight difference.

$\displaystyle
I:=\int_{\mathbb{R}^{2}}((g_{n})(x)-(g_{n})(y))^{2}\nu(\mathrm{d}x-y)\mathrm{d}y$

and we can also divide it into three parts.

The first part is similar to the Lemma 3.2 and we have

$\displaystyle I^{\prime}_{1}$
$\displaystyle=2\int_{[-N,N]\times(\mathbb{R}\backslash[-N,N])}(g_{n}(x)-g_{n}(y))^{2}\nu(\mathrm{d}x-y)\mathrm{d}y$
$\displaystyle\leq
2\int_{[-N,N]\times(\mathbb{R}\backslash[-N,N])}{\Big{(}c\frac{1}{2^{n}}\Big{)}^{2}}\nu(\mathrm{d}x-y)\mathrm{d}y=\dfrac{2c^{2}}{2^{2n}}\int_{\mathbb{R}}(|x|\wedge
2N)\nu(\mathrm{d}x).$

The second part is the integral on $[-N,N]\times[-N,N]\setminus D_{n}$ where

$D_{n}=\\{|x-y|>1/2^{n},|x|<N,|y|<N\\},$

and we have

$\displaystyle I^{\prime}_{2}$
$\displaystyle=\int_{[-N,N]\times[-N,N]\setminus
D_{n}}(g_{n}(x)-g_{n}(y))^{2}\nu(dx-y)dy$
$\displaystyle\leq(2c/2^{n})^{2}\int_{\frac{1}{2^{n}}\leq|x-y|\leq 2N,|y|\leq
N}\nu(\mathrm{d}x-y)dy\leq\dfrac{4c^{2}}{2^{2n}}\int_{\frac{1}{2^{n}}\leq|x|\leq
2N}2N\nu(dx)$
$\displaystyle=\dfrac{8Nc^{2}}{2^{n}}\int_{\frac{1}{2^{n}}\leq|x|\leq
2N}\dfrac{1}{2^{n}}\nu(dx)\leq\dfrac{8Nc^{2}}{2^{n}}\int_{-2N}^{2N}|x|\nu(dx).$

The third part is the integral on $D_{n}$ which is the union
$\displaystyle{\bigcup_{-N2^{n}\leq
i<N2^{n}}\Big{[}\dfrac{i}{2^{n}},\dfrac{i+1}{2^{n}}\Big{]}^{2}}$ and its
complement. It is essential for the former that

$|g_{n}(x)-g_{n}(y)|=|f(x)-f(y)|\leq c|x-y|\ \text{for}\
x,y\in\Big{[}\dfrac{i}{2^{n}},\dfrac{i+1}{2^{n}}\Big{]},$

and hence it follows that

$\displaystyle I^{\prime}_{3}$ $\displaystyle\leq\sum_{-N2^{n}\leq
i<N2^{n}}\int_{x,y\in[\frac{i}{2^{n}},\frac{i+1}{2^{n}}]}c^{2}(x-y)^{2}\nu(dx-y)dy$
$\displaystyle\ \ +2\sum_{-N2^{n}\leq
i<N2^{n}}\int_{x\in[\frac{i+1}{2^{n}},\frac{i+2}{2^{n}}],y\in[\frac{i}{2^{n}},\frac{i+1}{2^{n}}],x<y+\frac{1}{2^{n}}}(2c/2^{n})^{2}\nu(dx-y)dy$
$\displaystyle\leq
2N2^{n}\left(\dfrac{c^{2}}{2^{n}}\int_{|x|\leq\frac{1}{2^{n}}}x^{2}\nu(dx)+\dfrac{4c^{2}}{2^{2n}}\int_{|x|\leq\frac{1}{2^{n}}}|x|\nu(dx)\right).$

By the assumption, this concludes that
$I=I^{\prime}_{1}+I^{\prime}_{2}+I^{\prime}_{3}$ converges to $0$.

Finally, we prove (ii). Similarly, we shall compute the $\mathcal{E}$-norm of
$h_{\rho}$. It follows from the inequality (3.6) that

$|\hat{h}_{\rho}(\xi)|\leq 2(|\xi|^{-1}\wedge 1).$

Combining $0\leq 1-\cos y\leq 2(1\wedge y^{2})$, we have

$\displaystyle\mathcal{E}(h_{\rho},h_{\rho})=$
$\displaystyle\int_{\mathbb{R}}|\hat{h}_{\rho}(\xi)|^{2}|\psi(\xi)|\mathrm{d}\xi$
$\displaystyle\leq$ $\displaystyle 4\int_{\mathbb{R}^{2}}(\xi^{-2}\wedge
1)(1-\cos(\xi x))\mathrm{d}\xi\nu(\mathrm{d}x)$ $\displaystyle\leq$
$\displaystyle 8\int_{\mathbb{R}}\nu(dx)\int_{\mathbb{R}}(\xi^{-2}\wedge
1)(1\wedge(\xi x)^{2})\mathrm{d}\xi$ $\displaystyle=$ $\displaystyle
8\int_{\mathbb{R}}\left((2-\dfrac{2}{3|x|})1_{\\{|x|\geq
1\\}}+(2-\dfrac{2}{3}|x|)|x|1_{\\{|x|<1\\}}\right)\nu(\mathrm{d}x)$
$\displaystyle\leq$ $\displaystyle
16\int_{\mathbb{R}}(1\wedge|x|)\nu(\mathrm{d}x),$

which is a finite number independent of $\rho$. Now the other parts of the
proof are exactly the same. We mention that since (ii) already shows
$h_{\rho}\in\mathcal{F}$, so we do not need the ‘Step 1’ part in the proof of
Lemma 2.1. ∎

We next generalize the other part of Theorem 1.1. In this part, we assume
$\sigma\geq 0$.

######  Corollary 7.3.

If there exist some positive constants $c$ and $M$ such that $\psi(\xi)\geq
c\left\lvert\xi\right\rvert^{\alpha}$,$\forall\,\left\lvert\xi\right\rvert\geq
M$ for some $\alpha\in[1,2]$, then the Lévy process with characteristic
exponent $\psi$ has proper regular subspaces. In particular, when the
parameter of quadratic term $\sigma>0$, the corresponding process has proper
regular subspaces.

###### Proof.

We again consider the scale function in the form of (6.1). It is easy to see
that $\overline{\mathcal{F}_{s}}^{\mathcal{E}_{1}}$ is a regular subspace of
$\mathcal{F}$. Considering
$\overline{\mathcal{F}_{s}}^{\alpha/2}\subset\overline{\mathcal{F}_{s}}^{\mathcal{E}_{1}}$,
we can check that (b)$\Leftrightarrow$(c)$\Leftrightarrow$(d)$\Rightarrow$(a)
still holds in this case with $\alpha$ given by the statement of the
corollary. Then the conclusion follows from the proof in Section 6. ∎

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