TITLE: On solvable quintics and septics
QUESTION [14 upvotes]: Here is a nice sufficient (but not necessary) condition on whether a quintic is solvable in radicals or not.  Given,
$x^5+10cx^3+10dx^2+5ex+f = 0\tag{1}$
If there is an ordering of its roots such that,
$x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0\tag{2}$
or alternatively, its coefficients are related by the quadratic in f,
$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2
\tag{3}$
then (1) is solvable.  This also implies that if $c\neq0$, then it has a solvable twin,
$x^5+10cx^3+10dx^2+5ex+f' = 0\tag{4}$
where $f'$ is the other root of (3). The Lagrange resolvent are the roots of,
$z^4+fz^3+(2c^5-5c^3e-4d^2e+ce^2+2cdf)z^2-c^5fz+c^{10} = 0\tag{5}$
so,
$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag{6}$
Two questions though:
I. Does the septic (7th deg) analogue,
$x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 – (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0\tag{7}$
imply such a septic is solvable?
II. The septic has a $5! = 120$-deg resolvent. While this is next to impossible to explicitly construct, is it feasible to construct just the constant term? Equating it to zero would then imply a family of solvable septics, just like (3) above.
More details and examples for (2) like the Emma Lehmer quintic in my blog.

REPLY [1 votes]: It turns out there is an infinite number of such septics, such as the Hashimoto-Hoshi septic,
$$\small x^7 - (a^3 + a^2 + 5a + 6)x^6 +
  3(3a^3 + 3a^2 + 8a + 4)x^5 + (a^7 + a^6 + 9a^5 - 5a^4 - 15a^3 - 22a^2 -
        36a - 8)x^4 - a(a^7 + 5a^6 + 12a^5 + 24a^4 - 6a^3 + 2a^2 - 20a - 16)x^3 + a^2(2a^6 + 7a^5 + 19a^4 + 14a^3 + 2a^2 + 8a - 8)x^2 - a^4(a^4 + 4a^3 + 8a^2 + 4)x + a^7=0$$
For example, let $a=1$ so,
$$1 - 17 x + 44 x^2 - 2 x^3 - 75 x^4 + 54 x^5 - 13 x^6 + x^7=0$$
which is the equation involved in $\cos\frac{\pi k}{43}$ and order its roots as,
$$x_1,\,x_2,\,x_3,\,x_4,\,x_5,\,x_6,\,x_7 =\\
r_1,\,r_2,\,r_5,\,r_6,\,r_3,\,r_7,\,r_4 =
\\ -0.752399,\; 0.0721331,\; 2.63744,\; 3.62599,\; 0.480671,\; 6.29991,\; 0.636246$$
where the $r_i$ is the root numbering in Mathematica. Then,
$$
x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 - (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0$$


text:
{\bf Problem.} Simplify $\frac{10a^3}{55a^2}$ when $a=3$.
{\bf Level.} Level 4
{\bf Type.} Prealgebra
{\bf Solution.} The 10 in the numerator and the 55 in the denominator have a common factor of 5. Similarly, the $a^3$ and $a^2$ have a common factor of $a^2$. So we get \[
\frac{10a^3}{55a^2} = \frac{2\cdot 5\cdot a^2\cdot a^1}{11\cdot 5 \cdot a^2} = \frac{2\cdot \cancel{5}\cdot \cancel{a^2}\cdot a^1}{11\cdot \cancel{5} \cdot \cancel{a^2}} = \frac{2a}{11}.
\]Substituting $a=3$ gives $\boxed{\frac{6}{11}}$.

set_name: MATH


TITLE: Defining a list using set theory?
QUESTION [0 upvotes]: EDIT: Changed set
If I have the following set of numbers: $\{1, 2, 4, 8, 16,...\}$ where the universe of discourse is natural numbers $\{0, 1, 2, ...\}$
How can I define this? I note that definitions are usually in the form of {x | P(X)}, but in this case, is it valid to write $\{2^x | x \in \mathbb{N}\}$

REPLY [2 votes]: Yes $\{2^x\, |\, x \in \mathbb{N}\}$ is fine. It's a shorthand for $\{y\in\mathbb{N} \,|\, \exists x\in \mathbb{N}, 2^x = y\}$.
If you actually meant $\{1,2,4,8,16\}$ and no more elements, then just write that. Alternatively, use $\{y\in\mathbb{N} \,|\, \exists x\in \mathbb{N}, 2^x = y, x < 5\}$


TITLE: Probability Question ~Beta(1,2)
QUESTION [0 upvotes]: In the next hour, child C will
a) play video games for 40 minutes, and
b) watch TV for 20 minutes,
but not necessarily in that order (order could be a and then b, or b then a)
The two orders are equally possible.
The child's mother will call the child C at some random time t into the hour 0 <= t <= 1 ("the next hour" mentioned above starts at t=0, and t is measured in hours.)
This time variable has a ~beta(1,2) distribution.
What is the probability that child C will watch TV when his/her mother calls?
My thought was to calculate the expectation for beta distribution. 1 / (1 + 2)
expectation is 1/3 - mother is expected to call when t = 20 (time = 20 minutes)
Based on given info, child will either:
a) play video games 0 <= t <= 40 and then watch TV 40 <= t <= 60
b) watch TV 0 <= t <= 20 and then play video games 20 <= t <= 60
If mother calls at 20 minutes (expectation), then probability of child watching TV when mother calls is 0 since the child C will be playing video games in both cases.
Have i taken the correct approach with this question? I am not convinced / comfortable with my answer.
Thank you very much

REPLY [0 votes]: I think the problem areas are t=20 and t=40.  In situation a, the child is playing video games and watching tv when t = 40, and in situation b, the child is playing video games and watching tv when t = 20.
So, we either need to set the problem up with a) TV: $0\le t<40$ or $0\le t\le40$, Games: $40\le t\le60$ or $40<t\le60$ and similarly for b, or we let him be doing two things at once when $t=20$ or $t=40$.
If we accept two simultaneous activities, then, when his mother calls (at exactly 20 minutes), it will be a 50% chance that he is watching TV, and a 100% chance that he will be playing video games.
If we don't allow the overlap, and reword the problem, then it will depend on where the strict inequality is.


TITLE: Expectation of maximum of a function whose expectation is concave
QUESTION [1 upvotes]: An analysis of a data structure yields a property of the form
$\qquad \mathbb{E}[ f(k) ] = H_k + H_{n-k+1} - 1$
for some natural $n$ and all $1 \leq k \leq n$. Note that the $f(k)$ are not independent.
Now I am interested in the quantity
$\qquad \mathbb{E}[\max_{k=1,\dots,n} f(k)]$.
Since $\mathbb{E}[f(k)]$ is concave on $[1,k]$ if continued on the reals and the maximum is "nice", I think it should be possible to apply (the symmetric version of) Jensen's inequality to conclude
$\qquad \mathbb{E}[\max_{k=1,\dots,n} f(k)] \leq \max_{k=1,\dots,n} \mathbb{E}[f(k)] = \mathbb{E}[f(n/2)] = 2H_{n/2} + \frac{1}{n/2 + 1} - 1$,
glossing over odd $n$.
Does this really hold? If so, how can I make the argument formal? If not, is there any handle on the desired quantity? Right now, I don't know more about the distribution for $f(k)$.
Note that I can expect a logarithmic bound, so we should be able to make a connection.

REPLY [4 votes]: Jensen's inequality states that, if $\phi$ is a convex function
\begin{equation}
   \phi( \mathbb{E}X) \leq \mathbb{E}\phi(X)
\end{equation}
In your particular case, consider
\begin{equation}
   \begin{aligned}
      \phi_n : x = (x_1,...,x_n) \mapsto \max_{i \in [1, n]} \{x_i\}
   \end{aligned}
\end{equation}
For all $n$, $\phi_n$ is convex since $\forall t \in [0,1]$, $\max \{ tx^1_i + (1-t)x^2_i\}  \leq t \max\{x^1_i\} + (1-t) \max\{x^2_i\} $, therefore (unfortunately)
\begin{equation}
   \max_i \mathbb E X_i \leq \mathbb{E} (\max_i X_i)
\end{equation}


TITLE: Which mathematical ideas have done most to change history?
QUESTION [68 upvotes]: I'm planning a course for the general public with the general theme of "Mathematical ideas that have changed history" and I would welcome people's opinions on this topic.  What do you think have been the most influential mathematical ideas in terms of what has influenced science/history or changed the way humans think, and why?
I won't expect my audience to have any mathematical background other than high-school.
My thoughts so far are: non-Euclidean geometry, Cantor's ideas on uncountability, undecidability, chaos theory and fractals, the invention of new number systems (i.e. negative numbers, zero, irrational, imaginary numbers), calculus, graphs and networks, probability theory, Bayesian statistics.
My apologies if this has already been discussed in another post.

REPLY [7 votes]: Probably it can be viewed as a variant on already posted answers (cryptography etc.), but the study of permutation groups and its application in cracking the Enigma code literally changed history (namely, the outcome of World War II). Here is an article by Marian Rejewski, one of the people involved in the code-cracking, explaining what was done and how:
http://www.impan.pl/Great/Rejewski/article.html
Rejewski and his achievements were also mentioned in answers to the following MO questions:
Real-world applications of mathematics, by arxiv subject area?
Notable mathematics during World War II

REPLY [0 votes]: The normal distribution in probability theory. It is everywhere in moderen science.


TITLE: If $X$ is inductive, then the set $\{ x\in X : x$ is transitive}  is inductive. Hence every $n\in N$ is transitive
QUESTION [3 upvotes]: First some definitions:
$X$ is inductive if $\emptyset\in X$ and $x\in X\implies x\cup\{x\}\in X$
Let $N = \cap\{X : X \text{ is inductive }\}$.
Let us use the following notation:
$0 = \emptyset$, $1 = \{0\}$, $2 = \{0 , 1\}$, ...
if $n\in N$ let $n + 1 = n \cup \{n\}$. Let us define $<$ (on $N$) by $n < m$ iff $n\in m$.
A set $T$ is transitive if $x\in T$ implies $x\subset T$.
Then I have :
If $X$ is inductive, then the set $\{ x\in X : x$ is transitive} is inductive. Hence every $n\in N$ is transitive.
Solution. Suppose $X$ is inductive, it follows that $\emptyset\in X$. The empty set is also vacuously transitive, hence $\emptyset\in \{ x\in X : x$ is transitive}. Now suppose that $x\in\{ x\in X : x$ is transitive} it follows that $x\in X$ and hence $x\cup\{x\} \in X$. Similarly it follows that $x$ is transitive. Now, let $a\in x\cup\{x\}$, then there are two cases either $a\in x$ or $a\in \{x\}$. In the first case it follows that since $a\in x$ and $x$ is transitive $a\subset x$, and hence $a\subset x\cup\{x\}$. In the second case since $a\in\{x\}$ and $\{x\}$ is a singleton set it follows that $a = x$, thus $a \subset x\cup\{x\}$. Thus $x\cup\{x\}$ is transitive, and hence $x\cup\{x\} \in\{ x\in X : x$ is transitive}. Therefore we can conclude that $\{ x\in X : x$ is transitive} is inductive. $\Box$
Now the problem, I really cannot see how the corollary "Hence every $n\in N$ is transitive." follows from this.
But I do note that I know $N$ is transitive, and for each $n\in N$, $n = \{m\in N : m < n\}$ by the previous question.
So I was wondering if someone could help shed light on how the corollary follows from the initial question, or point out the flaw in my proof thus suggesting a reason it doesn't currently follow.
Thank you very much.

REPLY [7 votes]: First verify that $N$ is inductive. This is straightforward: if $n \in N$, and let $X$ be any inductive set. Then $n \in X$ by the definition of $N$, and since $X$ is inductive, $n \cup \{n\} \in X$. This is true for all inductive $X$, so $n \cup \{n\} \in \bigcap \{X:X\text{ is inductive}\} = N$. Now let $T = \{n \in N:n\text{ is transitive}\}$; $T$ is inductive by the theorem. But then $N \subseteq T$, since $N$ is the intersection of all inductive sets, and since by construction $T \subseteq N$, we must have $T=N$, i.e., each $n \in N$ is transitive.
Added: Taking a closer look at your proof of the theorem, I see one significant infelicity. It’s a bad idea to write ‘suppose that $x \in \{x \in X:x\text{ is transitive}\}\text{’}$: you're using the letter $x$ simultaneously to refer to a specific object and as the dummy variable in the set definition. Make it $$x \in \{y \in X:y\text{ is transitive}\}$$ instead. The rest is okay, though I think that parts could be worded a bit more clearly. For instance, after defining/choosing $x$, I’d continue something like this:

Clearly $x$ is transitive, and since $X$ is inductive, $x \cup \{x\} \in X$. If $a \in x \cup \{x\}$, then either $a \in x$, or $a \in \{x\}$. In the first case ... .


text:
\chapter{Strictly local Dirichlet spaces with doubling measure and 2-Poincar\'e inequality}\label{Sec:Metric-Curvature}

In this chapter we define the space  $BV(X)$ of functions of
bounded variation when $\mathcal{E}$ is strcitly local (see the exact assumptions in Section 4.1).
We shall then prove that under a weak Bakry-\'Emery curvature dimension condition,
one actually has $BV(X)=\mathbf{B}^{1,1/2}(X)$ with comparable seminorms. One of the key tools used there is the co-area
formula for BV functions; we establish this formula in Lemma~\ref{lem:Co-area} of the present chapter. We will also study  in that framework  the  spaces $\B^{p,\alpha}(X)$ introduced in Chapter 1 and relate them to Besov spaces previously considered in the literature. Finally, applications to Sobolev and isoperimetric inequalities are given.

\section{Local Dirichlet spaces and standing assumptions}\label{SubSec:StrongLocal}

We will assume in this section that $(X,\mu)$ is a topological space equipped with a Radon measure $\mu$,
and that there is a strongly local Markovian Dirichlet form $\mathcal{E}$ on $X$.
As $\mathcal{E}$ is strongly local, it is represented by a signed Radon measure
$\Gamma(u,v)$ for each $u,v\in \mathcal{F}$ such that whenever $\varphi$ is a bounded function in $\mathcal{F}$ and
$u,v\in\mathcal{F}$, we have
\[
\mathcal{E}(\varphi u,v)+\mathcal{E}(\varphi v,u)-\mathcal{E}(\varphi, uv)=2\int_X\varphi\, d\Gamma(u,v).
\]
We then also have $\mathcal{E}(u,v)=\int_X d\Gamma(u,v)$, see for example~\cite{St-I}.
With respect to $\mathcal{E}$ we can define the following \emph{intrinsic metric} $d_{\mathcal{E}}$:
\[
d_{\mathcal{E}}(x,y)=\sup\{u(x)-u(y)\, :\, u\in\mathcal{F}\cap C(X)\text{ and } \Gamma(u,u)\le \mu\}.
\]
Here by $\Gamma(u,u)\le \mu$ we mean that for each Borel set $A\subset X$ we have
$\Gamma(u,u)(A)=\int_Ad\Gamma(u,u)\le \mu(A)$.
In general there is no reason why $d_{\mathcal{E}}$ is a metric on $X$ (for it could be infinite for a given pair of points $x,y$
or zero for some distinct pair of points), but \emph{suppose that} $d_{\mathcal{E}}$ is a metric on $X$ such that the topology
generated by $d_{\mathcal{E}}$ coincides with the topology on $X$ and that balls with respect to $d_{\mathcal{E}}$ have
compact closures in this topology. Then by~\cite[Lemma~1, Lemma~$1^\prime$]{St-I}, functions $\varphi$ of the form
$x\mapsto d_{\mathcal{E}}(x,y)$ for each $y\in X$, and maps $\varphi$ of the form $x\mapsto (r-d_{\mathcal{E}}(x,y))_+$ for each
$r>0$ and $y\in X$ are in $\mathcal{F}_{loc}(X)$ with $d\Gamma(\varphi,\varphi)/d\mu\le 1$.
So we will also assume in this section that $d_\mathcal{E}$ gives a metric on $X$ that is commensurate with the topology
on $X$, and that with respect to this metric the measure $\mu$ is \emph{doubling}, that is, there is a constant $C>0$ such that
whenever $x\in X$ and $r>0$, we have $\mu(B(x,2r))\le C\, \mu(B(x,r))$.

\begin{lemma}\label{lem:diff-of-Lip}
Let $f:X\to\R$ be locally Lipschitz continuous with respect to the metric $d_{\mathcal{E}}$. Then $f\in\mathcal{F}_{loc}(X)$
with $\Gamma(f,f)\ll\mu$. If $f$ is locally $L$-Lipschitz, then $\Gamma(f,f)\le L\, \mu$.
\end{lemma}

\begin{proof}
Let $Q$ be a countable dense subset of $X$.
Let $U\subset X$ be a bounded open set and let $\{q_i\}_{i\in I\subset\mathbb{N}}$ be an enumeration of $Q\cap U$.
Note that $Q\cap U$ is dense in $U$. For each $i\in I$ let $\psi_i(x)=d_{\mathcal{E}}(x,q_i)$. Then as explained above,
$\psi_i\in\mathcal{F}(U)$ with $\Gamma(\psi_i,\psi_i)\le \mu$. For $j\in I$ set
\[
f_j(x):=\inf\{f(q_i)+L\, \psi_i(x)\, :\, i\in I\text{ with }i\le j\},
\]
where $L\ge 0$ is the Lipschitz constant of $f$ in $U$. The above functions are inspired by the proof of the McShane extension
theorem (see for example~\cite{Hei}). By the lattice properties of Dirichlet form, it is seen that each
$f_j\in\mathcal{F}(U)$ with $d\Gamma(f_j,f_j)\le L$. Furthermore, $f_j$ are $L$-Lipschitz in $U$ with $f_j(q_i)=f(q_i)$ for
$i\in I$ with $i\le j$.

