Proof. Let \(\varepsilon > 0\) be arbitrary but fixed. First we use Lemma 6 to argue that \[\begin{align} - \Re \int_{\partial \Omega} B\gamma(u) \overline{\gamma(u)} \,\mathrm{d}\sigma(x) & \leq \left| \int_{\partial \Omega} B\gamma(u) \overline{\gamma(u)} \,\mathrm{d}\sigma(x) \right| \leq \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \| \gamma(u) \|_{\mathrm{L}^{2}(\partial \Omega)}^{2} & \\ \;\\ & \leq \varepsilon \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \| \nabla u \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2} + \beta(\varepsilon) \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \| u \|_{\mathrm{L}^{2}(\Omega)}^{2} & \end{align}\] is true for every \(u \in H^{1}(\Omega)\). Furthermore \[\Re \int_{\Omega} A \nabla u \cdot \overline{\nabla u} \,\mathrm{d}x \;\geq \;\alpha \| \nabla u \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2}\] holds for \(u \in H^{1}(\Omega)\) which proves the claim. ◻

Proof.

i. Let us prove the continuity of \(\tilde{a}(B)\) first. Note that \(\| u \|_{\mathrm{L}^{2}\left( \Omega \right)} \leq \| u \|_{\tilde{a}(B)}\) is true for every \(u \in D \left( \tilde{a}(B) \right)\). For \(u,v \in D \left( \tilde{a}(B) \right)\) we easily conclude that \[\left| \tilde{a}(B)(u,v) \right| \leq \left| \langle A \nabla u, \nabla v \rangle_{\mathrm{L}^{2}\left( \Omega \right)^{d}} \right| + \left| \langle B \gamma(u), \gamma(v) \rangle_{\mathrm{L}^{2}\left( \partial \Omega \right)} \right| + \lambda_{0} \left| \langle u, v \rangle \right|\] is true. Now we focus on \[\begin{align} \left| \langle A \nabla u, \nabla v \rangle_{\mathrm{L}^{2}\left( \Omega \right)^{d}} \right| & = \left| \int_{\Omega} \left( A(x) (\nabla u)(x) \right)^{T} \overline{(\nabla v)(x)} \,\mathrm{d}x \right| & \\ & \leq \sum_{l=1}^{d} \sum_{k=1}^{d} \int_{\Omega} \left| a_{lk}(x) \right| \;\left| (\partial_{k}u)(x) \right| \;\left| (\partial_{l}u)(x) \right| \;\,\mathrm{d}x & \\ & \leq \max_{l,k} \left\| a_{l,k} \right\|_{\mathrm{L}^{\infty}\left( \Omega \right)} \sum_{l=1}^{d} \sum_{k=1}^{d} \left\| u \right\|_{H^{1}\left( \Omega \right)} \left\| v \right\|_{H^{1}\left( \Omega \right)} & \\ & \leq d^{2} \left\| A \right\|_{\mathrm{L}^{\infty}\left( \Omega \right)^{d \times d}} \left\| u \right\|_{H^{1}\left( \Omega \right)} \left\| v \right\|_{H^{1}\left( \Omega \right)}. & \end{align}\] Therefore we conclude to \[\begin{align} \left| \tilde{a}_{B}(u,v) \right| & \leq \left| \langle A \nabla u, \nabla v \rangle_{\mathrm{L}^{2}\left( \Omega \right)^{d}} \right| + \left| \langle B \gamma(u), \gamma(v) \rangle_{\mathrm{L}^{2}\left( \partial \Omega \right)} \right| + \lambda_{0} \left| \langle u, v \rangle \right| & \\ \;\\ & \leq d^{2} \left\| A \right\|_{\mathrm{L}^{\infty}\left( \Omega \right)^{d \times d}} \left\| u \right\|_{H^{1}\left( \Omega \right)} \left\| v \right\|_{H^{1}\left( \Omega \right)} + \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \; \| \gamma (u) \|_{\mathrm{L}^{2}\left( \partial \Omega \right)} \; \| \gamma (v) \|_{\mathrm{L}^{2}\left( \partial \Omega \right)} & \\ & + \lambda_{0} \| u \|_{\mathrm{L}^{2}\left( \Omega \right)} \| v \|_{\mathrm{L}^{2}\left( \Omega \right)} & \\ \;\\ & \leq \left( \;d^{2} \;\| A \|_{\mathrm{L}^{\infty}(\Omega)^{d \times d}} + \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \; \| \gamma \|_{H^{1} \to \mathrm{L}^{2}}^{2} \;\right) \| u \|_{H^{1}\left( \Omega \right)} \| v \|_{H^{1}\left( \Omega \right)} + \lambda_{0} \| u \|_{\mathrm{L}^{2}\left( \Omega \right)} \| v \|_{\mathrm{L}^{2}\left( \Omega \right)} & \\ \;\\ & \leq C \| u \|_{\tilde{a}(B)} \| v \|_{\tilde{a}(B)} & \end{align}\] for a sufficiently large constant \(C>0\) independent of \(u\) and \(v\).

