TITLE: Which notation to use to express summing probabilities with increasing conditionals?
QUESTION [0 upvotes]: How to write this:
$$
\Pr(x_1) + \Pr(x_2|x_1) + \Pr(x_3|x_1,x_2) + \ldots + \Pr(x_n | x_1,x_2,\ldots,x_{n-1})
$$
but by using the $\sum$ notation?
I'm trying to write an entropy equation with growing conditionals in $\ldots$:
$$
- \sum_{i\in\{1,2,\ldots,n\}} \Pr(x_i | \ldots) \ln[\Pr(x_i | \ldots)]
$$
but I don't know how to do express it.

REPLY [2 votes]: How about this?  Not ideal, but seems clear enough:
$$\sum\limits_{i=1}^n {\rm Pr}(x_i|x_1, \ldots, x_{i-1})$$


text:
/-
Copyright (c) 2019 Kevin Kappelmann. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kevin Kappelmann
-/
import algebra.continued_fractions.basic
import algebra.continued_fractions.translations
import algebra.continued_fractions.continuants_recurrence
import algebra.continued_fractions.terminated_stable
import algebra.continued_fractions.convergents_equiv
import algebra.continued_fractions.computation
/-!
# Default Exports for Continued Fractions
-/

subset_name: curated


text:
\chapter{Besov spaces related to a Dirichlet space}\label{Sec:Dirichlet}

\section{Definitions}

Let $(X,\mu,\mathcal{E},\mathcal{F}=\mathbf{dom}(\mathcal{E}))$ be a symmetric Dirichlet space,
that is, $X$ is a topological
measure space equipped with the Radon measure $\mu$, $\mathcal{E}$ a closed Markovian bilinear form on
$L^2(X,\mu)$, and $\mathcal{F}$ the collection of all functions $u\in L^2(X,\mu)$ with $\mathcal{E}(u,u)$ finite. The book~\cite{FOT} is a standard reference on the theory of Dirichlet forms.
Let $\{P_{t}\}_{t\in[0,\infty)}$ denote the Markovian semigroup associated with
$(X,\mu,\mathcal{E},\mathcal{F})$. From classical theory (see for instance Theorems 1.4.1 and 1.4.2 in \cite{EBD}), the semigroup $P_t$ lets $L^1(X,\mu) \cap L^\infty (X,\mu)$ invariant and may be extended to a positive, contraction semigroup on $L^p (X,\mu)$, $1 \le p \le +\infty$, that we shall still denote by $P_t$. Moreover, for $1 \le p <+\infty$, $P_t$ is strongly continuous and for $1<p<+\infty$, $P_t$ is a bounded analytic semigroup on $L^p (X,\mu)$ with angle $\theta_p=\frac{\pi}{2} \left(  1- \left| \frac{2}{p}-1 \right| \right)$.  In this monograph, we always assume that $P_t$ is conservative, i.e. $P_t 1=1$.

Our basic definition of the Besov seminorm is the following:
\begin{definition}
Let $p \ge 1$ and $\alpha \ge 0$. For $f \in L^p(X,\mu)$, we define the Besov  seminorm:
\[
\| f \|_{p,\alpha}= \sup_{t >0} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}.
\]
\end{definition}

Observe that if $P_t$ admits a heat kernel $p_t(x,y)$, then
\[
 \int_X P_t (|f-f(y)|^p)(y) d\mu(y)= \int_X \int_X |f(x)-f(y) |^p p_t (x,y) d\mu(x) d\mu(y) ,
\]
but the seminorm $\| \cdot \|_{p,\alpha}$ is well defined even if $P_t$ does not have a heat kernel.

Our goal in this work is to study the Besov  space
\begin{align}\label{eq:def:Besov}
\mathbf{B}^{p,\alpha}(X)=\{ f \in L^p(X,\mu)\, :\,  \| f \|_{p,\alpha} <+\infty \},
\end{align}
and specially $\mathbf{B}^{1,\alpha}(X)$ and its connection to the notion of bounded variation function. The norm on $\mathbf{B}^{p,\alpha}(X)$ is defined as:
\[
\| f \|_{\mathbf{B}^{p,\alpha}(X)} =\| f \|_{L^p(X,\mu)} + \| f \|_{p,\alpha}.
\]

\begin{example}
If $X=\mathbb{R}^n$ and $\mathcal{E}$ is the standard Dirichlet form on $\mathbb{R}^n$, that is, for $f,g \in W^{1,2}(\mathbb{R}^n)$
we have
\[
\mathcal{E}(f,g) =\int_{\mathbb{R}^n} \langle \nabla f (x) ,\nabla g (x) \rangle dx, \quad f,g \in W^{1,2}(\mathbb{R}^n),
\]
then $\mathbf{B}^{p,\alpha}(X)$ coincides with the  Besov-Nikol'skii space $B^{2\alpha}_{p,\infty} (\mathbb{R}^n)$ that consists of functions $f \in L^p(\mathbb{R}^n,dx)$ such that
\[
\sup_{h \in \mathbb{R}^n, h \neq 0} \frac{ \| f( \cdot+h)-f(\cdot) \|_p}{h^{2\alpha}} <+\infty.
\]
We refer for instance to \cite{AS} and \cite{Taibleson} (Theorems 4 and 4*)  for several equivalent descriptions of those spaces. In particular, we observe therefore that $\mathbf{B}^{1,1/2}(\mathbb{R}^n)$ is the space of bounded variation functions $BV(\mathbb{R}^n)$ and that $\mathbf{B}^{2,1/2}(\mathbb{R}^n)=W^{1,2}(\mathbb{R}^n)$ is the domain of $\mathcal{E}$. The observation $\mathbf{B}^{1,1/2}(\mathbb{R}^n)=BV(\mathbb{R}^n)$ will be generalized in the framework of metric spaces with a weak Bakry-\'Emery type non-negative curvature condition and the observation $\mathbf{B}^{1,1/2}(\mathbb{R}^n)=\mathcal{F}$ is a general fact, see Proposition \ref{prop:energyasbesov}
\end{example}

\section{A preliminary class of examples}

Before discussing general properties of the spaces $\mathbf{B}^{p,\alpha}(X)$, it may be instructive to quickly look at a preliminary large class of examples in a smooth setting.