We can see that $f_j\to f$ monotonically and hence (as $f$ and $f_j$ are bounded in $U$ because $U$ is bounded)
$f_j\to f$ in $L^2(U)$, with $d\Gamma(f,f)/d\mu\le L$ on $U$.
\end{proof}

Thus locally Lipschitz functions $f$ with respect to $d_{\mathcal{E}}$ satisfy
\begin{enumerate}
\item $f\in \mathcal{F}_{loc}(X)$,
\item $\Gamma(f,f)$ is absolutely continuous with respect to the underlying measure $\mu$; we denote its Radon-Nikodym
derivative by $|\nabla f|^2$, that is, $|\nabla f|$ is the square-root of its Radon-Nikodym derivative.
\end{enumerate}
We say that $(X,\mu,\mathcal{E},\mathcal{F})$ supports a $1$-Poincar\'e inequality if there are constants
$C,\lambda>0$ such that whenever $B$ is a ball in $X$ (with respect to the metric $d_{\mathcal{E}}$)
and $u\in\mathcal{F}$, we have
\[
\frac{1}{\mu(B)}\int_B|u-u_B|\, d\mu\le C\, \text{rad}(B)\, \frac{1}{\mu(\lambda B)}\, \int_{\lambda B}|\nabla u|\, d\mu.
\]
Of course, the $1$-Poincar\'e inequality does not make sense if $\mathcal{E}$
does not satisfy the condition of strong locality.
We say that $(X,\mu,\mathcal{E},\mathcal{F})$ supports  $2$-Poincar\'e
inequality if there are constants $C,\lambda>0$ such that
whenever $B$ is a ball in $X$ (with respect to the metric $d_{\mathcal{E}}$)
and $u\in\mathcal{F}$, we have
\[
\frac{1}{\mu(B)}\int_B|u-u_B|\, d\mu\le C\, \text{rad}(B)\, \left(\frac{1}{\mu(\lambda B)}\, \int_{\lambda B}d\Gamma(u,u)\right)^{1/2}.
\]
The requirement that $\mathcal{E}$ supports a $1$-Poincar\'e inequality is significantly
a stronger requirement than that of $2$-Poincar\'e inequality. Much of the current
theory on functions of bounded variation in the metric setting require a
$1$-Poincar\'e inequality. In this chapter we will \emph{not} require the support
of $1$-Poincar\'e inequality, only the weaker $2$-Poincar\'e inequality, but in
some of the analysis we would need an additional requirement called the weak
Bakry-\'Emery curvature condition, see the discussion below.

Should the 2-Poincar\'e inequality be satisfied, a standard argument, due to Semmes, tells us that locally Lipschitz continuous functions
forms a dense subclass of $\mathcal{F}$, where $\mathcal{F}$ is equipped with the norm
\[
\Vert u\Vert_{\mathcal{F}}:=\Vert u\Vert_{L^2(X)}+\sqrt{\mathcal{E}(u,u)},
\]
see for example~\cite[Chapter~8]{HKST}.

\

\noindent {\bf Standing assumptions for this chapter:} For this chapter we will assume that $\mathcal{E}$ is a strongly local
Markovian Dirichlet form that induces a metric $d=d_\mathcal{E}$ with respect to which $\mu$ is doubling and supports a
$2$-Poincar\'e inequality. We will also assume that the class of locally Lipschitz continuous functions on $X$ forms a dense
subclass of $L^1(X)$ and that $X$ is complete with respect to $d_\mathcal{E}$. As a consequence of
the doubling property of $\mu$ it follows that closed and bounded subsets of $X$ are compact.

The work of Saloff-Coste~\cite{Sal-Cos} tells us that the above standing assumptions are equivalent to the property that there is an associated
heat kernel function $p_t(x,y)$ on $[0,\infty)\times X\times X$ for which there are constants $c_1, c_2, C >0$ such that
whenever $t>0$ and $x,y\in X$,
\begin{equation}\label{eq:heat-Gauss}
 \frac{1}{C}\, \frac{e^{-c_1 d(x,y)^2/t}}{\sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}\le p_t(x,y)
   \le C\, \frac{e^{-c_2 d(x,y)^2/t}}{\sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}.
\end{equation}
The above inequality is called the \emph{Gaussian bounds} for the heat kernel. Examples of spaces that support such a Dirichlet
form include complete Riemannian manifolds with non-negative Ricci curvature, Carnot groups and other complete sub-Riemannian
manifolds, and doubling metric measure spaces that support a $2$-Poincar\'e inequality with respect to the upper gradient
structure of Heinonen and Koskela (see~\cite{HKST}).

\

In some parts of this chapter we need a condition that guarantees that the space is not negatively curved with respect to the
Dirichlet form. To do so, we will assume that the metric space $(X,d_\mathcal{E},\mu)$
satisfies a weak Bakry-\'Emery curvature condition,
namely that whenever $u\in \mathcal{F}\cap L^\infty(X)$,
\begin{equation}\label{eq:weak-BE}
\Vert |\nabla P_tu| \Vert_{L^\infty(X)}^2\le \frac{C}{t}\Vert u\Vert_{L^\infty(X)}^2.
\end{equation}
Here $P_tu$ denotes the heat extension of $u$ via the heat semigroup associated with the Dirichlet form $\mathcal{E}$, see~\cite{FOT}.

\

\begin{example}
\em The weak Bakry-\'Emery curvature condition is satisfied in the following examples:
\begin{itemize}
\item Complete Riemannian manifolds with non-negative Ricci curvature and more generally, the $RCD(0,+\infty)$ spaces (\cite{Jiang15}).
\item Carnot groups (see \cite{BB2})
\item Complete sub-Riemannian manifolds with generalized non-negative Ricci curvature (see \cite{BB,BK14})
\item Metric graphs with finite number of edges (see \cite{BK})
\end{itemize}
Several statements equivalent to the weak Bakry-\'Emery curvature condition are given in Theorem~1.2 in \cite{CJKS}.  There are some metric measure spaces equipped with a doubling measure supporting a $2$-Poincar\'e inequality but without the above Bakry-\'Emery condition, see for example~\cite{KRS}.  For instance, it should be noted, that in the setting of complete sub-Riemannian manifolds with generalized non-negative Ricci curvature in the sense of \cite{BG},  while the weak Bakry-\'Emery curvature condition is known to be satisfied (see \cite{BB,BK14}), the 1-Poincar\'e inequality has not been proven yet (though the 2-Poincar\'e is known, see \cite{BBG}).
\end{example}

\section{Construction of BV class using the Dirichlet form}\label{subsec:BV}

In this section we use the Dirichlet form and the associated family $\Gamma(\cdot,\cdot)$ of measures to construct a BV class
of functions on $X$. To do so, we only need $\mu$ to be a doubling measure on $X$ for $d_\mathcal{E}$ and that the class of
locally Lipschitz functions to be dense in $L^1(X)$. So in this section we do \emph{not} need the Poincar\'e inequality
nor do we need the weak Bakry-\'Emery curvature condition.

We now set the \emph{core} of the Dirichlet form,
$\mathcal{C}(X)$, to be the class of all $f\in\mathcal{F}_{loc}(X)$ such that $\Gamma(f,f)\ll \mu$.
In this case, we can also set $u\in L^1(X)$ to be in $BV(X)$ if there is a sequence of local Lipschitz functions $u_k\in L^1(X)$
such that $u_k\to u$ in $L^1(X)$ and
\[
\liminf_{k\to\infty}\int_X|\nabla u_k|\, d\mu<\infty.
\]
For $u\in BV(X)$ and open sets $U\subset X$, we set
\[
\Vert Du\Vert(U)=\inf_{u_k\in\mathcal{C}(U), u_k\to u\text{ in }L^1(U)}\liminf_{k\to\infty} \int_U|\nabla u_k|\, d\mu,
\]
and then for sets $A\subset X$ we define
\[
\Vert Du\Vert(A)=\inf\{\Vert Du\Vert(O)\, :\, A\subset O\text{ and }O\text{ is open in }X\}.
\]

\begin{lemma}\label{lem:BV-Leibniz}
If $u,v\in BV(X)$ and $\eta$ is a Lipschitz continuous function on $X$ with $0\le \eta\le 1$ on $X$, then
$\eta u+(1-\eta)v\in BV(X)$ with
\[
\Vert D(\eta u+(1-\eta)v)\Vert(X)\le \Vert Du\Vert(X)+\Vert Dv\Vert(X) +\int_X|u-v|\, |\nabla \eta|\, d\mu.
\]

\end{lemma}

\begin{proof}
From Lemma~\ref{lem:diff-of-Lip} we already know that given such $\eta$,
$\eta\in\mathcal{F}_{loc}$ with $|\nabla \eta|\in L^\infty(X)$.

We can choose sequences $u_k,v_k\in \mathcal{F}(X)$ such that $u_k\to u$ and $v_k\to v$ in $L^1(X)$ and
$\int_X |\nabla u_k|\, d\mu\to \Vert Du\Vert(X)$ and $\int_X |\nabla v_k|\, d\mu\to \Vert Dv\Vert(X)$ as $k\to\infty$.
Now an application of the properties of Dirichlet form (see~\cite[page~190]{St-I} as well
as the references therein) to the functions
$\eta u_k+(1-\eta)v_k$ tells us that
\begin{align*}
\Vert D(\eta u+(1-\eta)v)\Vert(X)&\le \liminf_{k\to\infty}\int_X |\nabla[\eta u_k+(1-\eta)v_k]|\, d\mu\\
 & \le \liminf_{k\to \infty}\left(\int_X\eta |\nabla u_k|\, d\mu+\int_X(1-\eta) |\nabla v_k|\, d\mu
 +\int_X |u_k-v_k|\, |\nabla \eta|\, d\mu\right).
\end{align*}
Now using the facts that $0\le \eta\le 1$ and that $u_k-v_k\to u-v$ in $L^1(X)$ we obtain the required inequality.
\end{proof}

We first list some elementary properties of $\Vert Du\Vert$.

\begin{lemma}\label{lem:elementary}
Let $U$ and $V$ be two open subsets of $X$. Then
\begin{enumerate}
\item $\Vert Du\Vert(\emptyset)=0$;
\item $\Vert Du\Vert(U)\le \Vert Du\Vert(V)$ if $U\subset V$,
\item $\Vert Du\Vert(\bigcup_i U_i)=\sum_i\Vert Du\Vert(U_i)$ if $\{U_i\}_i$ is a pairwise disjoint subfamily of open
subsets of $X$.
\end{enumerate}
\end{lemma}

\begin{proof}
We will only prove the third property here, as the other two are quite direct consequences of the definition of $\Vert Du\Vert$.
Since any function $f\in \mathcal{F}(\bigcup_i U_i)$ has restrictions $u_i=f\vert_{U_i}\in \mathcal{F}(U_i)$ with
$\int_{\bigcup_i U_i}|\nabla f|\, d\mu=\sum_i\int_{U_i}|\nabla u_i|\, d\mu$, it follows that
\[
\Vert Du\Vert(\bigcup_i U_i)\ge \sum_i\Vert Df\Vert(U_i).
\]
In the above we also used the fact that as $f$ gets closer to $u$ in the $L^1(\bigcup_i U_i)$ sense,
$u_i$ gets closer to $u$ in the $L^1(U_i)$ sense.

To prove the reverse inequality, for $\eps>0$ we can choose locally Lipschitz continuous
$u_i\in \mathcal{F}(U_i)$ for each $i$ such that
\[
\int_{U_i}|u-u_i|\, d\mu<2^{-i-2}\eps
\]
and
\[
\int_{U_i}|\nabla u_i|\, d\mu<\Vert Du\Vert(U_i)+2^{-i-2}\eps.
\]
Now the function $f_\eps=\sum_i u_i\mathbf{1}_{U_i}$ is in $\mathcal{F}(\bigcup_i U_i)$ because
the $U_i$ are pairwise disjoint open sets, and $\mathcal{E}$ is local.

Therefore
\[
\int_{\bigcup_i U_i}|u-f_\eps|\, d\mu\le \sum_i\int_{U_i}|u-u_i|\, d\mu\le \frac{\eps}{2}
\]
and
\[
\int_{\bigcup_i U_i}|\nabla f_\eps|\, d\mu=\sum_i\int_{U_i}|\nabla u_i|\, d\mu\le \frac{\eps}{2}
+\sum_i\Vert Df\Vert(U_i).
\]
From the first of the above two inequalities it follows that $\lim_{\eps\to 0^+}f_\eps=u$ in
$L^1(\bigcup_i U_i)$,
and therefore
\[
\Vert Du\Vert(\bigcup_i U_i)\le \liminf_{\eps\to 0^+}\left(\frac{\eps}{2}+\sum_i\Vert Du\Vert(U_i)\right)
=\sum_i\Vert Du\Vert(U_i).
\]
\end{proof}

We use the above definition of $\Vert Du\Vert$ on open sets to consider the following Caratheodory construction.

\begin{definition}
For $A\subset X$, we set
\[
\Vert Du\Vert^*(A):=\inf\{\Vert Du\Vert(O)\, :\, O\text{ is open subset of }X, A\subset O\}.
\]
\end{definition}

By the second property listed in the above lemma, we note that if $A$ is an open subset of $X$, then
$\Vert Du\Vert^*(A)=\Vert Du\Vert(A)$. With this observation, we re-name $\Vert Du\Vert^*(A)$ as
$\Vert Du\Vert(A)$ even when $A$ is not open.