ii. Now, let us show the closedness of \(\tilde{a}(B)\). Therefore let \((u_{n}) \subset D \left( \tilde{a}(B) \right)\) be a Cauchy sequence with respect to \(\| \cdot \|_{\tilde{a}(B)}\). From \(( \ref{H195inequ} )\) we conclude that \((u_{n})\) is Cauchy in \(H^{1}(\Omega)\). Hence there exists a limit \(u \in H^{1}(\Omega) = D \left( \tilde{a}(B) \right)\) of \((u_{n})\) with respect to \(\| \cdot \|_{H^{1}(\Omega)}\). Using the inequality \[\begin{align} & \left| \tilde{a}(B)(u_{n}-u, u_{n}-u) \right| \leq & \\ & \left( \;d^{2} \;\| A \|_{\mathrm{L}^{\infty}(\Omega)^{d \times d}} + \| B \|_{\mathrm{L}^{2} \to \mathrm{L}^{2}} \; \| \gamma \|_{H^{1} \to \mathrm{L}^{2}}^{2} \;\right) \| u_{n}-u \|_{H^{1}\left( \Omega \right)}^{2} + \lambda_{0} \| u_{n}-u, \|_{\mathrm{L}^{2}\left( \Omega \right)}^{2} & \end{align}\] we saw earlier in the proof of Lemma 10, we conclude that \(u\) is also a limit of \((u_{n})\) with respect to \(\| \cdot \|_{\tilde{a}(B)}\).

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Proof. The claim is also implied by the inequality \[\left| \tilde{a}(B)(u,v) \right| \leq C \| u \|_{H^{1}\left( \Omega \right)} \| v \|_{H^{1}\left( \Omega \right)} + \lambda_{0} \| u \|_{\mathrm{L}^{2}\left( \Omega \right)} \| v \|_{\mathrm{L}^{2}\left( \Omega \right)}\] for a costant \(C > 0\) from the proof of Lemma 10.
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Proof.

i. The claim is a direct implication of \[\begin{align} a(B)(u,v) & = \int_{\Omega} A \nabla u \cdot \overline{\nabla v} \,\mathrm{d}x + \int_{\partial \Omega} B \gamma(u) \;\overline{\gamma(v)} \,\mathrm{d}\sigma(x) \\ & = \tilde{a}(B)(u,v) - \lambda_{0} \int_{\Omega} u \;\overline{v} \,\mathrm{d}x. \end{align}\]

ii. Note that \(\mathrm{e}^{t\lambda_{0}}\mathrm{e}^{-t\tilde{L}(B)}\) indeed is a strongly continuous operator semigroup in \(\mathrm{L}^{2}\left( \Omega \right)\). For \(u \in D \left( L(B) \right)\) we conclude to \[\frac{\partial}{\partial t} \mathrm{e}^{-tL(B)} u = \frac{\partial}{\partial t} \mathrm{e}^{t\lambda_{0}}\mathrm{e}^{-t\tilde{L}(B)} u = \lambda_{0} \mathrm{e}^{t\lambda_{0}}\mathrm{e}^{-t\tilde{L}(B)} u - \tilde{L}(B) \mathrm{e}^{t\lambda_{0}}\mathrm{e}^{-t\tilde{L}(B)} u = - L(B) \mathrm{e}^{-tL(B)} u.\]