\begin{proposition}\label{Besov Rn}
Assume that $X$ is a smooth manifold. Let $L=V_0+\sum_{i=1}^d V_i^2$ be a H\"ormander's type operator on $X$, where the $V_i$'s are smooth vector fields. Let us assume that $L$ is essentially self-adjoint on $C_0^\infty(X)$ in $L^2(X,\mu)$ for some Radon measure $\mu$ on $X$. Consider the Dirichlet space $(X,\mu, \mathcal{E},\mathcal{F})$ obtained by closing the pre-Dirichlet form
\[
\mathcal{E}(f,g)=\int_X \Gamma(f,g) d\mu(x), \quad f,g \in C_0^\infty(X),
\]
where $\Gamma(f,g)$ is the carr\'e du champ operator defined by $\Gamma(f,g)=\frac{1}{2}( L(fg)-fLg -gLf)$. Assume that the associated semigroup $P_t$ is conservative.
Then, for every $p \ge 1$, $$C_0^\infty (X) \subset \mathbf{B}^{p,1/2}(X)$$ and one has for every $f \in C_0^\infty (X) $, and open set $A \subset X$,
\[
\lim_{t \to 0} t^{-1/2} \left( \int_A P_t (|f-f(x)|^p)(x) d\mu(x) \right)^{1/p} = 2\left( \frac{ \Gamma \left( \frac{1+p}{2} \right) }{\sqrt
{\pi}}\right)^{1/p} \left( \int_A \Gamma ( f,f) (x)^{p/2} d\mu(x)\right)^{1/p}.
\]
\end{proposition}
\begin{remark}
In the previous setting one has therefore for $f \in C_0^\infty(X)$,
\[
\left( \int_X \Gamma ( f,f) (x)^{p/2} d\mu(x)\right)^{1/p} \le C_p \| f \|_{p,1/2}.
\]
If $P_t$ satisfies moreover the  Bakry-\'Emery estimate $ \sqrt{\Gamma(P_t f)} \le C P_t \sqrt{\Gamma( f)} $, then we will see (Chapter \ref{Sec:Metric-Curvature}, Section 4.5) that we have a converse inequality
\[
\| f \|_{p,1/2} \le c_p \left( \int_X \Gamma ( f,f) (x)^{p/2} d\mu(x)\right)^{1/p},
\]
for $p=1$.
\end{remark}

\begin{remark}
Proposition \ref{Besov Rn} indicates that at a high level of generality, one may  expect the  Besov spaces $ \mathbf{B}^{p, 1/2}(X)$, $1 \le p <+\infty$  to be closely related to the various notions of Sobolev spaces that have been defined on metric measure spaces (see for instance \cite{Shan2000}). While in this work we shall only be concerned with the study of all the Besov  spaces $\mathbf{B}^{p,\alpha}(X)$, the comparison between Sobolev spaces and Besov spaces will be made in Chapter 4 in the framework of Dirichlet spaces with absolutely continuous energy measures. In the framework of Chapter 5,  it will be interesting to compare our results with the recent preprint \cite{HinzKochMeinert} on Sobolev spaces and calculus of variations on fractals. Such a comparison will be the subject of future study.
\end{remark}

\begin{example}
If $X=\mathbb{R}^n$ and $\mathcal{E}$ is the standard Dirichlet form on $\mathbb{R}^n$, it is natural to expect that for every $p \ge 1$, and every $f \in \mathbf{B}^{p,1/2}(X)$
\[
 \lim_{t \to 0} t^{-1/2} \left( \int_{\mathbb{R}^n} P_t (|f-f(x)|^p)(x) dx \right)^{1/p}=2\left( \frac{ \Gamma \left( \frac{1+p}{2} \right) }{\sqrt
{\pi}}\right)^{1/p} \left( \int_{\mathbb{R}^n} | \nabla f (x)|^p dx\right)^{1/p}.
\]
The case $p=1$ is proved in \cite{MPPP}, but we did not find it in the literature for $p>1$, $p \neq 2$.
\end{example}

\begin{proof} [Proof of Proposition \ref{Besov Rn}.]
We use here a probabilistic argument. For $x \in X$, we denote by $(B_t^x)_{t \ge 0}$ the $L$-Brownian motion on $X$ started from $x$, that is the diffusion with generator $L$. It can be constructed as the solution of a stochastic differential equation in Stratonovich form:
\[
dB_t^x=V_0(B_t^x)dt +\sqrt{2}\sum_{i=1}^d V_i(B_t^x)\circ d\beta^i_t
\]
where $\beta$ is a $d$-dimensional Brownian motion.
Let $f \in C_0^\infty(X)$. The process
\[
M_t^f =f(B_t^x) -f(x) -\int_0^t L f(B_s^x) ds
\]
is a square integrable martingale that can be written
\[
M_t^f=\sqrt{2} \sum_{i=1}^d \int_0^t (V_i f)(B_s^x)d\beta^i_s.
\]
We have then
\begin{align*}
P_t (|f-f(x)|^p)(x) & =\mathbb{E} \left( | f(B_t^x) -f(x) |^p\right) \\
 &=\mathbb{E} \left( \left| M_t^f +\int_0^t L f(B_s^x) ds \right|^p \right).
\end{align*}
Observe now that $\frac{1}{\sqrt{t}} \int_0^t L f(B_s^x) ds$ almost surely converges to 0 when $t \to 0$. Therefore, $\frac{1}{\sqrt{t}} M_t^f$ converges in all $L^p$'s to the Gaussian random variable $\sqrt{2} \sum_{i=1}^d (V_i f) (x) \beta^i_1$. Since $f$ has a compact support, one deduces that
\[
\lim_{t \to 0} t^{-1/2} \left( \int_A P_t (|f-f(x)|^p)(x) d\mu(x) \right)^{1/p} = C_p \left( \int_A \Gamma ( f,f) (x)^{p/2} d\mu(x)\right)^{1/p},
\]
with $C_p=\sqrt{2} \mathbb{E}(| N|^p)^{1/p}=2\left( \frac{ \Gamma \left( \frac{1+p}{2} \right) }{\sqrt
{\pi}}\right)^{1/p} $, where $N$ denotes a Gaussian random variable with mean 0 and variance 1.
\end{proof}

\section{Basic  properties of the Besov spaces}

\begin{proposition}
 For $p\ge 1$ and $\alpha \ge 0$, $\mathbf{B}^{p,\alpha}(X)$ is a Banach space.
 \end{proposition}

\begin{proof}
Let $f_n$ be a Cauchy sequence in $\mathbf{B}^{p,\alpha}(X)$. Let $f$ be the $L^p$ limit of $f_n$. From Minkowski inequality and conservativeness of $P_t$, one has
\begin{align*}
 &  \left| \left(\int_X P_t (|f_n-f_n(y)|^p)(y) d\mu(y) \right)^{1/p} - \left(\int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \right| \\
 \le &  \left(\int_X P_t (|(f_n-f)-(f_n(y)-f(y))|^p)(y) d\mu(y) \right)^{1/p} \\
 \le &\left(\int_X P_t (|f_n-f|^p)(y) d\mu(y) \right)^{1/p}+\left(\int_X P_t (|f_n(y)-f(y)|^p)(y) d\mu(y) \right)^{1/p} \\
 \le &2 \| f-f_n \|_{L^p(X,\mu)} .
\end{align*}
Therefore
\[
\lim_{n \to +\infty}  \left(\int_X P_t (|f_n-f_n(y)|^p)(y) d\mu(y) \right)^{1/p}=  \left(\int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p},
\]
from which we deduce that
\[
\left(\int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \le  t^\alpha \lim_{n \to +\infty} \| f_n \|_{p,\alpha}.
\]
Therefore $f \in \mathbf{B}^{p,\alpha}(X)$ and $\| f \|_{p,\alpha} \le \lim_{n \to +\infty} \| f_n \|_{p,\alpha}$. Similarly,
\[
\| f-f_m \|_{p,\alpha} \le  \lim_{n \to +\infty} \| f_n -f_m \|_{p,\alpha}
\]
and taking the limit $m \to +\infty$ finishes the proof.