\subsection{Radon measure property of BV energy}

The following lemma is due to De Giorgi and Letta~\cite[Theorem~5.1]{DeLe}, see
also~\cite[Theorem~1.53]{AFP}.

\begin{lemma}\label{lem:DeGLett}
If $\nu$ is a non-negative function on the class of all open subsets of $X$ such that
\begin{enumerate}
\item $\nu(\emptyset)=0$, \label{DeLett1}
\item if $U_1\subset U_2$ and they are both open sets, then $\nu(U_1)\le \nu(U_2)$, \label{DeLett2}
\item $\nu(U_1\cup U_2)\le \nu(U_1)+\nu(U_2)$ for open sets $U_1,U_2$ in $X$, \label{DeLett3}
\item if $U_1\cap U_2$ is empty and $U_1,U_2$ are open, then $\nu(U_1\cup U_2)=\nu(U_1)+\nu(U_2)$, \label{DeLett4}
\item for open sets $U$ \label{DeLett5}
\[
\nu(U)=\sup\{\nu(V)\, :\, V\text{ is bounded and open in }X\text{ with }\overline{V}\subset U\}.
\]
\end{enumerate}
Then the Caratheodory extension of $\nu$ to all subsets of $X$ gives a Borel regular outer measure on $X$.
\end{lemma}

The proof of the following theorem is based on that of~\cite{Mr}.

\begin{theorem}\label{thm:outer-measure}
If $X$ is  complete and
$f\in BV(X)$, then $\Vert Df\Vert$ is a Radon outer measure on $X$.
\end{theorem}

\begin{proof}
For simplicity of notation we will assume that $X$ is itself bounded.
Thanks to the lemma of De Giorgi and Letta (Lemma~\ref{lem:DeGLett}),
it suffices to verify that $\Vert Du\Vert$ satisfies the five conditions
set forth in Lemma~\ref{lem:DeGLett}. By Lemma~\ref{lem:elementary}, we know that $\Vert Du\Vert$ satisfies
Conditions~\ref{DeLett1},~\ref{DeLett2}, and Condition~\ref{DeLett4}. Thus it suffices for us to verify
Condition~\ref{DeLett3} and Condition~\ref{DeLett5}. We will first show the validity of Condition~\ref{DeLett5},
and use it (or rather, it's proof) to show that Condition~\ref{DeLett3} holds. We will do so for bounded open subsets of $X$;
a simple modification (by truncating $U_\delta$ by balls) would complete the proof for unbounded
sets; we leave this part of the extension as
an exercise.

\noindent {\bf Proof of Condition~\ref{DeLett5}:} From the monotonicity condition~\ref{DeLett2}, it suffices to prove
that
\[
\Vert Df\Vert(U)\le  \sup\{\Vert Df\Vert(V)\, :\, V\text{ is open in }X, \overline{V}\text{ is a compact subset of }U\}.
\]
For $\delta>0$ we set
\[
U_\delta=\{x\in U\, :\, \text{dist}(x,X\setminus U)>\delta\}.
\]
For $0<\delta_1<\delta_2<\text{diam}(U)/2$, let $V=U_{\delta_1}$ and $W=U\setminus\overline{U_{\delta_2}}$.
Then $V$ and $W$ are open subsets of $U$, and the closure of $V$ is a compact
subset of $U$. Note also that
$U=V\cup W$ and that
$\partial V\cap \partial W$ is empty. Thus we can find a Lipschitz function $\eta$ on $U$ that can be
used as a ``needle+thread" to stitch Sobolev functions on $V$ to Sobolev functions on $W$ as follows to obtain a Sobolev
function on $U$:  $0\le \eta\le 1$ on $U$, $\eta=1$ on $V\setminus W=\overline{U_{\delta_2}}$,
$\eta=0$ on $W\setminus V=U\setminus U_{\delta_1}$, and
\[
\text{Lip}\, \eta\le \frac{2}{\delta_2-\delta_1}\, \mathbf{1}_{V\cap W}.
\]
Now, for $v\in \mathcal{F}(V)$ and $w\in \mathcal{F}(W)$ we set $u=\eta v+(1-\eta)w$. As
we have the Leibniz rule (see~\cite{St-I}), we can see that $u\in \mathcal{F}(U)$ and
\begin{equation}\label{eq:B}
\int_U |\nabla u|\, d\mu\le \int_V |\nabla v|\, d\mu+\int_W |\nabla w|\, d\mu+ \frac{2}{\delta_2-\delta_1}\int_{V\cap W}|v-w|\, d\mu.
\end{equation}
Furthermore, whenever $h\in L^1(U)$, we can write $h=\eta h+(1-\eta)h$ to see that
\begin{equation}\label{eq:A}
\int_U|u-h|\, d\mu\le \int_V|v-h|\, d\mu+\int_W|w-h|\, d\mu.
\end{equation}
Now, we take $v_k$ from $\mathcal{F}(V)$ such that $v_k\to f$ in $L^1(V)$ and $\lim_{k\to\infty}\int_V|\nabla {v_k}|\, d\mu=\Vert Df\Vert(V)$,
and take $w_k\in \mathcal{F}(W)$ analogously. We then follow through by stitching together $v_k$ and $w_k$ into
the function $u_k$ as prescribed above. By~\eqref{eq:A} with $h=f$, we have that
\[
\int_U|f-u_k|\, d\mu\le \int_V|v_k-f|\, d\mu+\int_W|w_k-f|\, d\mu\to 0\text{ as }k\to\infty.
\]
It follows from~\eqref{eq:B} and the fact $\int_{V\cap W}|v_k-w_k|\, d\mu\to 0$ as $k\to\infty$  that
\[
\Vert Df\Vert(U)\le \liminf_{k\to\infty}\int_U|\nabla {u_k}|\, d\mu\le \Vert Df\Vert(V)+\Vert Df\Vert(W).
\]
Remembering again that the closure of $V$ is a compact subset of $U$, we see that
\[
\Vert Df\Vert(U)\le \sup\{\Vert Df\Vert(V)\, :\, V\text{ is open in }X, \overline{V}\text{ is compact subset of }U\}+
  \Vert Df\Vert(U\setminus \overline{U_{\delta_2}}).
\]
So now it suffices to prove that
\begin{equation}\label{eq:est-annuli}
\lim_{\delta\to 0^+}\Vert Df\Vert(U\setminus \overline{U_{\delta}})=0.
\end{equation}
To prove this, we note first that the above limit exists as
$\Vert Df\Vert(U\setminus \overline{U_{\delta}})$ decreases as
$\delta$ decreases. We fix a strictly monotone decreasing sequence of real numbers $\delta_k$ with
$\lim_{k\to\infty}\delta_k=0$, and for $k\ge 2$ we set
$V_k:=U_{\delta_{2k+3}}\setminus\overline{U_{\delta_{2k}}}$.
Observe that the family $\{V_{2k}\}_k$ is a pairwise disjoint family of open subsets of $X$ and that
the family $\{V_{2k+1}\}_k$ is also a pairwise disjoint family of open subsets of $X$.

By Lemma~\ref{lem:elementary}, we know that
\[
\infty>\Vert Df\Vert(U)\ge \Vert Df\Vert(\bigcup_{k\ge 1} V_{2k})=\sum_{k=1}^\infty \Vert Df\Vert(V_{2k}),
\]
and
\[
\infty>\Vert Df\Vert(U)\ge \Vert Df\Vert(\bigcup_{k\ge 1} V_{2k+1})=\sum_{k=1}^\infty \Vert Df\Vert(V_{2k+1}).
\]
It follows that for $\eps>0$ there is some positive integer $k_\eps\ge 2$ such that
\[
\sum_{k=k_\eps}^\infty \Vert Df\Vert(V_{2k})+\sum_{k=k_\eps}^\infty \Vert Df\Vert(V_{2k+1})<\eps.
\]

Now we stitch together approximations on $V_{2k}$ to approximations on $V_{2k+1}$, and from there to $V_{2k+2}$ and so
on. For each $k$ we choose a ``stitching function" $\eta_k$ as a Lipschitz function on $\bigcup_{j=k_\eps}^{k+1}V_j$
such that $0\le \eta_k\le 1$, with $\eta_k=1$ on $V_{k}\setminus V_{k-1}$, $\eta_k=0$ on
$\bigcup_{j=k_\eps}^{k-1} V_j\setminus V_{k}$, and $|\nabla {\eta_k}|\le C_k\mathbf{1}_{V_{k}\cap V_{k-1}}$.

Next, for each $k$ we can find $v_{k,j}\in \mathcal{F}(V_k)$ such that
\[
 \int_{V_k}|v_{k,j}-f|\, d\mu\le \frac{2^{-k-j}}{3(1+C_k)}
\]
and
\[
\int_{V_k}|\nabla {v_{k,j}}|\, d\mu\le \Vert Df\Vert(V_k)+2^{-j-k}.
\]
We now inductively stitch the functions together. To do so, we first fix $i\in\mathbb N$.

Starting with $k=k_\eps$, we stitch $u_{k,i}$ to $u_{k+1,i}$ using $\eta_{k+1}=\eta_{k_\eps+1}$ to obtain
$w_{i,k}\in \mathcal{F}(V_{k_\eps}\cup V_{k_\eps+1})$ so that we have
\[
\int_{V_{k_\eps}\cup V_{k_\eps+1}}|w_{i,k}-f|\, d\mu\le \frac{2^{-i-k_\eps}}{1+C_{k_\eps+1}}
\]
and
\[
\int_{V_{k_\eps}\cup V_{k_\eps+1}}|\nabla {w_{i,k}}|\, d\mu\le \sum_{j=k_\eps}^{k_\eps+1}\Vert Df\Vert(V_j)+2^{1-i-k_\eps}.
\]
Suppose now that for some $k\in\mathbb N$ with $k\ge k_\eps+1$
we have constructed $w_{i,k}\in \mathcal{F}(\bigcup_{j=k_\eps}^k V_j)$ such that
\[
\int_{\bigcup_{j=k_\eps}^k V_j}|w_{i,k}-f|\, d\mu\le \sum_{k=k_\eps}^k\frac{2^{-i-j}}{1+C_j}
\]
and
\[
\int_{\bigcup_{j=k_\eps}^k V_j}|\nabla {w_{i,k}}|\, d\mu\le \sum_{j=k_\eps}^{k}[\Vert Df\Vert(V_j)+2^{1-i-j}].
\]
Then we stitch $u_{k+1,i}$ to $w_{i,k}$ using $\eta_{k+1}$ to obtain $w_{i,k+1}$ satisfying inequalities analogous to the
above two. Note that $w_{i,k+1}=w_{i,k-1}$ on $V_{k-1}$ for $k\ge k_\eps+2$.
Thus, in the limit, we obtain a function $w_i=\lim_k w_{i,k}\in \mathcal{F}(\bigcup_{k=k_\eps}^\infty V_k)$ satisfying
\[
\int_{\bigcup_{j=k_\eps}^\infty V_j}|w_{i}-f|\, d\mu\le \sum_{k=k_\eps}^k\frac{2^{-i-j}}{1+C_j}<2^{1-i},
\]
\[
\int_{\bigcup_{j=k_\eps}^\infty V_j}|\nabla {w_{i}}|\, d\mu\le \sum_{j=k_\eps}^\infty\Vert Df\Vert(V_j)+2^{2-i}
   <\eps+2^{2-i}.
\]
From the first of the above two inequalities, we see that $w_i\to f$ in $L^1(\bigcup_{j=k_\eps}^\infty V_j)$
as $i\to\infty$,
and so from the second of the above two inequalities we obtain
\[
\Vert Df\Vert(\bigcup_{j=k_\eps}^\infty V_j)=\Vert Df\Vert(U\setminus \overline{U_{\delta_{k_\eps}}})
\le \liminf_{i\to\infty}\int_{\bigcup_{j=k_\eps}^\infty V_j}|\nabla {w_{i}}|\, d\mu\le \eps.
\]
This last inequality above tells us that the claim we set out to prove, namely
\[
\lim_{\delta\to 0^+}\Vert Df\Vert(U\setminus \overline{U_{\delta}})=0.
\]
this completes the proof of Condition~\ref{DeLett5}.

\noindent {\bf Proof of Condition~\ref{DeLett3}:}
By Condition~\ref{DeLett5} (which we have now proved above, so no circular argument
here!), for each $\eps>0$ we can find relatively compact open subsets $U_1^\prime\Subset U_1$ and
$U_2^\prime\Subset U_2$ such that
$\Vert Df\Vert(U_1\cup U_2)\le \Vert Df\Vert(U_1^\prime\cup U_2^\prime)+\eps$.
We then choose a Lipschitz ``stitching function" $\eta$ on $X$ such that $0\le \eta\le 1$ on $X$,
$\eta=1$ on $U_1^\prime$, $\eta=0$ on $X\setminus U_1$,  and
\[
|\nabla \eta|\le \frac{1}{C_{U_1,U_1^\prime}} \, \mathbf{1}_{U_1\setminus U_1^\prime}.
\]
For $u_1\in \mathcal{F}(U_1)$ and $u_2\in \mathcal{F}(U_2)$, we obtain the stitched function
$w=\eta u_1+(1-\eta)u_2$ and note that $w\in \mathcal{F}(U_1^\prime\cup U_2^\prime)$.
Observe that we cannot in general have $w\in \mathcal{F}(U_1\cup U_2)$ as $w$ is not defined in
$U_1\setminus (U_1^\prime\cup U_2)$ because $1-\eta$ is non-vanishing there and $u_2$ is not defined there, for example.
Then we have
\[
\int_{U_1^\prime\cup U_2^\prime}|\nabla w|\, d\mu\le \int_{U_1}|\nabla {u_1}|\, d\mu+\int_{U_2}|\nabla u_2\, d\mu
 +\frac{1}{C_{U_1,U_1^\prime}}\, \int_{U_1\cap U_2}|u_1-u_2|\, d\mu
\]
and
\[
\int_{U_1^\prime\cup U_2^\prime}|w-f|\, d\mu\le \int_{U_1}|u_1-f|\, d\mu+\int_{U_2}|u_2-f|\, d\mu.
\]
As before, choosing $u_{1k}$ to be the optimal approximating sequence for  $f$ on $U_1$ and
$u_{2,k}$ correspondingly for $f$ on $U_2$, we see from the first of the above two inequalities
that the stitched sequence $w_k$ approximates
$f$ on $U_1^\prime\cup U_2^\prime$. Therefore we obtain
\[
\Vert Df\Vert(U_1\cup U_2)\le \eps+\Vert Df\Vert(U_1^\prime\cup U_2^\prime)
  \le \eps+\liminf_{k\to\infty}\int_{U_1\cup U_2}|\nabla {w_k}|\, d\mu
  \le \Vert Df\Vert(U_1)+\Vert Df\Vert(U_2)+\eps.
\]
Letting $\eps\to 0$ now gives the desired Condition~\ref{DeLett3}.
\end{proof}

\subsection{Co-area formula}

\begin{definition}
We say that a measurable set $E\subset X$ is of \emph{finite perimeter} if $\mathbf{1}_E\in BV_{loc}(X)$ with
$\Vert D\mathbf{1}_E\Vert(X)<\infty$. We denote by $P(E,A):=\Vert D\mathbf{1}_E\Vert(A)$ the perimeter measure of $E$.
\end{definition}

\begin{lemma}\label{lem:Co-area}
The co-area formula holds true, that is, for Borel sets $A\subset X$ and $u\in L^1_{loc}(X)$,
\[
\Vert Du\Vert(A)=\int_{\R}P(\{u>s\}, A)\, ds.
\]
\end{lemma}

\begin{proof}
We first prove the formula for open sets $A$.