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Proof. Let \(u \in \mathrm{L}^{2}\left( \Omega \right)\) and note that \(\mathrm{L}^{2}\left( \Omega \right) \subset \mathrm{L}^{1}\left( \Omega \right)\) is true since \(\Omega\) is a bounded Lipschitz domain in \(\mathbb{R}^{d}\). Then \(\mathrm{e}^{-tL(B)^{\ast}}u \in \mathrm{L}^{2}\left( \Omega \right)\) is contained in \(\mathrm{L}^{1}\left( \Omega \right)\). Using the Hahn-Banach extension theorem there exists a \(v \in \mathrm{L}^{\infty}( \Omega ) = \left( \mathrm{L}^{1}( \Omega ) \right)^{\ast}\) with \(\| v \|_{\mathrm{L}^{\infty}(\Omega)} = 1\) such that \[\langle \mathrm{e}^{-tL(B)^{\ast}}u, v \rangle \;= \;\left\| \mathrm{e}^{-tL(B)^{\ast}}u \right\|_{\mathrm{L}^{1}(\Omega)}\] is true. Note that \(v \in \mathrm{L}^{2} \left( \Omega \right)\) is implied by \(\mathrm{L}^{\infty} \left( \Omega \right) \subset \mathrm{L}^{2} \left( \Omega \right)\) since \(\Omega\) is bounded. Hence we conclude to \[\begin{align} \left\| \mathrm{e}^{-tL(B)^{\ast}}u \right\|_{\mathrm{L}^{1}(\Omega)} & = \left| \langle \mathrm{e}^{-tL(B)^{\ast}}u, v \rangle \right| = \left| \langle u, \mathrm{e}^{-tL(B)}v \rangle \right| \\ \;\\ & \leq \left\| \mathrm{e}^{-tL(B)}v \right\|_{\mathrm{L}^{\infty}(\Omega)} \;\| u \|_{\mathrm{L}^{1}(\Omega)} \leq \left\| \mathrm{e}^{-tL(B)} \right\|_{ \mathrm{L}^{\infty} \to \mathrm{L}^{\infty} } \; \left\| v \right\|_{\mathrm{L}^{\infty}(\Omega)} \;\| u \|_{\mathrm{L}^{1}(\Omega)} \\ \;\\ & = \left\| \mathrm{e}^{-tL(B)} \right\|_{ \mathrm{L}^{\infty} \to \mathrm{L}^{\infty} } \;\| u \|_{\mathrm{L}^{1}(\Omega)} \end{align}\] is true where we used that \(\mathrm{e}^{-tL_{B}}\) is continuous in \(\mathrm{L}^{\infty}\left( \Omega \right)\) and \(\mathrm{e}^{-tL(B)^{\ast}}\) is the adjoint operator of \(\mathrm{e}^{-tL_{B}}\) in \(\mathrm{L}^{2}\left( \Omega \right)\). Therefore \(\mathrm{e}^{-tL(B)^{\ast}}\) is continuously extendable to a bounded operator in \(\mathrm{L}^{1}(\Omega)\) with \(\left\| \mathrm{e}^{-tL(B)^{\ast}} \right\|_{\mathrm{L}^{1} \to \mathrm{L}^{1}} \;\leq \; \left\| \mathrm{e}^{-tL(B)} \right\|_{ \mathrm{L}^{\infty} \to \mathrm{L}^{\infty} }\) since \(\mathrm{L}^{2}(\Omega)\) is dense in \(\mathrm{L}^{1}(\Omega)\).
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Proof. We use Theorem 2.21 on page 60 in [3] for our argumentation. Therefore we have to use the accretive forms \(\tilde{a}(B)\) and \(\tilde{a}(\overline{B})\). We chose \(\lambda_{0}\) sufficiently large that both forms are accretive.