\end{proof}

The following basic observation shall be useful in the sequel:

\begin{lemma}\label{Lemma limsup debut}
Let $p\ge1, \alpha \ge 0$.
\[
\mathbf{B}^{p,\alpha}(X)
=\left\{ f \in L^p(X,\mu)\, :\,  \limsup_{t \to 0} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} <+\infty \right\}.
\]
In particular if $\beta > \alpha$, $\mathbf{B}^{p,\beta}(X) \subset \mathbf{B}^{p,\alpha}(X) $.
Moreover, for $f \in \mathbf{B}^{p,\alpha}(X)$, one has for every $t >0$,
\[
\| f \|_{p,\alpha} \le \frac{2}{t^\alpha} \| f \|_{L^p(X)} +\sup_{s\in (0,t]} s^{-\alpha} \left( \int_X P_s (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}.
\]
\end{lemma}

\begin{proof}
Obviously, if $f \in \mathbf{B}^{p,\alpha}(X)$, then
\[
\limsup_{t \to 0} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \le \| f \|_{p,\alpha}.
\]
Conversely, if $\limsup_{t \to 0} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} <+\infty$, then, for some $\ve >0$,
\[
\sup_{t \in (0,\ve]} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} <+\infty
\]
For $t >\ve$, since $|f(x)-f(y)|^p \le 2^{p-1} (|f(x)|^p+|f(y)|^p)$ and the semigroup is conservative (and hence
$P_t1(x)=1$ for all $x\in X$), one has then
\begin{align*}
 t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \le 2 \ve^{-\alpha} \| f \|_{L^p(X)}.
\end{align*}
\end{proof}

Both the following lemma and Proposition~\ref{prop:energyasbesov} are well-known, but we do not know where in the literature they appear in this form. In the standard source~\cite{FOT} the corresponding results rely on a topological assumption, see footnote on page~ of~\cite{FOT}. The argument given below is similar to that in Lemma~2.3.2.1 of~\cite{BouleauHirsch}.
\begin{lemma}\label{lem:energyasbesov}
For $t>0$, $f\in L^2(X,\mu)$ and a bounded function $g$, writing $\langle\cdot,\cdot\rangle$ for the $L^2(X,\mu)$ inner product
\begin{equation*}
 \frac1t \int_X g(y)P_t(|f-f(y)|^2)(y)\,d\mu(y) = \frac2t\langle (I-P_t)f, fg \rangle - \frac1t\langle (I-P_t) f^2, g\rangle.
\end{equation*}
\end{lemma}
\begin{proof}
Consider a simple function $f=\sum_{j=1}^n a_j \mathbf{1}_{A_j}$ where the $A_j$ are pairwise disjoint and measurable and compute (using pairwise disjointness and the symmetry of $P_t$ on $L^2$) that
\begin{align*}
 \lefteqn{\int_X g(y) P_t(|f-f(y)|^2) \, d\mu(y)}\quad&\\
 &= \sum_{j,k,l,m} \int \biggl( a_j a_l P_t \bigl(\mathbf{1}_{A_j} \mathbf{1}_{A_l}\bigr)(y) - a_j a_m P_t\bigl(\mathbf{1}_{A_j}\bigr)(y)\mathbf{1}_{A_m}(y) \\
 &\qquad- a_k a_l \mathbf{1}_{A_k}(y) P_t\bigl(\mathbf{1}_{A_l}\bigr)(y) + a_k a_m\mathbf{1}_{A_k}(y)\mathbf{1}_{A_m}(y) \biggr) g(y)\,d\mu(y)\\
 &= \sum_{j} a_j^2 \biggl( \langle P_t \mathbf{1}_{A_j},g \rangle +\langle \mathbf{1}_{A_j},g\rangle \biggr)- 2\sum_{j,l} a_j a_l \langle P_t\mathbf{1}_{A_j},g\mathbf{1}_{A_l}\rangle\\
 &= \sum_{j,l} a_j a_l \Bigl( 2\langle (I-P_t)\mathbf{1}_{A_j},g\mathbf{1}_{A_l}\rangle - \langle (I-P_t)(\mathbf{1}_{A_j}\mathbf{1}_{A_l}),g\rangle \Bigr)\\
 &=2\langle (I-P_t)f, fg \rangle - \langle (I-P_t) f^2, g\rangle.
 \end{align*}
The result then follows by density of simple functions in $L^2(X,\mu)$ and $L^p$ boundedness of $P_t$ for $t>0$ for $p=1,2$; for the former see~\cite[page 33 ]{FOT}.
\end{proof}

\begin{proposition}\label{prop:energyasbesov}
$2\DF(f)=\|f\|_{2,1/2}^2$ and hence\/ $\mathbf{B}^{2,1/2}(X)=\mathcal{F}$.
\end{proposition}
\begin{proof}
Apply Lemma~\ref{lem:energyasbesov} with $g\equiv1$. Since $(I-P_t)$ is symmetric and $(I-P_t)1=0$ this simplifies to
\begin{equation}\label{prop:energyasbesoveq}
 \frac1{2t} \int_X P_t(|f-f(x)|^2)\, d\mu(x)= \frac1t \langle (I-P_t)f,f\rangle.
 \end{equation}
However it is standard (see~\cite{BouleauHirsch} Proposition~1.2.3) that the right side of~\ref{prop:energyasbesoveq} is positive and decreasing in $t$ and has limit $\DF(f)$ as $t\downarrow0$ if $f\in\mathcal{F}$, from which this limit is the supremum and the result follows.
\end{proof}

\begin{proposition}\label{prop:BesovCEp}
Assume that $\mathcal{E}$ has the property that $\mathcal{E}(f,f)=0$ implies $f$ constant. Then, any $f \in \mathbf{B}^{p,\alpha }(X)$ with $1 \le p \le 2$ and $\alpha >1/p$ is constant.
\end{proposition}

\begin{proof}
Let $f \in \mathbf{B}^{p,\alpha }(X)$ with $1 \le p \le 2$. For $n \ge 0$, one considers $f_n(x) =f(x)$ if $|f(x)| \le n$ and $f_n(x)=0$ otherwise. Since $|f_n(x) -f_n(y)| \le |f(x) -f(y)|$ for every $x,y \in X$, it is clear that $f_n \in \mathbf{B}^{p,\alpha }(X)$. One has then
\[
P_t(|f_n-f_n(x)|^2) =P_t(|f_n-f_n(x)|^{2-p} |f_n-f_n(x)|^p)  \le 2^{2-p} \| f_n \|^{2-p}_{\infty} P_t(|f_n-f_n(x)|^p).
\]
Therefore
\[
 \frac1{2t} \int_X P_t(|f_n-f_n(x)|^2) d\mu(x) \le  2^{1-p} t^{\alpha p -1}\| f_n \|^{2-p}_{\infty} \| f_n\|_{p,\alpha}^p.
 \]
 This implies
 \[
 \lim_{t \to 0} \frac1{2t} \int_X P_t(|f_n-f_n(x)|^2) d\mu(x)  =0.
 \]
 Thus $f_n \in \mathcal{F}$ and $\mathcal{E}(f_n,f_n)=0$. This implies that $f_n$ is constant for every $n$, thus $f$ is constant.

