# Short time $L^1$ bounds for semigroups obtained from elliptic operators

Let $\Omega$ be a domain in $\mathbb{R}^n$ with smooth boundary, and let $L$ be a negative definite second order elliptic differential operator defined with $\mathcal{D}(L) \subset H^2(\Omega)$, given by either Dirichlet or Neumann boundary conditions. My question is, for $f \in L^1(\Omega)$, do we have the following for small time $t$: $$\Vert e^{tL}f \Vert_{L^1(\Omega)} \leq C\Vert f\Vert_{L^1(\Omega)}?$$

The reason why I expect this to hold is, I expect something like $e^{tL}f \to f$ pointwise almost everywhere. Is this true? In particular, does $\Omega$ need to be bounded for this? Thanks in advance.

I can give an answer for divergence-type operators. Let $$Q(u,v)=\int_\Omega a\nabla u\cdot\nabla v\,dx$$ for $a=(a_{jk})$ with $a_{jk}\in L^\infty_{\mathrm{loc}}(\Omega)$ and $\sum_{j,k}a_{jk}(x)\xi_j\xi_k\geq c|\xi|^2$ for some $c>0$ and all $x\in\Omega$, $\xi\in\mathbb{R}^n$ (indeed, it suffices to assume the strong ellipticity condition only locally). Then $Q$ is a closed form on $D(Q)=\{u\in L^2(\Omega)\mid \int a\nabla u\cdot\nabla u<\infty\}$. If $a_{jk}\in L^\infty(\Omega)$, then $D(Q)=H^1(\Omega)$.

Furthermore, the generator of $Q$ is the divergence-type operator $$Lu=\mathrm{div}(a\nabla u)$$ with Neumann boundary conditions whenever you can make sense of that expression (say, if $a_{jk}\in C^\infty$).

If one restricts $Q$ to $D(Q_0)=\overline{C_c^\infty(\Omega)}^{\lVert\cdot\rVert_Q}$, then the generator is the corresponding divergence-type operator with Dirichlet boundary conditions.

The important property of $Q$ is that it is a Dirichlet form (see [MR92], Sectiion II.2), that is, if $u\in D(Q)$, then $(u\vee 0)\wedge 1\in D(Q)$ and $Q((u\vee 0)\wedge 1)\leq Q(u)$. The same is true for $Q_0$. Essentially, one only needs a sufficiently strong chain rule to see that $$Q((u\vee 0)\wedge 1)=\int_\Omega 1_{0\leq u\leq 1} a\nabla u\cdot \nabla u\,dx\leq Q(u).$$ By abstract theory, this property of $Q$ is equivalent to $0\leq e^{tL}u\leq 1$ for all $0\leq u\leq 1$ (such semigroups are called sub-Markovian). This equivalence was probably first stated in [BD58], but without proof. A proof can be deduced from the more general main theorem in [Ouh96] (of course, there are also direct proofs, I just don't have a reference at hand).

Since $(e^{tL})$ is sub-Markovian, it extends to a strongly continuous contraction semigroup on $L^\infty(\Omega)$, and by taking adjoints one also gets a strongly continuous contraction semigroup on $L^1(\Omega)$. Since $^{tL}$ is symmetric on $L^2$, this construction yields the original semigroup on $L^1\cap L^2$.

Yes, the inequality holds, you can choose $C=1$, you do not even have to restrict to small times, $e^{tL}f$ converges to $f$ in $L^1$-norm as $t\to 0$, and it does not matter whether $\Omega$ is bounded or not.
• Very nice answer, thanks! One question, if I may: if we were on a compact setting, by the Hille-Yosida theorem, we would have $e^{tL} f \to f$ in $L^2$-norm, which would imply, $e^{tL}f \to f$ in $L^1$ norm. In that case, we would not need divergence form, correct? – user95820 Aug 6 '16 at 9:02
• @user95820 Yes, as long as you can assure that $L$ generates a strongly continuous semigroup. I should add that divergence form is not really necessary for all I wrote above, you get a Dirichlet form under much more general conditions. Take a look at "Feller semigroups, Lp-sub-Markovian semigroups, and applications to pseudo-differential operators with negative definite symbols" by Farkas, Jacob and Schilling. Their main interest lies in pseudo-differential operators, but they also have an theorem dealing explicitely with quite general elliptic differential operators. – MaoWao Aug 6 '16 at 14:30
• @MaoWao Does the same hold true (with $C=1$ and in $L^\infty$ norm) also for other general boundary conditions (for example, Robin)? Also, do you have a reference that states and proves such maximum principle? – user60665 Mar 6 at 18:08