Suppose first that $u\in BV_{loc}(X)$ with $\Vert Du\Vert(A)<\infty$.
For $s\in\R$ we set $E_s:=\{x\in X\, :\, u(x)>s\}$. Consider the function $m:\R\to\R$ given by
\[
m(t)=\Vert Du\Vert(A\cap E_t).
\]
Then $m$ is a monotone decreasing function, and hence is differentiable almost everywhere. Let $t\in\R$ such that
$m^\prime(t)$ exists. Then
\[
m^\prime(t)=\lim_{h\to 0^+} \frac{\Vert Du\Vert(A\cap E_t\setminus E_{t+h})}{h}.
\]
Note that the functions
\[
u_{t,h}:=\frac{\max\{t,\min\{t+h,u\}\}-t}{h}
\]
converge in $L^1(X)$ to $\mathbf{1}_{E_t}$ as $h\to 0^+$. It follows that as $A$ is open,
\[
P(E_t, A)\le \liminf_{h\to 0^+}\Vert Du_{t,h}\Vert(A)=\liminf_{h\to0^+}\frac{\Vert Du\Vert(A\cap E_t\setminus E_{t+h})}{h}=m^\prime(t).
\]
Note also that by this lower semicontinuity of BV energy, $t\mapsto P(E_t, A)$ is a lower semicontinuous function, and hence
is measurable; and as it is non-negative, we can talk about its integral, whether that integral is finite or not.
Therefore, by the fundamental theorem of calculus for monotone functions,
\[
\int_{\R}P(E_t, A)\, dt\le \int_{\R} m^\prime(t)\, dt\le \lim_{s,\tau\to\infty}m(s)-m(-\tau)=\Vert Du\Vert(A).
\]
The above in particular tells us that if $u\in BV_{loc}(X)$ then almost all of its superlevel sets $E_t$ have finite perimeter.
If $u$ is not a BV function on $A$, then $\Vert Du\Vert(A)=\infty$, and hence
we also have
\begin{equation}\label{eq:1}
\int_{\R}P(E_t, A)\, dt\le \Vert Du\Vert(A).
\end{equation}
In particular, it also follows that $\int_{\R} P(E_t,A)\, dt<\infty$ if $u\in BV(X)$.

We still continue to assume that $A$ is open, and prove the reverse of the above inequality. If
$\int_{\R}P(E_t,A)\, dt=\infty$, then trivially
\[
\Vert Du\Vert(A)\le \int_{\R}P(E_t, A)\, dt.
\]
So we may assume without loss of generality that $\int_{\R}P(E_t,A)\, dt$ is finite. Note also by the Markovian property of
Dirichlet forms, filtered down to the level of the measure $|\nabla f|$, we have that
\[
\Vert Du\Vert(A)=\lim_{s,\tau\to\infty} \Vert Du_{s,\tau}\Vert(A),
\]
where $u_{s,\tau}=\max\{-\tau, \min\{u,s\}\}$. So without loss of generality we may assume that $0\le u\le 1$ for some
finite $a,b\in\R$. For positive integers $X$ we can divide $[0,1]$ into $X$ equal sub-intervals $[t_i,t_{i+1}]$, $i=0,\cdots, k$
with $t_{i+1}-t_i=1/k$. Then we can find $\rho_{k,i}\in (t_i, t_{i+1})$ such that
\[
\frac{1}{k}\, P(E_{\rho_{k,i}},A)\le \int_{t_i}^{t_{i+1}}\, P(E_s,A)\, ds.
\]
We set
\[
u_X=\sum_{j=1}^k\frac{1}{k} \mathbf{1}_{E_{\rho_{k,i}}}.
\]
Then as $|u_k-u|\le 1/k$ on $X$, we have that $u_k\to u$ in $L^1(A)$ as $k\to\infty$, and so
\begin{equation}\label{eq:2}
\Vert Du\Vert(A)\le \liminf_{k\to\infty}\Vert Du_k\Vert(A)=\liminf_{k\to\infty}\sum_{j=1}^k\frac{1}{k}\, P(E_{\rho_{k,i}},A)
 \le \int_0^1 P(E_s,A)\, ds.
\end{equation}

Note now that by the proofs of inequalities~\eqref{eq:1} and~\eqref{eq:2}, if $A$ is an open set then
$u\in BV(A)$ if and only if $\int_{\R}P(E_t,A)\, dt$ is finite.

Finally, we remove the requirement that $A$ be open. By the above comment, it suffices to prove this for the case that
$u\in BV(X)$. In this case, the maps $A\mapsto\Vert Du\Vert(A)$ and $A\mapsto\int_{\R}P(E_t,A)\, dt$ are
both Radon measures on $X$ that agree on open subsets of $X$ (that is, they are equal for open $A$). Hence it follows
that they agree on Borel subsets of $X$.

This completes the proof of the coarea formula.
\end{proof}

\section{$\mathbf{B}^{1,1/2}(X)$ is equal to $BV(X)$}\label{weak BE metric space}

In this section we will assume that $(X,d_\mathcal{E},\mu)$ is doubling, supports a $2$-Poincar\'e inequality,
and satisfies a weak Bakry-\'Emery curvature
condition~\eqref{eq:weak-BE}. Under this condition, we will compare the BV class, constructed in Section~\ref{subsec:BV},
with the Besov class $\mathbf{B}^{1,1/2}(X)$ from~\eqref{eq:def:Besov}.
We will also assume that
for each $u\in\mathcal{F}$ the measure $\Gamma(u,u)$ is absolutely continuous with respect to the underlying measure
$\mu$ on $X$; as in the previous section we will denote by $|\nabla u|^2$ the Radon-Nikodym derivative of
$\Gamma(u,u)$ with respect to $\mu$.

In what follows, $\mathcal{F}_1(X)$ denotes the collection of all $u\in L^1(X)\cap\mathcal{F}_{loc}(X)$ for which

$|\nabla u|\in L^1(X)$.
This means that for each ball $B$ in $X$ there is a compactly supported Lipschitz function $\varphi$ with $\varphi=1$ on $B$
such that $u\varphi\in\mathcal{F}$; in this case we can set $|\nabla u|=|\nabla (u\varphi)|$ in $B$, thanks to the strong locality
property of $\mathcal{E}$.

\begin{lemma}\label{lem:L1-norm-control}
Suppose that~\eqref{eq:weak-BE} holds.
For $u\in \mathcal{F}\cap \mathcal{F}_1(X)$, we have that
\[
\Vert P_t u-u\Vert_{L^1(X)}\le C\, \sqrt{t}\, \int_X|\nabla u|\, d\mu.
\]
Hence, if $u\in BV(X)$, then
\[
\Vert P_t u-u\Vert_{L^1(X)}\le C\, \sqrt{t}\, \Vert Du\Vert(X).
\]
\end{lemma}

\begin{proof}
To see the first part of the claim, we note that for each $x\in X$ and $s>0$, $\tfrac{\partial}{\partial s}P_su(x)$ exists,
and so by the fundamental theorem of calculus, for $0<\tau<t$ and $x\in X$,
\[
P_tu(x)-P_\tau u(x)=\int_\tau^t \frac{\partial}{\partial s}P_su(x)\, ds.
\]
Thus for each compactly supported function $\varphi\in \mathcal{F} \cap L^\infty(X)$,
by the facts that $P_t u$ satisfies the heat equation and that $P_s$ commutes with the infinitesimal generator (Laplacian),
\begin{align*}
\bigg\lvert\int_X\varphi(x)[P_tu(x)-P_\tau u(x)]\, d\mu(x)\bigg\rvert
 &=\bigg\lvert-\int_X\int_\tau^t\varphi(x)\frac{\partial}{\partial s}P_su(x)\, ds\, d\mu(x)\bigg\rvert\\
 &=\bigg\lvert\int_\tau^t \int_X d\Gamma(\varphi,P_su)(x)\, ds\bigg\rvert\\
 &=\bigg\lvert\int_\tau^t \int_X d\Gamma(P_s \varphi,u)(x)\, ds\bigg\rvert\\

 &\le \int_\tau^t\int_X|\nabla P_s\varphi|\, |\nabla u|\, d\mu\, ds\\

 &\le \Vert|\nabla P_s\varphi|\Vert_{L^\infty(X)}\, \int_\tau^t\int_X|\nabla u|\, ds\, d\mu.

\end{align*}
An application of~\eqref{eq:weak-BE} gives
\begin{align*}
\bigg\lvert\int_X\varphi(x)[P_tu(x)-P_\tau u(x)]\, d\mu(x)\bigg\rvert
&\le \frac{C}{\sqrt{t}}\Vert\varphi\Vert_{L^\infty(X)}\int_X\int_\tau^t |\nabla u|\, ds\, d\mu\\

&=C\, \frac{t-\tau}{\sqrt{t}}\, \Vert\varphi\Vert_{L^\infty(X)}\, \int_X|\nabla u|\, d\mu.

\end{align*}
As the above holds for all $\varphi\in \mathcal{F}\cap L^\infty(X)$, and as
compactly supported Lipschitz functions are dense in $L^\infty(X)$ (note that $X$ is separable as closed and bounded
subsets of $X$ are compact because $X$ is complete and supports a doubling measure),

we obtain
\[
\Vert P_t u-P_\tau u\Vert_{L^1(X)}\le C\, \frac{t-\tau}{\sqrt{t}}\, \int_X|\nabla u|\, d\mu.
\]
Now by the fact that $P_\tau u\to u$ as $\tau\to 0^+$ in $L^2(X)$, and by the fact that $\{P_t\}_{t>0}$ has an
extension as a contraction semigroup to $L^1(X)$ (see~\cite[Section~1.5 of page~37, Section~4.7 of page~201]{FOT}), we have that
\[
\Vert P_tu-u\Vert_{L^1(X)}\le C\, \sqrt{t}\, \int_X|\nabla u|\, d\mu
\]
as desired.

Finally, if $u\in BV(X)$, then we can find a sequence $u_k\in \mathcal{F}\cap W^{1,1}(X)$
such that
$u_k\to u$ in $L^1(X)$ and $\lim_k\int_X|\nabla u_k|\, d\mu=\Vert Du\Vert(X)$.
As $u\in L^1(X)$ and for $t>0$ we have that the heat kernel $p_t(x,y)$ is bounded in $X\times X$,

it follows that
$P_tu\in L^1(X)$ is well-defined with $\Vert P_tu\Vert_{L^1(X)}\le C_t\Vert u\Vert_{L^1(X)}$, see~\cite[Section~1.5]{FOT}. Now,
\begin{align*}
\Vert P_tu-u\Vert_{L^1(X)}
 &\le \Vert P_t(u-u_k)\Vert_{L^1(X)}+\Vert P_t u_k-u_k\Vert_{L^1(X)}+\Vert u_k-u\Vert_{L^1(X)}\\
 &\le C_t\Vert u-u_k\Vert_{L^1(X)}+C\, \sqrt{t}\, \int_X|\nabla u_k|\, d\mu+\Vert u-u_k\Vert_{L^1(X)}\\
 &\to C\, \sqrt{t}\, \lim_{k\to\infty} \int_X|\nabla u_k|\, d\mu=C\, \sqrt{t}\, \Vert Du\Vert(X)\text{ as }k\to\infty.
\end{align*}
\end{proof}

Recall the definition of $\mathbf{B}^{1,1/2}(X)$ from~\eqref{eq:def:Besov}: $f$ is in this class if
\[
\Vert f\Vert_{B^{1,1/2}(X)}:=\Vert f\Vert_{L^1(X)}+\sup_{t>0}\frac{1}{\sqrt{t}}\, \int_X P_t(|f-f(x)|)(x)\, d\mu(x)<\infty.
\]
Equivalently,
\[
\Vert f\Vert_{B^{1,1/2}(X)}=\Vert f\Vert_{L^1(X)}+\sup_{t>0}\frac{1}{\sqrt{t}}\, \int_X\int_Xp_t(x,y)|f(y)-f(x)|\, d\mu(y)\, d\mu(x)<\infty.
\]

We now wish to prove the following theorem.

\begin{theorem}\label{thm:W=BV}
Under the hypotheses of this section, $\mathbf{B}^{1,1/2}(X)=BV(X)$ with comparable seminorms.  Moreover, there exist constants $c,C>0$ such that for every $u \in BV(X)$
\[
c \limsup_{s  \to 0 }  s^{-1/2}  \int_X P_s (|u-u(y)|)(y) d\mu(y)  \le \Vert Du\Vert(X) \le C\liminf_{s \to 0}  s^{-1/2}  \int_X P_s (|u-u(y)|)(y) d\mu(y) .
\]

\end{theorem}

Note from the results of~\cite[Theorem~4.1]{MMS} that if the measure $\mu$ is doubling and supports a $1$-Poincar\'e
inequality, then a measurable set $E\subset X$ is in the BV class if
\[
\liminf_{t\to 0^+} \frac{1}{\sqrt{t}} \int_{E^{\sqrt{t}}\setminus E} P_t\mathbf{1}_E\, d\mu<\infty.
\]
Here $E^\varepsilon=\bigcup_{x\in E}B(x,\varepsilon)$. Note that
\[
\int_{E^{\sqrt{t}}\setminus E} P_t\mathbf{1}_E\, d\mu\le \int_{X\setminus E}P_t\mathbf{1}_E\, d\mu
=\frac{1}{2}\Vert P_t\mathbf{1}_E-\mathbf{1}_E\Vert_{L^1(X)}.
\]
Thus if $\mu$ is doubling and supports a $1$-Poincar\'e inequality, and in addition
\[
\sup_{t>0}\frac{1}{\sqrt{t}}\, \Vert P_t\mathbf{1}_E-\mathbf{1}_E\Vert_{L^1(X)}<\infty,
\]
then $E$ is of finite perimeter. Note that we do not assume the validity of
$1$-Poincar\'e inequality, but the weaker
version of~\eqref{eq:weak-BE}.

\begin{proof}
First we assume that $u\in BV(X)$. Then we know that for almost every $t\in\R$ the set $E_t$ is of finite perimeter,
where
\[
E_t=\{x\in X\, :\, u(x)>t\},
\]
and by the co-area formula for BV functions (see for example Lemma~\ref{lem:Co-area} or~\cite{Mr}),
\[
\Vert Du\Vert(X)=\int_{\mathbb{R}}\Vert D\mathbf{1}_{E_t}\Vert(X)\, dt.
\]
For such $t$, by Lemma~\ref{lem:L1-norm-control} we know that
\[
\sup_{s>0}\frac{1}{\sqrt{s}}\int_X|P_s\mathbf{1}_{E_t}(x)-\mathbf{1}_{E_t}(x)|\, d\mu (x) \le C\, \Vert D\mathbf{1}_{E_t}\Vert(X).
\]
Now, setting $A=\{(x,y)\in X\times X\, :\, u(x)<u(y)\}$, we have for $s>0$,
\begin{align*}
  & \int_{X}\int_X p_s(x,y)  |u(x)-u(y)|\, d\mu(x)d\mu(y) \\
 &=2\iint_A p_s(x,y)|u(x)-u(y)|\, d\mu(x)d\mu(y) \\
 &=2\iint_A \int_{u(x)}^{u(y)}\, p_s(x,y)\, dt\, d\mu(x)d\mu(y)\\
 &=2\int_X\int_X \int_{\mathbb{R}}\mathbf{1}_{[u(x),u(y))}(t)\, \mathbf{1}_A(x,y)\, p_s(x,y)\, dt\, d\mu(x)d\mu(y) \\
 &=2\int_{\mathbb{R}} \int_X\int_X \mathbf{1}_{E_t}(y)[1-\mathbf{1}_{E_t}(x)]\, p_s(x,y)\, d\mu(x)d\mu(y)\, dt\\
 &=2\int_{\mathbb{R}}\int_X P_s\mathbf{1}_{E_t}(x)[1-\mathbf{1}_{E_t}(x)]\, d\mu(x)\, dt\\
 &=2\int_{\mathbb{R}}\int_{X\setminus E_t}P_s\mathbf{1}_{E_t}(x)\, d\mu(x)\, dt.
\end{align*}
Observe that
\[
\int_X|P_s\mathbf{1}_{E_t}(x)-\mathbf{1}_{E_t}(x)|\, d\mu(x)
 \ge \int_{X\setminus E_t}|P_s\mathbf{1}_{E_t}(x)-\mathbf{1}_{E_t}(x)|\, d\mu(x)=\int_{X\setminus E_t}P_s\mathbf{1}_{E_t}(x)\, d\mu(x).
\]
Therefore we obtain
\[
\int_X\int_X p_s(x,y) \ |u(x)-u(y)|\, d\mu(x) d\mu(y)\le 2\int_{\mathbb{R}}\Vert P_s\mathbf{1}_{E_t}-\mathbf{1}_{E_t}\Vert_{L^1(X)}\, dt.
\]
An application of Lemma~\ref{lem:L1-norm-control} now gives
\[
\int_X\int_X p_s(x,y) \ |u(x)-u(y)|\, d\mu(x)d\mu(y)\le C\sqrt{s}\, \int_{\mathbb{R}}\Vert D\mathbf{1}_{E_t}\Vert(X)\, dt,
\]
whence with the help of the co-area formula we obtain
\[
\Vert u\Vert_{1,1/2}\le C\, \Vert Du\Vert(X),
\]
that is, $u\in \mathbf{B}^{1,1/2}(X)$.
Thus $BV(X)\subset \mathbf{B}^{1,1/2}(X)$ boundedly.