 \end{proof}

\begin{proposition}
If $\DF$ is regular and $f\in\mathbf{B}^{p,1/2}(X)$ for $p>2$ then the energy measure $\nu_f$ (in the sense of Beurling-Deny) is absolutely continuous with respect to $\mu$.
\end{proposition}
\begin{proof}
Let $p>2$, $f \in \mathbf{B}^{p,1/2}(X)$ and $g \in L^{p/(p-2)}(X,\mu) \cap L^\infty (X,\mu)\cap \mathcal{F}$, $g \ge 0$.   From H\"older's inequality we have
\begin{align*}
 \frac1t\int_X |g(y)| P_t(|f-f(y)|^2)\,d\mu(y)
 &\leq \frac1t\int_X |g(y)| \bigl( P_t (|f-f(y)|^p)\bigr)^{2/p}\,d\mu(y) \\
 &\leq \frac1t\biggl( \int_X P_t(|f-f(y)|^p) \,d\mu(y) \biggr)^{2/p} \|g\|_{p/(p-2)}\\
 &\leq \|f\|_{p,1/2}^2\|g\|_{p/(p-2)}
 \end{align*}
Now use the result of Lemma~\ref{lem:energyasbesov} and take the limit as $t\downarrow0$ to obtain
\begin{equation*}
 \|f\|_{p,1/2}^2\|g\|_{p/(p-2)}
 \geq \lim_{t\downarrow0} \frac2t\langle (I-P_t)f, fg \rangle - \frac1t\langle (I-P_t) f^2, g\rangle
 = 2\DF(fg,f)-\DF(f^2,g)= \int_X  2g\,d\nu_f
 \end{equation*}
where, as in the previous result, the limit is by Proposition~1.2.3 of~\cite{BouleauHirsch}. The final equality is from the definition of $\nu_f$, Definition~4.1.2 of~\cite{BouleauHirsch}.

In particular, if $E_1\subset E_2$ are of finite $\mu$ measure and $\mathbf{1}_{E_1}\leq g\leq \mathbf{1}_{E_2}$ then we obtain
\begin{equation*}
\nu_{f}(E_1) \leq \int_X g\,d\nu_f \leq \frac{1}{\sqrt2} \| f\|_{p,1/2}^2 \bigl( \mu(E_2) \bigr)^{(p-2)/p}.
\end{equation*}
Regularity of $\DF$ then supplies sufficiently many functions $g\in\mathcal{F}$ to conclude that $\nu_f$ is absolutely continuous with respect to $\mu$, see  ~\cite{BouleauHirsch,FOT,ChenFukushima}.
\end{proof}

There are spaces on which the Kusuoka measure (and hence $\nu_f$ for any non-constant $f\in\mathcal{F}$) is singular to $\mu$, see~\cite{Kusuoka,BenBassatStrichartzTeplyaev}. For such spaces $\mathbf{B}^{p,1/2}(X)$ is therefore trivial when $p>2$.

\begin{corollary}\label{singular Kusuoka}
If the Kusuoka measure is singular to $\mu$ then $\mathbf{B}^{p,1/2}(X)$ contains only constant functions when $p>2$.
\end{corollary}

\section{Pseudo-Poincar\'e inequalities}

The following pseudo-Poincar\'e inequalities will play an important role in many parts of this monograph.

\begin{lemma}\label{pseudo-Poincare}
Let $ p \ge 1$ and $\alpha >0$. For every $f \in \mathbf{B}^{p,\alpha} (X)$, and $t \ge 0$,
\[
\| P_t f -f \|_{L^p(X,\mu)} \le t^\alpha \| f \|_{p,\alpha}.
\]
\end{lemma}

\begin{proof}
From conservativeness of the semigroup and H\"older's inequality, we have
\begin{align*}
\| P_t f - f \|_{L^p(X,\mu)} &=\left(\int_X | P_t f (x)-f(x)|^p d\mu(x)\right)^{1/p} \\
 & = \left(\int_X | P_t (f -f(x))(x)|^p d\mu(x)\right)^{1/p} \\
 & \le \left( \int_X P_t (|f-f(x)|^p)(x) d\mu(y) \right)^{1/p}.
\end{align*}
\end{proof}

\begin{remark}
Triebel \cite{Trie} (Section 1.13.6) introduced the interpolation spaces:
\[
(L^p(X,\mu), \mathcal{E} )_{\alpha,\infty}=\left\{ u \in L^p(X,\mu)\, :\, \sup_{t >0} t^{-\alpha} \| P_t u -u \|_{L^p(X,\mu )} <+\infty \right\}.
\]

From the previous lemma, it is therefore clear that $\mathbf{B}^{p,\alpha} (X) \subset (L^p(X,\mu), \mathcal{E})_{\alpha,\infty}.$ However, it may not be true that $\mathbf{B}^{p,\alpha} (X) = (L^p(X,\mu), \mathcal{E})_{\alpha,\infty}$, even when $X=\mathbb{R}^n$, see Remark 4.5 in \cite{MPPP} and \cite{Taibleson} (Theorems 4 and 4*).
\end{remark}

The following lemma will be useful:

\begin{lemma}
Let $L$ the generator of $\mathcal{E}$ . Let $p > 1$ and $\alpha >0$. Then, there exists a constant $C>0$ such that for every $f \in \mathbf{B}^{p,\alpha}(X)$ and $t \ge 0$,
\[
\| LP_t f \|_{L^p(X,\mu)} \le C \frac{\| f \|_{p,\alpha}}{t^{1-\alpha}}.
\]
\end{lemma}

\begin{proof}
Since $\lim_{t \to +\infty} \| LP_t f \|_{L^p(X,\mu)} =0$, we have
\begin{align*}
\| LP_t f \|_{L^p(X,\mu)} & =\left\| \sum_{k=1}^{+\infty}( LP_{2^k t} f -  LP_{2^{k-1} t} f )  \right \|_{L^p(X,\mu)} \\
 & \le  \sum_{k=1}^{+\infty} \left\| LP_{2^k t} f -  LP_{2^{k-1} t} f  \right \|_{L^p(X,\mu)} \\
 & \le \sum_{k=1}^{+\infty} \left\| LP_{2^{k-1} t} (P_{2^{k-1}t} f -  f)  \right \|_{L^p(X,\mu)} \\
 &\le \sum_{k=1}^{+\infty} \frac{1}{2^{k-1} t}  \left\| P_{2^{k-1}t} f -  f  \right \|_{L^p(X,\mu)} \\
 &\le C  \sum_{k=1}^{+\infty} \frac{(2^{k-1} t)^{\alpha}}{2^{k-1} t}  \| f \|_{p,\alpha}  \\
 &\le C \frac{\| f \|_{p,\alpha}}{t^{1-\alpha}},
\end{align*}
where we used in the proof the fact that since $P_t$ is an analytic semigroup, we have for any $g \in L^p(X,\mu) $, $\left\| LP_t g  \right \|_{L^p(X,\mu)} \le \frac{C}{t} \| g \|_{L^p(X,\mu)}$.
\end{proof}

