[${\bf NOTATIONS}$]
Let $M$ be a closed Riemannian manifold of $m$ dimensional, $p\in\{1,\cdots,m\}$. $A^p:=\{\text{smooth p-forms on }M\}$. $\delta:A^{p+1}\to A^p$ denotes the formally adjoint operator w.r.t. the $L^2$-metric on $A^p$ ,that is, \begin{eqnarray} \delta\alpha:=g^{ij}\iota\left(\frac{\partial}{\partial x^j}\right)\nabla_{\frac{\partial}{\partial x^i}}\alpha \end{eqnarray} , which satisfies \begin{eqnarray} \langle d\alpha, \beta\rangle_{L^2}+\langle\alpha, \delta\beta\rangle_{L^2}=0\ \ \ (\alpha\in A^p, \beta\in A^{p+1}). \end{eqnarray} We define $\Delta\alpha:=d\delta\alpha+\delta d\alpha\ \ \ (\alpha\in A^p)$.
Let $\alpha\in A^p$. We accept the fact that there is an unique smooth solution $\beta:[0,\infty)\to A^p$ of the equation \begin{align} \begin{cases} \dfrac{\partial\beta}{\partial t}(t)=\Delta\beta(t)\ \ \ (t\geq0)\\ \beta(0)=\alpha. \end{cases} \end{align} Then for each $t\geq0$ we can define a linear operator $T_t(=e^{t\Delta}):A^p\to A^p$ by \begin{eqnarray} T_t\alpha:=\beta(t). \end{eqnarray}
Since $\dfrac{d}{dt}||\beta(t)||_{L^2}^2\leq0$, $||T_t||\leq1$ therefore $T_t$ extends continuously to $T_t:L^2A^p\to L^2A^p$, where $L^2A^p:=\{\text{measurable p-forms on }M\text{ whose norms belong to }L^2(M)\}$.
[${\bf PROPERTIES}$] I was able to follow the proofs of the following properties:
$T_t\circ T_s=T_{s+t}\ \ \ (s,t\geq0)$.
$T_t$ is symmetric i.e. $\langle\alpha_1, T_t\alpha_2\rangle_{L^2}=\langle T_t\alpha_1,\alpha_2\rangle_{L^2}$.$\exists H\alpha\in A^p\ \ \ T_t\alpha\to H\alpha$ in $L^2(M)$.
[${\bf QUESTION}$]
I'm reading Milgram and Rosenbloom Harmonic Forms and Heat Conduction
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063329/?page=1
. Let $\alpha\in A^p$ be a closed form. I need to show the property that
there exists a sequence $(t_n)_{n\in\mathbb{N}}$ s.t. $\lim t_n=\infty$ and $T_{t_n}\alpha\to H\alpha$ uniformly on $M$.
But I cannot understand the explanation in the paper below. Please help me. Thank you.