\

Now we show that $\mathbf{B}^{1,1/2}(X)\subset BV(X)$; this inclusion holds
even when $\mathcal{E}$ does not support a Bakry-\'Emery curvature
condition; only a $2$-Poincar\'e inequality and the doubling condition on
$\mu$ are needed.
Now suppose that $u\in \mathbf{B}^{1,1/2}(X)$.

Then there is some $C\ge 0$ such that for each $t>0$,
\[
\int_X\int_Xp_t(x,y)|u(y)-u(x)|\, d\mu(y)\, d\mu(x)\le C\sqrt{t}.
\]
By~\eqref{eq:heat-Gauss}, we have a Gaussian lower bound for the heat kernel:
\[
 p_t(x,y)\ge \frac{e^{-cd(x,y)^2/t}}{C\, \sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}.
\]
Let $C_0=\Vert u\Vert_{\B^{1,1/2}(X)}-\Vert f\Vert_{L^1(X)}$.
Therefore, setting $\Delta_\varepsilon=\{(x,y)\in X\, :\, d(x,y)<\varepsilon\}$ for some $\varepsilon>0$, we get
\begin{align*}
C_0\, \sqrt{t}&\ge \int_X\int_X\frac{e^{-cd(x,y)^2/t}}{C\, \sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}|u(y)-u(x)|\, d\mu(y)\, d\mu(x)\\
&\ge C^{-1}\iint_{\Delta_\varepsilon} \frac{e^{-cd(x,y)^2/t}}{\sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}|u(y)-u(x)|\, d\mu(x)d\mu(y) \\
&\ge \frac{e^{-c\varepsilon^2/t}}{C}\,
 \iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\sqrt{t}))\mu(B(y,\sqrt{t}))}}\, d\mu(x)d\mu(y).
\end{align*}
With the choice of $\varepsilon=\sqrt{t}$, we now get
\[
C_0\, \varepsilon
\ge \frac{1}{C}\iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}}\, d\mu(x)d\mu(y).
\]
It follows that
\begin{equation}\label{eq:Ledoux1}
\liminf_{\varepsilon\to 0^+}\frac{1}{\varepsilon}
 \iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}}\, d\mu(x)d\mu(y) \le C\, C_0<\infty.
\end{equation}
Now an argument as in the second half of the proof of~\cite[Theorem~3.1]{MMS} tells us that
$u\in BV(X)$. We point out here that although~\cite[Theorem~3.1]{MMS} assumes that $X$ supports a $1$-Poincar\'e
inequality, this part of the proof of~\cite[Theorem~3.1]{MMS} does not need this assumption; the argument using
discrete convolution there is valid also in our setting.
We also then obtain that
\[
\Vert Du\Vert(X)\le \liminf_{\varepsilon\to 0^+}\frac{1}{\varepsilon}
 \iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}}\, d\mu(x)d\mu(y)
 \le
 \Vert u\Vert_{\B^{1,1/2}(X)}-\Vert f\Vert_{L^1(X)}.
\]
\end{proof}

\begin{remark}\label{chaining metric}
As a byproduct of this proof,  we also obtain that there exists a constant $C>0$ such that for every $u \in BV(X)$,
\begin{align*}
\sup_{\varepsilon >0} \frac{1}{\varepsilon}
 \iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}}\, d\mu(x)d\mu(y)
 \le C \liminf_{\varepsilon\to 0^+}\frac{1}{\varepsilon}
 \iint_{\Delta_\varepsilon}\frac{|u(y)-u(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}}\, d\mu(x)d\mu(y)
\end{align*}
because both sides are comparable to $\| Du(X) \|$. This property of the metric space $(X,d)$ can be viewed as
an interesting consequence of the weak Bakry-\'Emery estimate.
\end{remark}

We say that $X$ supports a $1$-Poincar\'e inequality if there are constants $C,\lambda>0$ such that
whenever $B$ is a ball in $X$ and $u\in W^{1,1}(X)$, we have
\[
\int_B|u-u_B|\, d\mu\le C\, \text{rad}(B)\, \int_{\lambda B}|\nabla u|\, d\mu.
\]

Given $A\subset X$ we set
\[
\mathcal{H}(A):=\lim_{\varepsilon\to 0^+} \inf\bigg\lbrace\sum_i\frac{\mu(B_i)}{\text{rad}(B_i)}\,
:\, A\subset\bigcup_iB_i, \text{ and }\forall i, \ \text{rad}(B_i)<\varepsilon\bigg\rbrace.
\]
In the event that $X$ supports a $1$-Poincar\'e inequality, from the results of
Ambrosio, Miranda and Pallara, we know that $P(E,X)\simeq \mathcal{H}(\partial_mE)$
whenever $E$ is of finite perimeter. Here $\partial_mE$ consists of points $x\in\partial E$
for which both
\[
\limsup_{r\to 0^+}\frac{\mu(B(x,r)\cap E)}{\mu(B(x,r))}>0
\]
and
\[
\limsup_{r\to 0^+}\frac{\mu(B(x,r)\setminus E)}{\mu(B(x,r))}>0.
\]
In this current work we do not assume the strong condition that $X$ supports a $1$-Poincar\'e inequality,
but we assume that $X$ supports a $2$-Poincar\'e inequality; in this case, we have
the double Gaussian bounds~\eqref{eq:heat-Gauss} for the heat kernel $p_t(x,y)$.
Note that even without the assumption of $2$-Poincar\'e inequality, we know that if
$\mathcal{H}(\partial E)<\infty$, then $E$ is of finite perimeter.

For $r_0>0$ and $0<\alpha<1$ we set $\partial_\alpha^{r_0}E$ to be the collection of all
points $x\in\partial E$ such that for all $0<r\le r_0$,
\[
\frac{\mu(B(x,r)\cap E)}{\mu(B(x,r))}>\alpha, \qquad
\frac{\mu(B(x,r)\setminus E)}{\mu(B(x,r))}>\alpha.
\]
Observe that $\partial_mE=\bigcup_{0<\alpha<1}\bigcup_{0<r_0<1}\partial_\alpha^{r_0}(E)$.
This union can be made a countable union by taking $\alpha$ and $r_0$ to be
rational numbers.

\begin{lemma}
Suppose that $\mu$ is doubling and supports a $2$-Poincar\'e inequality, and that
$E\subset X$ with $\Vert \mathbf{1}_E\Vert_{\mathbf{B}^{1,1/2}(X)}$ finite. Then for all
$r_0>0$ and $0<\alpha<1$,
\[
\mathcal{H}(\partial_\alpha^{r_0}E)\le \frac{C}{\alpha}\, P(E,X).
\]
Consequently, $\mathcal{H}\vert_{\partial_mE}$ is a $\sigma$-finite measure.
\end{lemma}

\begin{proof}
Note that the proof of Theorem~\ref{thm:W=BV} also tells us that even without the
Bakry-\'Emery condition~\eqref{eq:weak-BE}, we know that if $\mathbf{1}_E\in\mathbf{B}^{1,1/2}(X)$ then
$\mathbf{1}_E\in BV(X)$. By the definition of $\B^{1,1/2}(X)$, we know that
\[
\sup_{t>0}\frac{1}{\sqrt{t}}\int_X\int_Xp_t(x,y)|\mathbf{1}_E(x)-\mathbf{1}_E(y)|\, d\mu(x)\, d\mu(y)
\le C\, P(E,X).
\]
Fix $t<(r_0/3)^2$.
We cover $\partial_\alpha^{r_0}E$ by countably many balls $B_i$ of radius $\sqrt{t}$
such that the balls $5B_i$ have a bounded overlap (the bound depending solely on the
doubling constant of $\mu$, see~\cite{Hei} for example).
Then by the doubling property of $\mu$ and by the Gaussian lower bound for $p_t(x,y)$
(which follows from the $2$-Poincar\'e inequality together with the
doubling property of $\mu$),
\begin{align*}
C\, \sqrt{t}\, P(E,X)&\ge \sum_i\int_{B_i\cap E}\int_{B_i\setminus E}
p_t(x,y)\, d\mu(x)\, d\mu(y)\\
&\ge C^{-1}\sum_i\int_{B_i\cap E}\int_{B_i\setminus E}
 \frac{e^{-C}}{\mu(B_i)}\, d\mu(x)\, d\mu(y)\\
&\ge C^{-1}\sum_i\frac{\mu(B_i\cap E)\,\mu(B_i\setminus E)\, \mu(B_i)}{\mu(B_i)^2}.
\end{align*}
In the above computations, $C$ stands for various generic constants that depend only
on the doubling and Poincar\'e constants of the space, and the value of $C$ could change
at each occurrence.
Note that at least one of $\mu(B_i\cap E)/\mu(B_i)$ and $\mu(B_i\setminus E)/\mu(B_i)$
is larger than $1/2$.
Now by the definition of $\partial _\alpha^{r_0}E$
we obtain
\[
C\, P(E,X)\ge\alpha \sum_i\frac{\mu(B_i)}{\sqrt{t}}.
\]
Since $\sqrt{t}$ is the radius of each $B_i$, we get
\[
C\, P(E,X)\ge \alpha\limsup_{t\to 0^+}\sum_i\frac{\mu(B_i)}{\sqrt{t}}
\ge \alpha\, \mathcal{H}(\partial _\alpha^{r_0}E).
\]

If $0<r_1<r_0$, then
\[
\partial_\alpha^{r_0}E\subset\partial_\alpha^{r_1}E\subset\partial_mE.
\]
Setting
\[
\partial_\alpha E:=\bigcup_{0<r_0<1}\partial_\alpha^{r_0}E=\bigcup_{(0,1)\cap\mathbb{Q}}\partial_\alpha^{r_0}E,
\]
we now see by the continuity of measure that if  the sets $\partial_\alpha^{r_0}E$
are Borel sets, then
\[
\mathcal{H}(\partial_\alpha E)\le \frac{C}{\alpha}\, P(E,X).
\]
To see that $\partial_\alpha^{r_0}E$ is a Borel set we argue as follows. Recall that
we assume $\mu$ to be Borel regular. Therefore, given a $\mu$-measurable set $E$ and $r>0$, the function
\[
x\mapsto \mu(B(x,r)\cap E)
\]
is lower semicontinuous, the map
\[
\varphi_{E,r}(x):= \frac{\mu(B(x,r)\cap E)}{\mu(B(x,r))}
\]
is a Borel function. Hence the function $\Phi_{E,r_0}:=\inf_{r\in\mathbb{Q}\cap(0,r_0]}\varphi_{E,r}$ is also a Borel function,
and hence
\[
\partial_\alpha^{r_0}E=\{x\in X\, :\, \Phi_{E,r_0}(x)>\alpha\}\cap\{x\in X\, :\, \Phi_{X\setminus E,r_0}(x)>\alpha\}
\]
is a Borel set.
\end{proof}

Should $X$ support a $1$-Poincar\'e inequality, then by the results of~\cite{AMP},
there is a number $\gamma\in (0,1/2]$ such that
$\mathcal{H}(\partial_m E\setminus \partial_\gamma E)=0$; this number $\gamma$ depends
solely on the doubling and the $1$-Poincar\'e constants. Observe that the results of~\cite{AMP}
also tells us that if $E$ is of finite perimeter, then
$\mathcal{H}(\partial_m E)\simeq P(E,X)$.

\section{Metric characterization of $\mathbf{B}^{p,\alpha}(X)$}

Now we turn our attention to the study of Besov spaces in the metric setting. In this section,
we follow the notions given in~\cite{GKS}. For $0\le\alpha<\infty$, $1\le p<\infty$
and $p<q\le \infty$, we set $B^\alpha_{p,q}(X)$ to be the collection of all functions
$u\in L^p(X,\mu)$ for which
\begin{equation}\label{eq:BesovMetric-q<infty}
\Vert u\Vert_{B^\alpha_{p,q}(X)}:=\left(\int_0^\infty
\left(\int_X\int_{B(x,t)}\frac{|u(y)-u(x)|^p}{t^{\alpha p}\mu(B(x,t))}\, d\mu(y)\, d\mu(x)
\right)^{q/p}\, \frac{dt}{t}\right)^{1/q}
\end{equation}
is finite when $q<\infty$, and
\begin{equation}\label{eq:BesovMetric-q=infty}
\Vert u\Vert_{B^\alpha_{p,\infty}(X)}:=\sup_{t>0}
\left(\int_X\int_{B(x,t)}\frac{|u(y)-u(x)|^p}{t^{\alpha p}\mu(B(x,t))}\, d\mu(y)\, d\mu(x)
\right)^{1/p}
\end{equation}
is finite when $q=\infty$. In the setting considered in the next chapter,
with $\mu$ an Ahlfors $d_H$-regular measure (that is, $\mu(B(x,r))\simeq r^{d_H}$
for all $x\in X$ and $0<r<2\text{diam}(X)$)
$\mathfrak{B}^{\alpha}_{p}(X)=B^\alpha_{p,\infty}(X)$.

\begin{proposition}\label{prop:B=B}
Suppose that $\mu$ is doubling and supports a $2$-Poincar\'e inequality. Then for
$1\le p<\infty$ and $0\le \alpha<\infty$ we have
\[
\mathbf{B}^{p,\alpha/2}(X)=B^{\alpha}_{p,\infty}(X),
\]
with equivalent seminorms.
\end{proposition}

\begin{proof}
Since $\mu$ is doubling and supports a $2$-Poincar\'e inequality, we have the
Gaussian double bound for $p_t(x,y)$. Hence if $u\in \mathbf{B}^{p,\alpha}(X)$, we then
must have
\begin{align*}
\Vert u\Vert_{\mathbf{B}^{p,\alpha/2}(X)}^p
&\ge C^{-1}\sup_{t>0}\int_X\int_X\frac{|u(y)-u(x)|^p}{t^{\alpha p/2}}\,
 \frac{e^{-C\, d(x,y)^2/t}}{\mu(B(x,\sqrt{t}))}\, d\mu(y)\, d\mu(x)\\
&\ge C^{-1}\sup_{\sqrt{t}>0}\int_X \int_{B(x,\sqrt{t})}
 \frac{|u(y)-u(x)|^p}{t^{\alpha p/2}}\,
 \frac{e^{-C\, d(x,y)^2/t}}{\mu(B(x,\sqrt{t}))}\, d\mu(y)\, d\mu(x)\\
&\ge C^{-1} \sup_{\sqrt{t}>0}\int_X \int_{B(x,\sqrt{t})}
 \frac{|u(y)-u(x)|^p}{\sqrt{t}^{\alpha p}\, \mu(B(x,\sqrt{t}))}
 \, d\mu(y)\, d\mu(x)\\
&=C^{-1}\Vert u\Vert_{B^\alpha_{p,\infty}(X)}^p,
\end{align*}
and from this it follows that $\mathbf{B}^{p,\alpha/2}(X)$ embeds boundedly into
$B^\alpha_{p,\infty}(X)$.