In a very general framework, one can  resort to (Hille-Yosida) spectral theory to define the fractional powers of a closed operator $A$ on a Banach space  via the following formula
\[
(-A)^s x = \frac{\sin \pi s}{\pi} \int_0^\infty \lambda^{s-1} (\lambda I - A)^{-1} (-A)x\ d\lambda,
\]
for every $x\in D(A)$. In fact, using Bochner's subordination one can express the fractional powers of $A$ also in terms of the heat semi-group $P_t = e^{tA}$ via the following formula, see (5) on p. 260 in \cite{Yosida},
\begin{equation}\label{As}
(-A)^s x = - \frac{s}{\Gamma(1-s)} \int_0^\infty t^{-s-1} [P_t x - x]\ dt.
\end{equation}

Let us denote by $L$ the generator of $\mathcal{E}$ and, for $0< s \le 1$, by $\mathcal{L}_p^s$ the domain of the operator $(-L)^s$ in $L^p(X,\mu)$, $1 \le p <+\infty$. One has then the following proposition:

\begin{proposition}
Let $\alpha \in (0,1]$, $p \ge 1$ and $0<s<\alpha$. Then
\[
\B^{p,\alpha}(X) \subset \mathcal{L}_p^s,
\]
and there exists a constant $C=C_{p,s,\alpha}$ such that for every $\B^{p,\alpha}(X)$,
\[
\| (-L)^s f \|_{L^p(X,\mu)} \le C \|  f \|^{1-\frac{s}{\alpha}}_{L^p(X,\mu)} \| f \|_{p,\alpha}^{\frac{s}{\alpha}}.
\]
In particular, $(-L)^s: \B^{p,\alpha}(X) \to L^p(X,\mu)$ is bounded.
\end{proposition}

\begin{proof}

Let  $f \in \B^{p,\alpha}(X)$. One shall prove that the integral $\int_0^\infty t^{-s-1} [P_t f - f]\ dt$ is finite, and therefore that $f \in \mathcal{L}_p^s$. For $\delta >0$, one has
\begin{align*}
 \left\| \int_0^\infty t^{-s-1} [P_t f - f]\ dt \right\|_{L^p(X,\mu)} & \le  \int_0^\infty t^{-s-1}  \| P_t f - f \|_{L^p(X,\mu)}  dt \\
  & \le \int_0^\delta t^{-s-1}  \| P_t f - f \|_{L^p(X,\mu)}  dt +  \int_\delta^\infty t^{-s-1}  \| P_t f - f \|_{L^p(X,\mu)}  dt  \\
  & \le \| f \|_{p,\alpha} \int_0^\delta t^{-s-1 +\alpha}   dt +2 \| f \|_{L^p(X,\mu)} \int_\delta^\infty t^{-s-1}    dt \\
  & \le \| f \|_{p,\alpha} \frac{\delta^{\alpha-s}}{\alpha -s}+2 \| f \|_{L^p(X,\mu)} \frac{\delta^{-s}}{s}.
\end{align*}
The result follows then by optimizing $\delta$.
\end{proof}

\section{Reflexivity of the Besov spaces}

For any $p\ge 1$ and $\alpha>0$, recall the norm on $\mathbf{B}^{p,\alpha}(X)$:
\[
\| f \|_{\mathbf{B}^{p,\alpha}(X)} =\| f \|_{L^p(X,\mu)} + \| f \|_{p,\alpha},
\]
where
\[
\| f \|_{p,\alpha}= \sup_{t >0} t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}.
\]

In the following proof, for convenience we use an equivalent norm, still denoted by $\| \cdot \|_{\mathbf{B}^{p,\alpha}(X)}$, defined as follows
\[
\| f \|_{\mathbf{B}^{p,\alpha}(X)} = \brak{\| f \|_{L^p(X,\mu)}^p +\| f \|_{p,\alpha}^p }^{\frac1p}.
\]

\begin{lemma}[Clarkson type inequalities]
Let $f,g\in \mathbf{B}^{p,\alpha}(X)$. Let $1<p<\infty$ and $q$ be its conjugate. If $2\le p<\infty$, then
\begin{equation}\label{eq:CTIge2}
\norm{\frac{f+g}{2}}_{\mathbf{B}^{p,\alpha}(X)}^p + \norm{\frac{f-g}{2}}_{\mathbf{B}^{p,\alpha}(X)}^p
\le \frac12 \|f\|_{\mathbf{B}^{p,\alpha}(X)}^p +\frac12 \|g\|_{\mathbf{B}^{p,\alpha}(X)}^p.
\end{equation}
If $1<p\le 2$, then
\begin{equation}\label{eq:CTIle2}
\norm{\frac{f+g}{2}}_{\mathbf{B}^{p,\alpha}(X)}^{q} + \norm{\frac{f-g}{2}}_{\mathbf{B}^{p,\alpha}(X)}^{q}
\le \brak{\frac12 \|f\|_{\mathbf{B}^{p,\alpha}(X)}^p +\frac12 \|g\|_{\mathbf{B}^{p,\alpha}(X)}^p}^{q-1}.
\end{equation}
\end{lemma}
The proof is essentially the same as in the $L^p$ case (see for instance \cite[Theorem 2.38]{AR}), so we omit it for concision. As a corollary, we immediately obtain

\begin{corollary}\label{lem:reflexive}
For any $p > 1$ and $\alpha>0$, $\mathbf B^{p,\alpha}(X)$ is a  reflexive Banach space.
\end{corollary}
\begin{proof}
Since $\mathbf B^{p,\alpha}(X)$ is a Banach space, from Milman-Pettis theorem it suffices to show that $\mathbf B^{p,\alpha}(X)$ is uniformly convex. That is,
for every $0<\varepsilon \le 2$, there exists $\delta>0$ such that for any $f,g\in \mathbf B^{p,\alpha}(X)$ with $\|f\|_{\mathbf B^{p,\alpha}(X)}=\|g\|_{\mathbf B^{p,\alpha}(X)}=1$ and $\|f-g\|_{\mathbf B^{p,\alpha}(X)}\ge \varepsilon$, then
\[
\norm{\frac{f+g}{2}}_{\mathbf B^{p,\alpha}(X)}\le 1-\delta.
\]
This can be seem as follows.

If $2\le p<\infty$, then \eqref{eq:CTIge2} implies that
\[
\norm{\frac{f+g}{2}}_{\mathbf B^{p,\alpha}(X)}^p \le 1-\frac{\varepsilon^p}{2^p}.
\]
If $1< p\le 2$, then \eqref{eq:CTIle2} implies that
\[
\norm{\frac{f+g}{2}}_{\mathbf B^{p,\alpha}(X)}^q \le 1-\frac{\varepsilon^q}{2^q}.
\]
\end{proof}

\section{Interpolation inequalities}

We have the following basic interpolation inequalities.

\begin{proposition}\label{interpolation inequality}
Let $\theta\in [0,1]$, $1 \le p,q <+\infty$ and $\alpha,\beta >0$. Let us assume $\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{r}$ and $\alpha=\theta \beta+(1-\theta)\gamma$. Then, $ \B^{q,\beta}(X)\cap \B^{r,\gamma}(X) \subset \B^{p,\alpha}(X) $  and for any $f\in \B^{q,\beta}(X)\cap \B^{r,\gamma}(X)$,
\[
\|f\|_{p,\alpha} \le \|f\|_{q,\beta}^{\theta} \|f\|_{r,\gamma}^{1-\theta}.
\]
\end{proposition}

\begin{proof}
Let $f \in \B^{q,\beta}(X)\cap \B^{r,\gamma}(X) $. One has for every $t >0$
\begin{align*}
    t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}  = t^{-\theta \beta-(1-\theta)\gamma} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}.
\end{align*}
Then, from H\"older's inequality
\begin{align*}
\int_X P_t (|f-f(y)|^p)(y) d\mu(y) &=\int_X P_t (|f-f(y)|^{p\theta+p(1-\theta)})(y) d\mu(y)  \\
 & \le \left( \int_X P_t (|f-f(y)|^q)(y) d\mu(y) \right)^{\frac{p \theta}{q}} \left( \int_X P_t (|f-f(y)|^r)(y) d\mu(y)\right)^{\frac{p (1-\theta)}{r}}.
\end{align*}
One deduces
\begin{align*}
  & t^{-\alpha} \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}  \\
 \le & t^{-\theta \beta}  \left( \int_X P_t (|f-f(y)|^q)(y) d\mu(y) \right)^{\frac{ \theta}{q}}  t^{-(1-\theta)\gamma} \left( \int_X P_t (|f-f(y)|^r)(y) d\mu(y)\right)^{\frac{ 1-\theta}{r}}.
\end{align*}
Taking the supremum over $t>0$ finishes the proof.
\end{proof}

As a special case of the previous result, note that since $\B^{2,1/2}(X)=\mathcal F$ we easily deduce:

\begin{corollary}\label{duality Dirichlet}
Let $1<p \le 2$ and $q$ be its conjugate $(\frac{1}{p}+\frac{1}{q}=1)$. Let $0<\alpha<1$. Then, for any $f\in \mathcal F \cap \mathbf B^{p,\alpha}(X)$ and $g\in  \mathcal F \cap \mathbf B^{q,1-\alpha}(X)$, it holds that
\[
| \Ecal(f,g) | \le \|f\|_{p,\alpha} \|g\|_{q,1-\alpha}.
\]
\end{corollary}

\section{Continuity of the heat semigroup in the Besov spaces}

Our goal in this section is to study the continuity properties of the semigroup $P_t$ in the Besov spaces with range $1<p \le 2$ and parameter $\alpha=\frac{1}{2}$. As corollaries we will deduce several important properties of the Besov spaces themselves. The main result is the following:

\begin{theorem}\label{continuity Besov chapter 1}
Let $1<p\le 2$. There exists a constant $C_p>0$ such that for every $f \in L^p(X,\mu)$ and $t \ge 0$
\[
\| P_t f \|_{p,1/2} \le \frac{C_p}{t^{1/2}} \| f \|_{L^p(X,\mu)}.
\]
In particular $P_t: L^p(X,\mu) \to \B^{p,1/2}(X)$ is bounded for $t>0$.
\end{theorem}

A key intermediate result is the following inequality that will follow from nice ideas originally due to Nick Dungey \cite{Dungey} and then developed in \cite{LiChen,LiChen2}.

\begin{lemma}\label{Lemma interpolation}
Let $1<p \le 2$. There exists a constant $C_p>0$ such that for every non-negative $f \in L^p(X,\mu)$ and $t >0$
\[
 \left( \int_X P_t (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \le C_p \| f \|^{1/2}_{L^p(X,\mu)} \| P_t f -f \|^{1/2}_{L^p(X,\mu)}.
\]
\end{lemma}

\begin{proof}
Let $1 <p \le 2$ and $t>0$ be fixed in the following proof. The constant $C$ in the following will denote a positive constant depending only on $p$ that may change from line to line. For simplicity of notation, we shall assume  in this proof that $P_t$ admits a measurable  heat kernel $p_t(x,y)$, however the argument can readily be extended if this is not the case, by using the heat kernel measure instead. For $\alpha,\beta \ge 0$, set
\[
\gamma_p(\alpha,\beta):=p\alpha(\alpha-\beta)-\alpha^{2-p} (\alpha^p-\beta^p)
\]
and for a non-negative function $f \in L^p(X,\mu)$
\begin{align*}
\Gamma_p(f)(x)
&:=p f(x) \int_X p_t(x,y)(f(x)-f(y)) d\mu(y)  - f^{2-p}(x) \int_X p_t(x,y) \brak{f^p(x)-f^p(y)}d\mu(y)
\\&= \int_X p_t(x,y)\,\gamma_p(f(x),f(y))d\mu(y).
\end{align*}
Note that from Lemma 3.5 in \cite{LiChen}, one has for any $\alpha,\beta \ge 0$
\[
(p-1) (\alpha -\beta)^2 \le \gamma_p(\alpha,\beta)+\gamma_p(\beta,\alpha) \le p (\alpha-\beta)^2
\]
and that, similarly to \cite{Dungey}, page 122, one has $\Gamma_p(f) \ge 0$.
Then the same argument as in \cite{LiChen}, Lemma 3.6, gives
\begin{align*}
\int_X\int_X p_t(x,y)\, |f(x)-f(y)|^p d\mu(x)d\mu(y)
&\le
 C \int_X\int_X p_t(x,y) \brak{\gamma_p(f(x),f(y))+\gamma_p(f(y),f(x))}^{p/2} d\mu(x)d\mu(y)
\\&\le
C \int_X\int_X p_t(x,y) \brak{\gamma_p^{p/2}(f(x),f(y))+\gamma_p^{p/2}(f(y),f(x))} d\mu(x)d\mu(y)
\\&=
C\int_X\int_X p_t(x,y) \,\gamma_p^{p/2}(f(x),f(y))\, d\mu(x)d\mu(y)
\\&\le
C\int_X\brak{\int_X p_t(x,y)\, \gamma_p(f(x),f(y)) d\mu(y)}^{p/2} d\mu(x)
\\&=
C\int_X \Gamma_p^{p/2}(f)(x)\, d\mu(x).
\end{align*}
We then follow the proof of Theorem 1 in \cite{LiChen} (see also Theorem 1.3 in  \cite{Dungey}). Let $u(s,x)=e^{-s\Delta_t} f (x)$ where $\Delta_t=\mathbf{Id} -P_t$. Note that
\begin{align*}
\Gamma_p(u)& =pu (  u - P_t u) -u^{2-p} (u^p -P_t (u^p)) \\
 & =pu \Delta_t u  -u^{2-p} \Delta_t (u^p) \\
 &=-pu \partial_s u  -u^{2-p} \Delta_t (u^p) \\
 & =-u^{2-p} \left( \partial_s +\Delta_t \right)u^p.
\end{align*}