Now we focus on proving the converse embedding.
Since $X$ supports $2$-Poincar\'e inequality, it is connected. Therefore by the doubling
property of $\mu$ (see~\cite{Hei}), there are constants $0<Q\le s<\infty$
and $C\ge 1$
such that whenever $0<r\le R$, $x\in X$, and $y\in B(x,R)$,
\begin{equation}\label{eq:mass-bounds}
\frac{1}{C}\left(\frac{r}{R}\right)^s\le \frac{\mu(B(y,r))}{\mu(B(x,R))}
 \le C\left(\frac{r}{R}\right)^Q.
\end{equation}
Therefore,
\begin{align*}
&\frac{1}{t^{\alpha p/2}}\int_X\int_X|u(y)-u(x)|^p\, p_t(x,y)\, d\mu(y)\, d\mu(x)\\
&\le \frac{C}{t^{\alpha p/2}}\int_X\sum_{i=-\infty}^\infty\int_{B(x,2^i\sqrt{t})\setminus B(x,2^{i-1}\sqrt{t})}
 \frac{|u(y)-u(x)|^p\, e^{-C4^i}}{\sqrt{\mu(B(\sqrt{t}))\, \mu(B(y,\sqrt{t}))}}\, d\mu(y)\, d\mu(x)\\
 &\le \frac{C}{t^{\alpha p/2}}\int_X\sum_{i=-\infty}^\infty\int_{B(x,2^i\sqrt{t})}
 \frac{|u(y)-u(x)|^p\, e^{-C4^i}}{\mu(B(x,2^i\sqrt{t}))}\, \sqrt{\frac{\mu(B(x,2^i\sqrt{t}))}{\mu(B(x,\sqrt{t}))}}
 \, \sqrt{\frac{\mu(B(y,2^i\sqrt{t}))}{\mu(B(y,\sqrt{t}))}}\, d\mu(y)\, d\mu(x)\\
 &\le \frac{C}{t^{\alpha p/2}}\sum_{i=-\infty}^\infty e^{-C4^i}\,\max\{2^{is},2^{iQ}\}\, (2^i\sqrt{t})^{\alpha p}\,
 \int_X\int_{B(x,2^i\sqrt{t})}\frac{|u(y)-u(x)|^p}{(2^i\sqrt{t})^{\alpha p}\, \mu(B(x,2^i\sqrt{t}))}\, d\mu(y)\, d\mu(x)\\
 &\le C\, \Vert u\Vert_{B^\alpha_{p,\infty}(X)}^p\, \sum_{i=-\infty}^\infty e^{-C4^i}\, 2^{i\alpha p}\, \max\{2^{is},2^{iQ}\}.
\end{align*}
Since
\[
\sum_{i=-\infty}^\infty e^{-C4^i}\, 2^{i\alpha p}\, \max\{2^{is},2^{iQ}\}\le
\sum_{i\in\mathbb{N}}e^{-C4^i}\,  2^{i[\alpha p+s]}\, +\, \sum_{i=0}^\infty 2^{-i[\alpha p+Q]}<\infty,
\]
the desired bound follows.

\end{proof}

\begin{remark}
While the previous theorem gave us a way to control, from above, the $\mathcal{H}$-measure of
$\partial_mE$ for a set $E$ of finite perimeter, Proposition~\ref{prop:B=B} give us a way
to control, from below, the co-dimension $1$ Minkowski measure of $\partial E$. Given set $A\subset X$,
the co-dimension $1$-Minkowski measure of $A$ is given by
\[
\mathcal{M}_{-1}(A):=\liminf_{\varepsilon\to 0^+} \frac{\mu(A_\varepsilon)}{\varepsilon},
\]
where $A_\varepsilon=\bigcup_{x\in A}B(x,\varepsilon)$. As $\mu$ is doubling, we see
with the aid of Proposition~\ref{prop:B=B} that if
$E\subset X$ is a measurable set such that $\mathbf{1}_E\in \mathbf{B}^{1,1/2}(X)$, then
\[
\liminf_{t\to 0^+}\int_X\int_{B(x,t)}\frac{|\mathbf{1}_E(x)-\mathbf{1}_E(y)|}{t^{1/2}\mu(B(x,t))}\, d\mu(y)\, d\mu(x)
\le \mathcal{M}_{-1}(\partial E),
\]
with the limit infimum on the left-hand side a finite number.
\end{remark}

\begin{remark}
Another application of Proposition~\ref{prop:B=B} is the following. It is in general not true that
if $\Vert Du\Vert(X)=0$ then $u$ is constant almost everywhere in $X$, even if $X$ is connected.
Should $X$ support a $1$-Poincar\'e inequality, the fact that $\Vert Du\Vert(X)=0$ implies $u$ is constant
follows immediately. We can use the above proposition to show that even if we do not have $1$-Poincar\'e
inequality, if $X$ supports the Bakry-\'Emery curvature condition~\eqref{eq:weak-BE}, then
we still have the constancy of $u$ from $\Vert Du\Vert(X)=0$, for then
\[
\sup_{t>0} \int_X\int_{B(x,t)}\frac{|\mathbf{1}_E(x)-\mathbf{1}_E(y)|}{t^{1/2}\mu(B(x,t))}\, d\mu(y)\, d\mu(x)
 \simeq \Vert Du\Vert(X).
\]
\end{remark}

To conclude this section, we point out the following result that shows that $1/2$ is a  critical index for the spaces $\mathbf{B}^{1,\alpha}(X)$.

\begin{proposition}
Let $f \in \mathbf{B}^{1,\alpha}(X)$ with $\alpha >1/2$. Then, $f$ is constant.
\end{proposition}

\begin{proof}
Let $f \in \mathbf{B}^{1,\alpha}(X)$ with $\alpha >1/2$. Since $\mathbf{B}^{1,\alpha}(X) \subset \mathbf{B}^{1,1/2}(X)=BV(X)$, we deduce that $f$ is a BV function. Now since $f \in \mathbf{B}^{1,\alpha}(X)$, one has for every $ t >0$,
\[
\int_X \int_X p_t(x,y) |f(x)-f(y)| d\mu(x) d\mu(y) \le t^\alpha \| f \|_{1,\alpha}.
\]
By using the heat kernel lower bound as before, it implies
\[
\liminf_{\varepsilon\to 0^+}\frac{1}{\varepsilon}
 \iint_{\Delta_\varepsilon}\frac{|f(y)-f(x)|}{\sqrt{\mu(B(x,\varepsilon))\mu(B(y,\varepsilon))}} d\mu(x) d\mu(y) =0.
 \]
 Therefore $\| Df(X) \|=0$, so from the previous remark $f$ is constant.
\end{proof}

\section{Comparison between Sobolev and Besov seminorms}

In this section, we compare the Besov and Sobolev seminorms for $p>1$. The case $p=1$ was studied in detail in Section 4.3.

Throughout the section we will assume that $(X,d_\mathcal{E},\mu)$ is doubling, supports a $2$-Poincar\'e inequality.
As before, we will also assume that
for each $u\in\mathcal{F}$ the measure $\Gamma(u,u)$ is absolutely continuous with respect to the underlying measure
$\mu$ on $X$; as in the previous section we will denote by $|\nabla u|^2$ the Radon-Nikodym derivative of
$\Gamma(u,u)$ with respect to $\mu$.  We also assume that $\mathcal{E}$ is regular and strictly local.

\subsection{Lower bound for the Besov seminorm}

\begin{theorem}\label{thm:BesovLB}
Let $p >1$. There exists a constant $C>0$ such that for every $ u \in \B^{p,1/2}(X)\cap \mathcal{F}$,
\[
\| |\nabla u| \|_{L^p(X,\mu)} \le C \| u \|_{p,1/2}
\]
\end{theorem}

\begin{proof}
Let $u\in\B^{p,1/2}(X)\cap\mathcal{F}$. Then as in the derivation of~\eqref{eq:Ledoux1} we obtain that

for each $\eps>0$,
\[
\frac{1}{\eps^p}\iint_{\Delta_\eps}\frac{|u(x)-u(y)|^p}{\mu(B(x,\eps))}\, d\mu(y)\, d\mu(x)\le \| u \|_{p,1/2}<\infty.
\]
Fix $\eps>0$ and cover $X$ by family of balls $B_i^\eps=B(x_i^\eps,\eps)$ such that $\frac12 B_i^\eps$ are pairwise
disjoint, and let $\pip_i^\eps$ be a $C/\eps$-Lipschitz partition of unity subordinate to this cover: that is,
$0\le \pip_i^\eps\le 1$ on $X$, $\sum_i\pip_i^\eps=1$ on $X$, and $\pip_i^\eps=0$ in $X\setminus B_i^\eps$. We then
set
\[
u_\eps:=\sum_i u_{B_i^\eps}\, \pip_i^\eps.
\]
Then as each $\pip_i^\eps$ is Lipschitz, we know that $u_\eps$ is locally Lipschitz and hence is in $\mathcal{F}_{loc}(X)$.
For $x,y\in B_j^\eps$ we see that
\begin{align*}
|u_\eps(x)-u_\eps(y)|&\le \sum_{i:2B_i^\eps\cap 2B_j^\eps\ne\emptyset}|u_{B_i^\eps}-u_{B_j^\eps}||\pip_i^\eps(x)-\pip_i^\eps(y)|\\
 &\le \frac{C\, d(x,y)}{\eps} \sum_{i:2B_i^\eps\cap 2B_j^\eps\ne\emptyset}
    \left(\vint_{B_i^\eps}\vint_{B(x,2\eps)}|u(y)-u(x)|^p\, d\mu(y)\, d\mu(x)\right)^{1/p}.
\end{align*}
Therefore, by Lemma~4.1, we see that
\begin{align*}
|\nabla u_\eps|&\le \frac{C}{\eps}\sum_{i:2B_i^\eps\cap 2B_j^\eps\ne\emptyset}
    \left(\vint_{B_i^\eps}\vint_{B(x,2\eps)}|u(y)-u(x)|^p\, d\mu(y)\, d\mu(x)\right)^{1/p}\\
    &\le C\left(\vint_{2B_j^\eps} \vint_{B(x,2\eps)}\frac{|u(y)-u(x)|^p}{\eps^p}\, d\mu(y)\, d\mu(x)\right)^{1/p},
\end{align*}
and so by the bounded overlap property of the collection $2B_j^\eps$ (which is a consequence of the doubling
property of $\mu$),
\begin{align*}
\int_X|\nabla u_\eps|^p\, d\mu &\le \sum_j \int_{B_j^\eps}|\nabla u_\eps|^p\, d\mu\\
  &\le C\, \sum_j \int_{2B_j^\eps} \vint_{B(x,2\eps)}\frac{|u(y)-u(x)|^p}{\eps^p}\, d\mu(y)\, d\mu(x)\\
  &\le C\, \int_X \vint_{B(x,2\eps)}\frac{|u(y)-u(x)|^p}{\eps^p}\, d\mu(y)\, d\mu(x)\\
  &\le C\, \frac{1}{\eps^p}\int_{\Delta_{2\eps}}\frac{|u(x)-u(y)|^p}{\mu(B(x,\eps))}\, d\mu(y)\, d\mu(x)\le 2M.
\end{align*}
In a similar manner, we can also show that
\[
\int_X|u_\eps(x)-u(x)|^p\, d\mu(x)\le C \eps^p \int_{\Delta_{2\eps}}\frac{|u(x)-u(y)|^p}{\eps^p\, \mu(B(x,\eps))}\, d\mu(y)\, d\mu(x)
  \le CM\, \eps^p,
\]
that is, $u_\eps\to u$ in $L^p(X)$ as $\eps\to 0^+$. So we get that if $u\in\mathcal{F}$, then
\[
C\, \| u\|_{p,1/2}\ge \| |\nabla u| \|_{L^p(X)}.
\]
\end{proof}

\subsection{Upper bound for the Besov seminorm}

We now turn to the proof of the upper bound for the Besov seminorm in terms of the Sobolev seminorm. We will use an additional assumption: The strong Bakry-\'Emery curvature condition. As before, $P_tu$ will denote the heat extension of $u$ via the heat semigroup associated with the Dirichlet form $\mathcal{E}$. We recall that 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 still denote by $P_t$. Moreover, from the same reference, for $1 \le p <+\infty$, $P_t$ is strongly continuous  (i.e. $P_t u \to u$ in $L^p(X,\mu)$ when $t \to 0$). We will denote by $L$ the generator of $P_t$.

\

\textbf{Assumption:}
\textit{In this subsection, in addition to the assumptions stated at the beginning of Section 4.5, we will furthermore assume the strong Bakry-\'Emery curvature condition: There exists a constant $C>0$ such that for every $u \in \mathcal{F}$ and $t \ge 0$ we have $\mu$ a.e.
\begin{equation}\label{eq:strong-BE}
 |\nabla P_t u|  \le C P_t |\nabla u|  .
\end{equation}
}

\

We note that the strong Bakry-\'Emery curvature condition  implies the weak one, see the proof of Theorem 3.3 in  \cite{BK}. Examples where the strong Bakry-\'Emery estimate is satisfied include: Riemannian manifolds with non negative Ricci curvature and more generally $\mathbf{RCD}(0,+\infty)$ spaces (see \cite{EKS}), some metric graphs (see \cite{BK}), the Heisenberg group and more generally H-type groups (see \cite{BBBC, Eldredge})

A first important corollary of the strong Bakry-\'Emery estimate is the following Hamilton's type gradient estimate for the heat kernel. Such type of estimate is well-known on Riemannian manifolds with non-negative Ricci curvature (see for instance  \cite{Kotschwar}), but is new in our general framework.

\begin{theorem}\label{Hamilton estimate}
There exists a constant $C>0$ such that for every $t>0$, $x,y \in X$,
\[
| \nabla_x \ln p_t (x,y) |^2 \le \frac{C}{t} \left( 1 +\frac{d(x,y)^2}{t} \right)
\]
\end{theorem}

\begin{proof}
The proof proceeds in two steps.

\textbf{Step 1:}  We first collect a gradient bound for the heat kernel. We observe that \eqref{eq:strong-BE} implies a weaker $L^2$ version as follows
\[
 |\nabla P_t u|^2  \le C P_t (|\nabla u|^2),
\]
and hence the pointwise  heat kernel gradient bound (see \cite[Lemma 3.3]{ACDH})
\[
|\nabla_x p_t(x,y)| \le \frac{C}{\sqrt t} \frac{e^{-cd(x,y)^2/t}}{\sqrt{\mu(B(x,\sqrt t)) \mu(B(y,\sqrt t))}}.
\]
In particular, we note that $|\nabla_x p_t(x,\cdot)| \in L^p(X,\mu)$ for every $p \ge 1$.

\

\textbf{Step 2:} In a second step, we prove a reverse log-Sobolev inequality for the heat kernel. Let $\tau,\ve>0$ and $x\in X$ be fixed in this Step 2.

We denote $u=p_\tau (x,\cdot)+\ve$. One has, from the chain rule for stricly local forms \cite[Lemma 3.2.5]{FOT},

\begin{align*}
P_t (u \ln u)-P_t u \ln P_t u &=\int_0^t \partial_s \left( P_s (P_{t-s} u  \ln P_{t-s} u) \right) ds \\
&=\int_0^t LP_s (P_{t-s} u \ln P_{t-s} u) -P_s (LP_{t-s}u \ln P_{t-s} u) -P_s(LP_{t-s}u)  ds \\
&=\int_0^tP_s (L(P_{t-s} u \ln P_{t-s} u)) -P_s (LP_{t-s}u \ln P_{t-s} u) -P_s(LP_{t-s}u)  ds\\
&=\int_0^tP_s \left[ L(P_{t-s} u \ln P_{t-s} u)) -LP_{t-s}u \ln P_{t-s} u -LP_{t-s}u \right]  ds \\
 &=\int_0^t   P_s ((P_{t-s} u) | \nabla \ln P_{t-s} u|^2)  ds, \\
 \end{align*}
 where the above computations may be justified by using the Gaussian heat kernel estimates for the heat kernel and the Gaussian upper bound for the gradient of the heat kernel obtained in Step 1. In particular, we point out that the commutation $ LP_s (P_{t-s} u \ln P_{t-s} u) =P_s (L(P_{t-s} u \ln P_{t-s} u)$ is justified by noting that $P_{t-s} u \ln P_{t-s} u -\ve \ln \ve$ is in the domain of $L$ in $L^2(X,\mu)$.
Here, $L$ is the infinitesimal generator (the Laplacian operator) associated with $\mathcal{E}$.