Set now
\[
J(s,x)=- \left( \partial_s +\Delta_t \right)u^p(s,x),
\]
so that
\[
\Gamma_p(u)=u^{2-p} J.
\]
Note that since $u \ge 0$ and $\Gamma_p(u) \ge 0$, one has $J \ge 0$.
One has then from H\"older's inequality
\begin{align*}
\int_X \Gamma_p^{p/2}(u) d\mu & =\int_X u^{p(2-p)/2} J^{p/2} d\mu \\
 & \le \left( \int_X u^p d\mu \right)^{\frac{2-p}{2}} \left( \int_X J d\mu \right)^{p/2}.
\end{align*}
On computes then
\[
\int_X J d\mu =- \int_X \partial_s (u^p) d\mu=-p \int_X u^{p-1} \partial_s u d\mu=p \int_X u^{p-1} \Delta_t u d\mu.
\]
Thus, we have from H\"older's inequality
\[
\int_X J d\mu \le p \| u \|_{L^p(X,\mu)}^{p-1}  \| \Delta_t u \|_{L^p(X,\mu)}^{p} .
\]
From the definition of $\Delta_t$ one concludes therefore
\[
\left( \int_X\int_X p_t(x,y)\, |u(s,x)-u(s,y)|^p d\mu(x)d\mu(y)\right)^{1/p} \le C \| u(s,\cdot) \|^{1/2}_{L^p(X,\mu)} \| P_t u(s,\cdot) -u(s,\cdot) \|^{1/2}_{L^p(X,\mu)}.
\]
Letting $s \to 0^+$ yields from Fatou lemma
\[
\left( \int_X\int_X p_t(x,y)\, |f(x)-f(y)|^p d\mu(x)d\mu(y)\right)^{1/p} \le C \| f \|^{1/2}_{L^p(X,\mu)} \| P_t f -f \|^{1/2}_{L^p(X,\mu)}
\]
\end{proof}

We are now ready for the proof of Theorem \ref{continuity Besov chapter 1}.

\begin{proof}
Let $f \in L^p(X,\mu)$. We can assume $f \ge 0$. If not, it is enough to decompose $f$ as $f^+-f^-$ with $f^+=\max \{ f ,0 \}$ and $f^-=\max \{ -f ,0 \}$. Let $s,t >0$, applying Lemma \ref{Lemma interpolation} to $P_s f$, one obtains
\[
 \left( \int_X P_t (|P_sf-P_sf(y)|^p)(y) d\mu(y) \right)^{1/p} \le C_p \| P_s f \|^{1/2}_{L^p(X,\mu)} \| P_{t+s} f -P_sf \|^{1/2}_{L^p(X,\mu)}.
\]
Note that $\| P_s f \|_{L^p(X,\mu)} \le \|  f \|_{L^p(X,\mu)}$ and that
\begin{align*}
\| P_{t+s} f -P_sf \|_{L^p(X,\mu)} & = \left\| \int_0^t LP_{s+u} f  du \right\|_{L^p(X,\mu)} \\
 & =\left\| \int_0^t P_u LP_{s} f  du \right\|_{L^p(X,\mu)}  \\
 &  \le \int_0^t \left\|  P_u LP_{s} f \right\|_{L^p(X,\mu)} du \\
 & \le t  \left\|   LP_{s} f \right\|_{L^p(X,\mu)}  \\
 & \le \frac{t}{s}  \left\|    f \right\|_{L^p(X,\mu)},
\end{align*}
where in the last step we used analyticity of the semigroup. One concludes
\[
 \left( \int_X P_t (|P_sf-P_sf(y)|^p)(y) d\mu(y) \right)^{1/p} \le C_p \left(  \frac{t}{s} \right)^{1/2} \|  f \|_{L^p(X,\mu)}.
\]
Dividing both sides by $\sqrt{t}$ and taking the supremum over $t>0$ finishes the proof of Theorem \ref{continuity Besov chapter 1}.
\end{proof}

We now collect several corollaries of Theorem \ref{continuity Besov chapter 1}. The first one is the following:

\begin{corollary}
For $1 <  p \le 2$, $\B^{p,1/2}(X)$ is dense in $L^p(X,\mu)$.
\end{corollary}

\begin{proof}
Let $1<p \le 2$. Since for any $f \in L^p(X,\mu)$, $P_t f \in \B^{p,1/2}(X)$ and $P_t f \to f$ when $t \to 0$ (see for instance~\cite[Theorem 1.4.1]{EBD}),
one concludes that $\B^{p,1/2}(X)$ is dense in $L^p(X,\mu)$.
\end{proof}

We then have the following second  corollary of Theorem \ref{continuity Besov chapter 1}.

\begin{proposition}\label{Critical bound Chapter 1}
Assume that $\mathcal{E}$ is regular. Let $ 2 \le p <+\infty  $. For every $f \in L^p(X,\mu)$, and $t \ge 0$,
\[
\| P_t f -f \|_{L^p(X,\mu)} \le C_p   t^{1/2}  \liminf_{s \to 0}  s^{-1/2} \left( \int_X P_s (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}
\]
\end{proposition}

\begin{proof}
Let $f \in  L^p(X,\mu)$ and $g \in L^q(X,\mu) \cap \mathcal{F} $ where $q$ is the conjugate exponent of $p$. One has for $t \ge 0$,
\[
\int_X (P_t f -f ) g d\mu =  \int_0^t \mathcal{E}(P_s f ,g) ds  .
\]
The idea is now to bound $\mathcal{E}(g ,P_s f) $ by using an approximation of $\mathcal{E}$.
For  $\tau \in(0,\infty)$  we set
\begin{equation}\label{eq:energy-pt-Lp3}
\mathcal{E}_{\tau}(u,u):= \frac{1}{ \tau}  \langle (I-P_\tau)u,u\rangle.
\end{equation}
We have then $ \mathcal{E}(P_s f ,g) =\lim_{\tau \to 0} \mathcal{E}_{\tau}(P_s f ,g)$. Note now that $\mathcal{E}_{\tau}(P_s f ,g)=\mathcal{E}_{\tau}( f,P_s g)$ and that from H\"older inequality (applied as in the proof of Proposition \ref{interpolation inequality})
\begin{align*}
2 | \mathcal{E}_{\tau}(f,P_s g) | & \le \tau^{-1/2} \left( \int_X P_\tau  (|P_sg-P_sg(y)|^q)(y) d\mu(y) \right)^{1/q} \tau^{-1/2} \left( \int_X P_\tau (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \\
 & \le  \tau^{-1/2} \left( \int_X P_\tau (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p} \| P_s g \|_{q,1/2} \\
 &\le C_p  \tau^{-1/2} \left( \int_X P_\tau (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}  s^{-1/2} \| g \|_{L^q(X,\mu)}.
\end{align*}
One has therefore
\begin{align*}
\left| \int_X (P_t f -f ) g d\mu \right| \le  C_p t^{1/2} \| g \|_{L^q(X,\mu)} \liminf_{s \to 0}  s^{-1/2} \left( \int_X P_s (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p},
\end{align*}
and we conclude by $L^p-L^q$ duality.
\end{proof}

\begin{corollary}\label{corollary 1.25}
Assume that $\mathcal{E}$ is regular. Let $ 2 \le p <+\infty  $ and $\alpha >1/2$. If $ f \in \B^{p,\alpha} (X)$ then $\mathcal{E}(f,f)=0$.
\end{corollary}

\begin{proof}
Indeed, for $ f \in \B^{p,\alpha} (X)$ with $\alpha >1/2$ one has
\[
 \liminf_{s \to 0}  s^{-1/2} \left( \int_X P_s (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}=0,
 \]
 so that for every $t \ge 0$, $P_t f=f$, and thus $\mathcal{E}(f,f)=0$.
\end{proof}

We finally point out the following other corollary of Theorem \ref{continuity Besov chapter 1}.