Thus,  from the strong Bakry-\'Emery estimate and the Cauchy-Schwarz inequality in the form of $P_s\left( \frac{f^2}{g}\right) \ge \frac{(P_sf)^2}{P_s g}$, we obtain
\begin{align*}
P_t (u \ln u)-P_t u \ln P_t u  &=\int_0^t   P_s\left(\frac{| \nabla P_{t-s}u |^2 }{P_{t-s} u } \right)   ds \\
 & \ge \int_0^t  \frac{(P_s | \nabla P_{t-s} u |)^2}{  P_s(P_{t-s} u) }    ds \\
 &\ge \frac{1}{C} \frac{1}{P_t u} \int_0^t | \nabla P_t u|^2 ds \\
  &\ge \frac{t}{C}   \frac{1}{P_t u}  | \nabla P_t u|^2
\end{align*}

Coming back to the definition of $u$, noting that $P_t p_\tau(x,\cdot)=P_{t +\tau}(x,\cdot)$ and applying the previous inequality with $t=\tau$ one deduces
\[
| \nabla_{y} \ln (p_{2t} (x,y)+\ve)|^2\le \frac{C}{t} P_t \left[  \ln \left( \frac{M_t(x)+\ve }{p_{2t} (x,\cdot) +\ve} \right) \right] (y)
\]
where $M_t(x)=\sup_{y \in X} p_t (x,y) $. Therefore, by letting $\ve \to 0$ and using the Gaussian heat kernels, one concludes
\[
| \nabla_y \ln p_{2t} (x,y) |^2 \le \frac{C}{t} \left( 1 +\frac{d(x,y)^2}{t} \right)
\]
and thus
\[
| \nabla_y \ln p_{t} (x,y) |^2 \le \frac{C}{t} \left( 1 +\frac{d(x,y)^2}{t} \right).
\]
Finally, we note that the inequality
\[
| \nabla_x \ln p_{t} (x,y) |^2 \le \frac{C}{t} \left( 1 +\frac{d(x,y)^2}{t} \right).
\]
is obtained similarly be exchanging the roles of $x$ and $y$ in our proof.
\end{proof}

\begin{corollary}
Let $p>1$. There exists a constant $C>0$ such that for every $u \in L^p(X,\mu)$,
\[
| \nabla P_t u| \le \frac{C}{\sqrt{t}} P_t( |u|^p)^{1/p}.
\]
\end{corollary}

\begin{proof}
Let $p>1$, $q$ the conjugate exponent and $u \in L^p(X,\mu)$. One has from H\"older's inequality
\begin{align*}
| \nabla P_t u| (x) & \le \int_X |\nabla _x p_t (x,y)| u(y) d\mu(y) \\
& \le \left( \int_X \frac{|\nabla _x p_t (x,y)|^q}{p_t(x,y)^{q/p}}  d\mu(y)\right)^{1/q} P_t( |u|^p)^{1/p} \\
&\le  \left( \int_X | \nabla_x \ln p_t(x,y)|^q p_t(x,y)  d\mu(y)\right)^{1/q} P_t( |u|^p)^{1/p}
\end{align*}
The proof follows then from Theorem \ref{Hamilton estimate} and the Gaussian upper bound for the heat kernel.
\end{proof}

Note that, by integrating over $X$ the previous proposition immediately yields:
\begin{lemma}\label{lem:LpGradient}
Let $p>1$. There exists a constant $C>0$ such that for every $u \in L^p(X,\mu)$
\[
\Vert |\nabla P_tu| \Vert_{L^p(X,\mu)}^2\le \frac{C}{t}\Vert u\Vert_{L^p(X,\mu)}^2.
\]
\end{lemma}

We deduce then:

\begin{lemma}\label{lem:HSminusF1}
Let  $p > 1$. There exists a constant $C>0$ such that for every $u \in L^p(X,\mu)\cap\mathcal{F}$ with
$|\nabla u|\in L^p(X,\mu)$
\[
\| P_t u -u \|_{L^p(X,\mu)} \le C \sqrt{t} \| |  \nabla u | \|_{L^p(X,\mu)}
\]
\end{lemma}

\begin{proof}
With the previous lemma in hand, the proof is similar to the one in Lemma~4.10, with $\varphi$ in
$\mathcal{F}\cap L^q(X,\mu)$ where $p^{-1}+q^{-1}=1$ As compactly supported Lipschitz functions form a
dense subclass of $L^p(X,\mu)$ and also belong to $\mathcal{F}$ from the discussion in Chapter~4, we recover
the $L^p$-norm of $P_tu-u$ by taking the supremum over all such $\varphi$ with $\int_X|\varphi|^q\, d\mu<\infty$.
\end{proof}

\begin{lemma}\label{lem:HSminusF2}
Let $p >1$, then
\[
\left( \int_X \int_X | P_tu(x)-u (y) |^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p} \le C \sqrt{t}  \| |  \nabla u | \|_{L^p(X)}.
\]
\end{lemma}

\begin{proof}
 Let $u\in L^p(X)$ and $t >0 $ be fixed in the above argument. By an application of Fubini's theorem we have
\[
\left( \int_X \int_X | P_tu(x)-u (y) |^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p}=\left( \int_X P_t ( | P_t u(x) -u|^p)(x) d\mu(x)\right)^{1/p}.
\]

The main idea now is to  adapt the proof of~\cite[Theorem~6.2]{BBBC}.

As above, let $q$ be the conjugate of $p$. Let $x \in X$ be fixed. Let $g$ be a function in $L^\infty(X,\mu)$  such that $P_t (|g|^q)(x) \le 1$.

We first note that from the chain rule:

\begin{align*}
& \partial_s \left[ P_s  ((P_{t-s} u) (P_{t-s} g))(x) \right]  \\
=& LP_s  ((P_{t-s} u) (P_{t-s} g))(x) -P_s  ((LP_{t-s} u) (P_{t-s} g))(x) -P_s  ((P_{t-s} u) (LP_{t-s} g))(x) \\
=&P_s  (L(P_{t-s} u) (P_{t-s} g))(x) -P_s  ((LP_{t-s} u) (P_{t-s} g))(x) -P_s  ((P_{t-s} u) (LP_{t-s} g))(x)  \\
=&2P_s (\Gamma(P_{t-s}u,P_{t-s}g))
\end{align*}

Therefore, we have

\begin{align*}
P_t ( ( u-P_t u(x) ) g)(x) & =P_t (ug)(x)-P_t u(x) P_t g (x) \\
 &=\int_0^t \partial_s \left[ P_s  ((P_{t-s} u) (P_{t-s} g))(x) \right] ds \\
 &=2 \int_0^t P_s \left( \Gamma(P_{t-s} u,P_{t-s} g)  \right) (x) ds \\
 &\le 2 \int_0^t P_s \left( | \nabla P_{t-s} u | | \nabla P_{t-s} g| )  \right) (x) ds \\
 &\le 2 \int_0^t P_s \left( | \nabla P_{t-s} u |^p \right)^{1/p}(x) P_s\left(  | \nabla P_{t-s} g|^q  \right)^{1/q} (x) ds
\end{align*}
Now from the strong BE estimate and H\"older's inequality we have
\[
P_s \left( | \nabla P_{t-s} u |^p \right)^{1/p}(x) \le  CP_s  \left( P_{t-s}( | \nabla  u |^p) \right)^{1/p}(x) =C P_t ( | \nabla  u |^p)^{1/p}(x)
\]
On the other hand
\[
 | \nabla P_{t-s} g|^q\le \frac{C}{(t-s)^{q/2}} P_{t-s} (|g|^q)
 \]
 Thus,
 \[
 P_s\left(  | \nabla P_{t-s} g|^q  \right)^{1/q} (x) \le \frac{C}{(t-s)^{1/2}} P_{t} (|g|^q)^{1/q}(x) \le \frac{C}{(t-s)^{1/2}} .
 \]
 One concludes
 \[
 P_t ( ( u-P_t u(x) ) g)(x) \le C \sqrt{t} P_t ( | \nabla  u |^p)^{1/p}(x)
 \]
 Thus by $L^p-L^q$ duality, one concludes
 \[
  P_t ( | u-P_t u(x)| ^p)(x)^{1/p} \le C \sqrt{t} P_t ( | \nabla  u |^p)^{1/p}(x)
 \]
 and finishes the proof by integration over $X$.
\end{proof}

We are now finally in position to prove the main result of the section.

\begin{theorem}\label{thm:BesovUB}
Let $p>1$. There exists a constant $C>0$ such that for every  $u \in L^p(X,\mu)\cap \mathcal{F}$ with $|\nabla u| \in L^p(X,\mu)$,
\[
\| u \|_{p,1/2} \le C \| |  \nabla u | \|_{L^p(X)}
\]
\end{theorem}

\begin{proof}
One has
\begin{align*}
 & \left( \int_X \int_X | u(x)-u(y)|^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p}  \\
 \le &  \left( \int_X \int_X | u(x)-P_tu (x) |^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p} + \left( \int_X \int_X | P_tu(x)-u(y)|^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p}   \\
 \le &  \| P_t u -u \|_{L^p(X)} +  \left( \int_X \int_X | P_tu(x)-u (y) |^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p}
\end{align*}
One has from the previous lemma
\[
 \| P_t u -u \|_{L^p(X)} \le C \sqrt{t} \| |  \nabla u | \|_{L^p(X)}
\]
and the term $\left( \int_X \int_X | P_tu(x)-u (y) |^p p_t (x,y) d\mu(x) d\mu(y) \right)^{1/p}$ can be controlled by the above lemma.
\end{proof}

In view of  Theorems \ref{thm:BesovLB} and \ref{thm:BesovUB}, as well as  \cite[Theorem 1.4]{ACDH},   we conclude this section by the following result on the Riesz transform.
\begin{corollary}\label{thm:RT}
Let $p>1$. Then for any $f\in L^p(X) \cap \mathcal F$,
\[
 \Vert  f\Vert_{p,1/2} \simeq \Vert \sqrt{-L} f \Vert_{L^p(X)}.
\]
Consequently,
\[
\mathbf B^{p,1/2}(X) =\mathcal L_p^{1/2},
\]
where $\mathcal{L}_p^{1/2}$ defined in Section 1.4 is the domain of the operator $\sqrt{-L}$ in $L^p(X,\mu)$.
\end{corollary}

\section{Critical exponents}

In this section, our assumptions are the same as Subsection 4.5.2., that is, the standing assumptions of this chapter as well as the strong Bakry-\'Emery curvature condition \eqref{eq:strong-BE}. We first notice that as a consequence of Lemma \ref{lem:LpGradient} and Theorem \ref{thm:BesovUB},
\begin{theorem}
Let $p>1$. There exists a constant $C>0$ such that for every $f\in L^p(X, \mu)$ and $t>0$
\[
\Vert P_t f \Vert_{p,1/2} \le \frac{C}{t^{1/2}} \Vert f \Vert_{L^p(X)}.
\]
\end{theorem}

\begin{remark}
The bound
\[
\Vert P_t f \Vert_{p,1/2} \le \frac{C}{t^{1/2}} \Vert f \Vert_{L^p(X)}, \quad p>1,
\]
likely holds if one only assumes the weak Bakry-\'Emery estimate. This may be seen  by adapting to this setting the argument yielding Theorem \ref{P:PtinBesovp4} in Chapter 5. For concision, we will not expand this remark.
\end{remark}

Similarly as in Section 1.7, we have several corollaries of the above result.

\begin{corollary}
Let $p > 1$. Let $L$ be the infinitesimal generator of $\mathcal{E}$ and $\mathcal{L}_p$ be the generator 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_p \norm{ Lf}_{L^p(X,\mu)} \| f\|_{L^p(X,\mu)}.
\end{equation}
\end{corollary}

Recall that $L$ plays the role of Laplacian in the theory of Dirichlet forms, see~\cite{FOT}.

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

\begin{corollary}
Let $p>1$. For every $f\in L^p(X,\mu)$, and $t>0$,
\[
\| P_t f -f \|_{L^p(X,\mu)} \le C_p   t^{1/2}  \liminf_{s \to 0}  s^{-1/2} \left( \int_X P_s (|f-f(y)|^p)(y) d\mu(y) \right)^{1/p}
\]
\end{corollary}

\begin{corollary}
Let $p>1$. If $f\in \mathbf B^{p,\alpha}(X)$ with $\alpha>1/2$ then $\mathcal E(f,f)=0$.
\end{corollary}

In particular, with the notation of  Section 1.8, one concludes:

\begin{proposition}
For every $p \ge 1$, $\alpha^*_p(X)=1/2$.
\end{proposition}

\section{Sobolev and isoperimetric inequalities}\label{section Sobolev local}

Combining the conclusions of this chapter with the results in Chapter \ref{section sobolev}, we immediately obtain the
following  results that encompass many of the known results found in the literature.

\begin{corollary}
Suppose that $\mu$ is doubling and
supports a $2$-Poincar\'e inequality. Assume moreover that  the volume growth condition
$\mu(B(x,r)) \ge C_1 r^Q$,  $r \ge 0$,  is satisfied for some $Q>0$,
 Then, one has the following weak type Besov space embedding. Let $0<\delta < Q $. Let $1 \le p < \frac{Q}{\delta} $.   There exists a constant $C_{p,\delta} >0$ such that for every $f \in \mathbf{B}^{p,\delta/2}(X) $,
\[
\sup_{s \ge 0} s \mu \left( \{ x \in X, | f(x) | \ge s \} \right)^{\frac{1}{q}} \le C_{p,\delta} \sup_{r>0} \frac{1}{r^{\delta+Q/p}}\biggl(\iint_{\{(x,y)\in X\times X\mid d(x,y)<r\}}|f(x)-f(y)|^{p}\,d\mu(x)\,d\mu(y)\biggr)^{1/p}
\]
where $q=\frac{p Q}{ Q -p \delta}$. Furthermore, for every $0<\delta <Q $, there exists a constant $C_{\emph{iso},\delta}$ such that for every measurable $E \subset X$, $\mu(E) <+\infty$,
\begin{align*}
\mu(E)^{\frac{d_H-\delta}{d_H}}
\le C_{\emph{iso},\delta} \sup_{r>0} \frac{1}{r^{\delta+Q}} (\mu \otimes \mu) \left\{ (x,y) \in E \times E^c\, :\, d(x,y) \le r\right\}
\end{align*}
\end{corollary}

\begin{proof}
From  the heat kernel upper bound, the  volume growth condition $\mu(B(x,r)) \ge C_1 r^Q$,  $r \ge 0$, implies the ultracontractive estimate
\[
p_t(x,y) \le \frac{C}{t^{Q/2}}.
\]
We are therefore in the framework of Chapter 2. From Theorem \ref{pol}, one deduces therefore the following. Let $0<\delta < Q $. Let $1 \le p < \frac{Q}{\delta} $.   There exists a constant $C_{p,\delta} >0$ such that for every $f \in \mathbf{B}^{p,\delta/2}(X) $,
\[
\sup_{s \ge 0}\, s\, \mu \left( \{ x \in X\, :\, | f(x) | \ge s \} \right)^{\frac{1}{q}} \le C_{p,\delta}  \| f \|_{p,\delta/2}
\]
where $q=\frac{p Q}{ Q -p \delta}$. We conclude then with Theorem \ref{prop:B=B}.

\end{proof}

In the case where $\delta=1/2$ and the weak Bakry-\'Emery estimate is satisfied, then one gets a strong Sobolev inequality.

\begin{theorem}
Suppose that $\mu$ is doubling and
supports a $2$-Poincar\'e inequality and that the weak Bakry-\'Emery estimate is satisfied. If the volume growth condition
$\mu(B(x,r)) \ge C_1 r^Q$,  $r \ge 0$,  is satisfied for some $Q>0$, then there exists a constant $C_2 >0$ such
that for every $f \in BV(X)$,
\[
\| f \|_{L^q(X,\mu)} \le C_2 \| Df \|(X)
\]
where $q=\frac{Q}{ Q-1}$. In particular, if $E$ is a set with finite perimeter in $X$, then
\[
\mu(E)^{\frac{Q-1}{Q}} \le C_2 P(E,X).
\]
\end{theorem}

\begin{proof}
We apply Theorem \ref{Sobolev}. Since $\| f \|_{1,1/2} \le C \| Df \|(X)$, it is enough to prove that $(P_{1,1/2})$ is satisfied. This follows from
\[
\sup_{s >0 }  s^{-1/2}  \int_X P_s (|f-f(y)|)(y) d\mu(y)  \le C\liminf_{s \to 0}  s^{-1/2}  \int_X P_s (|f-f(y)|)(y) d\mu(y) ,
\]
which is a consequence from Theorem \ref{thm:W=BV}.
\end{proof}

config: arxiv


TITLE: Definition of mean free time in the Drude model
QUESTION [6 upvotes]: In the Drude model they derive a formule for the conductivity of a conductor.
I wonder though how the main free time $\tau$ is defined in this formula.
Wikipedia says that it is "the average time between subsequent collisions".
But I have two possible interpretations of this:

the average time an electron travels before colliding (which it seems to imply).
The average amount of time electrons have been travelling at a given time $t$ ($\tau$ will be substantially smaller in this definition).

The first definition seems to me how they describe it, while the second definition seems to be implied by the formulas. I wonder the same for "mean free path", which seems to be analogous.

REPLY [11 votes]: This is a really good question, with a mind-bending answer. Check this out:
(A) Pick a random electron at a random time. How long (on average) do I need to wait until the next time it collides?
(B) Pick a random electron at a random time. How long (on average) has it been since the last time it collided?
(C) Pick a random electron that just collided. How long (on average) do I need to wait until the next time it collides?
(D) Pick a random electron that just collided. How long (on average) has it been since the last time it collided?
The answer is that all four of these are the same. (All four of them are what's called "mean free time".) It's mind-bending because it seems like (C) should be the sum of (A) plus (B), not the same as (A) and (B). But if you think about it, (C) and (A) have to be the same, because the fact that the electron just collided - something that happened in the past - cannot give any information about the statistics of what will happen to that electron in the future. We are assuming that collisions occur randomly! Similarly, (B) and (D) have to be the same. (A) and (B) have to be the same, obviously, because of time-reversal symmetry, and likewise (C) and (D) have to be the same.
Here's something that might help you come to peace with this apparent paradox (the so-called "Inspection Paradox"): An average where you pick a random electron at a random time is different than an average where you pick a random collision-to-collision free trajectory. (The former is relevant for (A-B), the latter is relevant for (C-D).) Compared to the former, the latter gives disproportionate weight to very short free trajectories. In other words, the ensemble of collision-to-collision free trajectories that you look at in (C) and (D) are, on average, shorter in duration than the ensemble of collision-to-collision free trajectories that you look at in (A) and (B).


TITLE: Given 4 integers, $a, b, c, d > 0$, does $\frac{a}{b} < \frac{c}{d}$ imply $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$?
QUESTION [9 upvotes]: We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$?

REPLY [2 votes]: Weighted average of two numbers with positive weight lies between the two numbers.
$$
x<y\phantom{x}\leftrightarrow\phantom{x}x<\frac{b}{b+d}\cdot x+\frac{d}{b+d}\cdot y<y
$$
Now substitute $x=\frac{a}{b}$ and $y=\frac{c}{d}$


TITLE: Is $-1$ a perfect square?
QUESTION [4 upvotes]: Is $-1$ a perfect square?

We know that $i^2 = -1$. Does that mean $-1$ is a perfect square because $i$ is not an irrational or decimal number?

REPLY [22 votes]: The term perfect square is typically reserved for squares of integers, unless further context is specified, so in the regular usage of the term, $-1$ is not a perfect square.
More generally, you might want to refer to perfect squares in a ring (especially the ring of integers of an algebraic number field)—in such a context, we might say that $-1$ is a perfect square, but with reference to the ring. For example, $-1$ is a perfect square in $\mathbb{Z}[i]$, but is not a perfect square in $\mathbb{Z}$.
This situation is similar to how $i$ is not an irrational number—the term irrational number refers to real numbers which are not rational, not just any old mathematical entity which is not a rational number.


TITLE: Given $\kappa$ infinite cardinal, $\{\lambda \in \text{Card } | \ \lambda^\kappa>\lambda\}$ is a proper class
QUESTION [2 upvotes]: I'm trying to do the following exercise

Show that for any infinite cardinal $\kappa$ the classes $\{\lambda \in \text{Card }| \ \lambda^\kappa = \lambda\}$ and $\{\lambda \in \text{Card }| \ \lambda^\kappa > \lambda\}$ are proper.

I think it suffices to show that both classes are unbounded in Card (which is proper).

For the first class $\{\lambda \in \text{Card }| \ \lambda^\kappa = \lambda\}$, I'd define the following functional: $$\begin{align}f : \text{Card} &\longrightarrow \text{Card} \\ \lambda &\longmapsto \lambda^\kappa \end{align}$$ this function is increasing and continuous, therefore the class of the fixed points is unbounded.
Regarding the second class $\{\lambda \in \text{Card }| \ \lambda^\kappa > \lambda\}$, which is the complement of the previous class, I would use the fact that $\kappa^{\text{cof}(\kappa)}>\kappa$ for every infinite cardinal $\kappa$. Therefore if I consider a cardinal $\lambda$ s.t. $\text{cof}(\lambda) = \text{cof}(\kappa)$ I'd have $$\lambda^\kappa \ge\lambda^{\text{cof}(\kappa)}=\lambda^{\text{cof}(\lambda)}>\lambda$$
So I'd proceed proving that $\{\lambda \in \text{Card }| \ \text{cof}(\lambda) = \text{cof}(\kappa)\}$ is unbounded (1).

I have some doubts about the correctness of my approach to the second part of the exercise. It seems a bit too convoluted.

If the approach is correct, I'd prove (1) by noticing that given an ordinal $\alpha$ and a regular cardinal $\kappa$, then $$\aleph_{\alpha+\kappa} > \aleph_\alpha \text{ and }\text{cof}(\aleph_{\alpha+\kappa})=\kappa$$
where the sum in the index of $\aleph$ is meant as an ordinal sum. Is this correct regardless of the main exercise? Thanks

REPLY [4 votes]: The function $\lambda\mapsto\lambda^\kappa$ is not continuous. Exactly because limits of $\kappa$-cofinality will witness that.
Instead, note that for every $\mu$, $\lambda=\mu^\kappa$ satisfies $\lambda^\kappa=\lambda$.
For the second part, the approach is indeed the correct one. We essentially want to show there is a proper class of cardinals of cofinality $\kappa$. Indeed, your suggestion is the correct one.

However, one can combine the two into a single approach. Note that $\beth_{\alpha+\kappa}^\kappa>\beth_{\alpha+\kappa}$, by the same argument as you pointed. But now notice that $\lambda=\beth_{\alpha+\kappa}^+$ satisfies $\lambda^\kappa=\lambda$.


TITLE: Temperature of gases
QUESTION [2 upvotes]: I can't find any law that states this (maybe the combined gas law does and I'm misinterpreting it?), but Feynman said that if you compress a gas, the temperature increases. This makes sense, for example, a diesel engine (or gas engine with insufficient octane or too high a compression ratio). Also, must thinking about a piston "hitting" particles as it is compressed makes sense that energy is imparted.
But he goes on to say that when the gas expands, there is a decrease in temperature. This used to make more sense to me, but the more I think about it, it doesn't at all. Why would the particles lose energy if the container expands?

REPLY [5 votes]: Just for completeness, when a gas expands its temperature does not necessarily change. The temperature of the gas only changes if it does work on something, for example its container as discussed in Danu's answer. If a gas is expanding into a vacuum it does no work and (to a first approximation) its temperature does not change. This type of expansion is known as a Joule expansion.
I used the qualifier to a first approximation above because the temperature is only guaranteed not to change if the gas is ideal. In real gases there are forces acting between the gas molecules and even in a Joule expansion the gas may do work against these forces and the kinetic energy of the gas molecules and therefore the temperature may change. This is known as the Joule-Thomson effect.


TITLE: Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$
QUESTION [1 upvotes]: If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$.
What should be the smart approach to this question. If I try equating area of triangle formed by three points equal to $0$, then it gets too complicated. Please provide some insight.

REPLY [1 votes]: HINT.- I do not  know if my answer could be “the smart approach to this question” as you want. Anyway here I give you the following.
Equalizing slopes
$$\frac{\frac{a^3-3}{a-1}-\frac{b^3-3}{b-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{\frac{3}{b-1}-\frac{3}{a-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{3}{ab(a+b)-(a^2+ab+b^2)}$$
So, by symmetry,  $$a^2b+ab^2-(a^2+ab+b^2)=a^2c+ac^2-(a^2+ac+c^2)=b^2c+bc^2-(b^2+bc+c^2)$$
It follows $$\begin{cases}\color{green}{(a^2b+ab^2)}+(a^2+ac+c^2)=a^2c+ac^2+\color{green}{(a^2+ab+b^2)}\\\color{green}{(a^2b+ab^2)}+(b^2+bc+c^2)=b^2c+bc^2+\color{green}{(a^2+ab+b^2)}\end {cases}\qquad(*)$$
Hence, by subtracting the equalities in (*),
$$(a^2+ac+c^2)- (b^2+bc+c^2)= (a^2c+ac^2)-(bc^2+bc^2)\Rightarrow(c-1)(a+b+c)=0$$
Thus, because of $c\ne 1$, $$\color {red}{a+b+c=0}$$
Consequently you have to prove the easier relation
$$abc-(ab+ac+bc)=0$$
There are several ways to solve now. Try to finish the question. If you cannot  (I believe you can), I will return.


TITLE: Cartier divisor on an open subscheme whose complement is of codim 2
QUESTION [2 upvotes]: 1- If a Cartier divisor is defined away from a codim 2 closed subset, when can we say that this must be a restriction of a Cartier divisor on the whole scheme?
2- If two Cartier divisor agree away from a codim 2 closed subset, when can we say that they must be the same?

REPLY [10 votes]: 1 - This is true for locally factorial schemes; in this case all Weil divisors are Cartier, and extending Weil divisors is not a problem.
2 - This is not true in general (for example, if you glue together two points of the affine plane over a field k, the Picard group of the resulting scheme is $k^*$, while the Picard group of the complement is trivial). It is true for schemes satisfying Serre's S2 condition.


TITLE: Compute: as $\displaystyle{\lim_{n\to \infty} \int_0^\pi \sin^n x dx}$
QUESTION [3 upvotes]: $$\lim_{n\to \infty} \int_0^\pi \sin^n x dx$$
I have been trying to compute this integral for months and have gotten nowhere. I seem to be going in circles. Any help at all would be seriously appreciated.

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}
\int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} =
\lim_{n \to \infty}
\int_{-\pi/2}^{\pi/2}\cos^{n}\pars{ x}\,\dd x =
2\lim_{n \to \infty}\int_{0}^{\pi/2}\cos^{n}\pars{ x}\,\dd x
\end{align}
The main contribution to the integration comes from values of $\ds{x \gtrsim 0}$. Then,
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}
\int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} =
2\lim_{n \to \infty}
\int_{0}^{\pi/2}\expo{n\ln\pars{\cos\pars{x}}}\,\,\,\dd x
\\[5mm] = &\
2\lim_{n \to \infty}
\int_{0}^{\infty}\expo{-nx^{2}\,/2}\,\,\dd x\qquad
\pars{~Laplace\,'s\ Method~}
\\[5mm] = &\
2\lim_{n \to \infty}\pars{\root{2 \over n}{\root{\pi} \over 2}} =
\bbx{0} \\ &
\end{align}


TITLE: Difference in weight while pulling weights
QUESTION [1 upvotes]: Let's say you are using a standard pulley gym machine to lift some amount of weight. If while you exercise you are standing on a weighing scale. What variations in weight if any would occur?
Does it depend on the position of the weights? For example would you increase in weight until it reaches what would essentially be your center of mass?
I think your weight increases while while the weights are going up (you are pulling) and decreases back to your original weight on it's weight down. Probably some scale jitter when you start pulling.

REPLY [1 votes]: It is actually the opposite, the weight in the balance will go down in the same amount than the force applied to the bar attached to the pulley. This is because the force this bar makes on the person is upwards. The only way the weight in the scale will mark more is if you are pushing something up (but not if slowly releasing it up, because you are still making a down force in such a case)

REPLY [1 votes]: It depends which type of machine it is, quite simply.
If it is the type that are pulled upwards (which have multiple pulleys), then it will add the amount of force you are applying to your weight.
If it is the type that are pulled downwards (which usually have only one pulley and therefore the force required is equal to the weight), then it will reduce your scale reading accordingly.
In regard as to when it adds how much, I am not sure. I would assume that as soon as the weights are relying entirely upon you to hold them aloft, their weight would be added/taken away, but as my old gym teacher used to say, to assume makes an ass out of you (u) and me...

REPLY [1 votes]: There are several possible scenarios as pointed out by other users, but I'll focus on the following one:
Consider the case where the weights are pulled downwards. For the sake of simplicity, let's assume we're just holding the weight (so we can ignore the shifting center of mass, since a human is not a rigid body).
Observe the diagram, where a human of mass $M$ is holding a weight of mass $m$ using 1 (assumed massless and frictionless) pulley:

Note that the magnitude of the normal force $\mathbf N$ is what the scale will read. As you can see, the upwards tension will decrease the required normal force for the human to remain stationary.
Edit: yes, the normal force will be exerted at the person's feet (so in two spots), but the way I drew it makes the magnitude of the normal force clearer.


TITLE: Proof by induction to prove an inequality help?
QUESTION [1 upvotes]: Prove by induction on m where m is an integer, such that m ≥ 2:
$$Pm:=\sum_{n=1}^{m} \frac{1}{\sqrt{n}} < 2\sqrt{m}-1$$
I know this holds for the base case since when m=2, P2 is:
$$4<3\sqrt2$$ which is clearly true. The inductive hypothesis is suppose Pn is true for all m ≥ 2, which we will use to show Pm+1 is true for m ≥ 2, ie:
$$\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}} < 2\sqrt{m+1}-1$$
The LHS of the inequality can be rewritten as:
$$x=\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}}= \sum_{n=1}^{m} \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{m+1}}$$
Thus, the induction hypothesis implies:
$$x <\frac{1}{\sqrt{m+1}}+2\sqrt{m}-1$$
So my question, would my proof my induction be valid if I show:
$$\frac{1}{\sqrt{m+1}}+2\sqrt{m}-1< 2\sqrt{m+1}-1$$?
Working out the algebra, I found this last inequality is equivalent to the statement, "0< m+1" which is clearly true if m is a positive integer. So does this pretty much complete the proof?

REPLY [0 votes]: $P$ a partition of $[1,m]$ $P=\{1,2,...m\}$ and $f(x)=\frac{1}{\sqrt{x}}$
$$Pm=I(f,P)<\int_{1}^{m} \frac {dx}{\sqrt{x}}=2\sqrt{m}-2< 2\sqrt{m}-1$$