\begin{prop}
Let $1<p \le 2$. Let $L$ be the generator of $\mathcal{E}$ and $\mathcal{L}_p$ be the domain of $L$ in $L^p(X,\mu)$.  Then
\[
\mathcal{L}_p \subset \B^{p,1/2}(X)
\]
and for every $f \in \mathcal{L}_p$,
\begin{equation}\label{eq:multi2}
\|f\|^2_{p,1/2} \le C \norm{ Lf}_{L^p(X,\mu)} \| f\|_{L^p(X,\mu)}.
\end{equation}
\end{prop}
\begin{proof}
Write for $\lambda >0$
\[
R_{\lambda}f=(L-\lambda)^{-1} f=\int_0^{\infty} e^{-\lambda t} P_tf dt.
\]
Consequently
\[
\|R_{\lambda}f\|_{p,1/2} \le \int_0^{\infty} e^{-\lambda t} \|P_tf\|_{p,1/2}  dt
\le \int_0^{\infty} e^{-\lambda t}  \frac{C}{t^{1/2}} \|f\|_{L^p(X,\mu)}  dt\le C \lambda^{-1/2} \|f\|_{L^p(X,\mu)}.
\]
It follows that
\[
\|f\|_{p,1/2} \le C  \lambda^{-1/2} \norm{(L-\lambda)f}_p  \le C ( \lambda^{-1/2} \|Lf\|_{L^p(X,\mu)}+ \lambda^{1/2} \|f\|_{L^p(X,\mu)}).
\]
Taking $\lambda=\|Lf\|_{L^p(X,\mu)} \|f\|_{L^p(X,\mu)}^{\,-1}$, we then get the result.

\end{proof}

\section{Besov critical exponents}

Let $p \ge 1$. We define the $L^p$ Besov  critical exponent of $X$ as
\[
\alpha^*_p(X)=\inf \{ \alpha >0\, :\, \mathbf{B}^{p,\alpha}(X) \text{ is trivial} \}.
\]
 $\mathbf{B}^{p,\alpha}(X)$ trivial means that any $f \in \mathbf{B}^{p,\alpha}(X)$ is constant. We have then the following result:

\begin{proposition}\label{Besov critical exponents}
Assume that $\mathcal{E}$ is regular and irreducible, i.e. $\mathcal{E}$ has the property that $\mathcal{E}(f,f)=0$ implies $f$ constant, then
\begin{enumerate}
\item $ \alpha^*_2(X)=\frac{1}{2}$;
\item $p \to \alpha^*_p(X)$ is non increasing;
\item For $p \ge 2$, $\alpha^*_p (X) \le \frac{1}{2}$;
\item For $1 \le p \le 2$, $ \frac{1}{2} \le \alpha^*_p(X) \le \frac{1}{p}$.
\end{enumerate}
\end{proposition}

\begin{proof}
\

\begin{enumerate}

\item Since $\mathbf{B}^{2,1/2}(X)=\mathcal{F}$, one has $\alpha^*_2(X) \ge \frac{1}{2}$. Then, from Corollary \ref{corollary 1.25}, one has $\alpha^*_2(X) \le \frac{1}{2}$.
\item

For simplicity of notation, we assume in the proof that $P_t$ has a measurable heat kernel $p_t(x,y)$, but the proof easily extends to the general case. It suffices to show that $\alpha^*_p(X) \le \alpha^*_q (X)$ if $q\le p$. Indeed, first notice that  for any $a,b>0$ such that $a\neq b$, there holds
\[
|a^p-b^p||a^q-b^q|^{-1}\leq \frac{p}{q} \max\{a^q,b^q\}^{\frac{p}{q}-1}.
\]
Equivalently,
\[
|a^p-b^p|^q \leq \brak{\frac{p}{q}}^q \max\{a^q,b^q\}^{p-q} |a^q-b^q|^{q}.
\]
Next by H\"older's inequality and Fubini's theorem, we have
 \begin{align*}
	 & \lefteqn{\int_X \int_X p_t(x,y)\bigl| |f(x)|^p-|f(y)|^p\bigr|^q \,d\mu(x)\,d\mu(y) } \quad&\\
	&\leq \brak{\frac{p}{q}}^q \int_X \int_X p_t(x,y) \bigl( |f(x)|^{p-q}+|f(y)|^{p-q}\bigr)\bigl| |f(x)|^q-|f(y)|^q\bigr|^q \,d\mu(x)\,d\mu(y) \\
	&\leq 2\brak{\frac{p}{q}}^q \int_X  |f(x)|^{p-q} \int_X p_t(x,y) \bigl| |f(x)|^q-|f(y)|^q\bigr|^q \,d\mu(y)\,d\mu(x) \\
	&\leq 2\brak{\frac{p}{q}}^q \int_X |f(x)|^{p-q} \Bigl(\int_X p_t(x,y) \bigl| |f(x)|^q-|f(y)|^q\bigr|^p \,d\mu(y)\Bigl)^{q/p} \, d\mu(x)\\
	&\leq 2\brak{\frac{p}{q}}^q \||f|^q\|_p^{p-q} \Bigl( \int_X \int_X p_t(x,y) \bigl| |f(x)|^q-|f(y)|^q\bigr|^p \,d\mu(x)\, d\mu(y) \Bigl)^{q/p}\\
	&\leq 2\brak{\frac{p}{q}}^q \||f|^q\|_p^{p-q} t^{\alpha q} \|| f|^q\|_{p,\alpha}^q,
	\end{align*}
which implies that
\[
\| |f|^p\|_{q,\alpha}^q \leq 2\brak{\frac{p}{q}}^q  \||f|^q\|_p^{p-q} \|| f|^q\|_{p,\alpha}^q.
\]
Hence  $\alpha^*_p (X)\le \alpha^*_q(X)$.

\item This is Corollary \ref{corollary 1.25}.
\item This follows from Theorem \ref{continuity Besov chapter 1} and Proposition \ref{prop:BesovCEp}.
\end{enumerate}
\end{proof}

\begin{remark}\label{remark conjecture 1}
In view of the duality given by Corollary \ref{duality Dirichlet}, it is natural to conjecture that under suitable conditions one may have
\[
\alpha^*_p(X)+\alpha_q^*(X)=1
\]
if $p$ and $q$ are conjugate, i.e. satisfy $\frac{1}{p}+\frac{1}{q}=1$.
\end{remark}

\begin{remark}
We will see in Chapters 4 and 5  that for local Dirichlet forms the limit $$\alpha_\infty(X)=\lim_{p \to +\infty} \alpha_p(X)$$ is closely related to a H\"older regularity property in space of the heat semigroup. If the conjecture in Remark \ref{remark conjecture 1} is true, then classical interpolation theory suggests that it is reasonable to expect that for every $p \ge 1$:
\[
\alpha^*_p (X) =\frac{1}{p} +\left( 1-\frac{2}{p} \right) \alpha_\infty(X).
\]
 \end{remark}

 \begin{example}
 For strongly local Dirichlet  forms with absolutely continuous energy measures, we will see in Chapter 4 that one generically has $\alpha^*_p(X)=\frac{1}{2}$ for every $p \ge 1$.
 \end{example}

config: arxiv