i) First we have to show that \(\mathrm{e}^{-t\tilde{L}(\overline{B})}\) are all positive operators. Let \(u \in C^{1}( \overline{\Omega} )\). Then \(\partial_{k}(\Re u)^{+} \partial_{l} (\Re u)^{-} = 0\) is true almost everywhere in \(\Omega\) due to disjoint supports where we remember that \((\Re u)^{+}, (\Re u)^{-} \in H^{1}(\Omega)\). So \[\langle A\nabla (\Re u)^{+}, \nabla (\Re u)^{-} \rangle_{\mathrm{L}^{2}\left( \Omega \right)^{d}} = 0\] is implied. Hence we infer \[\begin{align} & \tilde{a}(\overline{B}) \left( (\Re u)^{+}, (\Re u)^{-} \right) = \int_{\partial \Omega} \overline{B} (\Re u)^{+} \;(\Re u)^{-} \,\mathrm{d}\sigma(x) & \\ & = - \int_{\partial \Omega} \underbrace{B_{pos} (\Re u)^{+}}_{\geq 0} \;\underbrace{(\Re u)^{-}}_{\geq 0} \,\mathrm{d}\sigma(x) - \int_{\partial \Omega} \underbrace{\left( \beta h - B_{pos} h \right)}_{\geq 0} \;\underbrace{(\Re u)^{-}}_{\geq 0} \,\mathrm{d}\sigma(x) \leq 0. & \end{align}\] Using Theorem 2.6 on page 50 in [3] we conclude that \(\mathrm{e}^{-t\tilde{L}(\overline{B})}\) are positive operators since \(C^{1}( \overline{\Omega} )\) is a core of \(\tilde{a}(\overline{B})\).

ii) First note that \(H^{1}(\Omega)\) is an ideal in itself by Proposition 2.20 on page 59 in [3] since \(\mathrm{e}^{-t\tilde{L}(\overline{B})}\) are positive. Now let \(u,v \in H^{1}(\Omega)\) be such that \(u\overline{v} \geq 0\). In the following we argue that \[\tilde{a}(\overline{B})(|u|, |v|) \;\leq \;\Re \tilde{a}(B)(u, v)\] is implied. First we conclude that \[\begin{align} - \Re \langle B\gamma(u), \gamma(v) \rangle_{\mathrm{L}^{2}( \partial \Omega )} & \leq \left| \langle B\gamma(u), \gamma(v) \rangle_{\mathrm{L}^{2}( \partial \Omega )} \right| \leq \langle \left| B\gamma(u) \right|, |\gamma(v)| \rangle_{\mathrm{L}^{2}( \partial \Omega )} & \\ \;\\ & \leq \langle B_{pos} \left| \gamma(u) \right|, |\gamma(v)| \rangle_{\mathrm{L}^{2}( \partial \Omega )} & \\ \;\\ & \leq \langle B_{pos} \left| \gamma(u) \right| + \left( \beta h - B_{pos} h \right), |\gamma(v)| \rangle_{\mathrm{L}^{2}( \partial \Omega )} & \\ \;\\ & = \langle - \overline{B} \left| \gamma(u) \right|, \left| v \right| \rangle_{\mathrm{L}^{2}( \partial \Omega )} & \end{align}\] is true. Hence \(\langle \overline{B} \left| \gamma(u) \right|, \left| v \right| \rangle_{\mathrm{L}^{2}( \partial \Omega )} \leq \Re \langle B\gamma(u), \gamma(v) \rangle_{\mathrm{L}^{2}( \partial \Omega )}\) is implied.
We need similar results for the remaining part of the forms. Note that \(\mathrm{e}^{-t\tilde{L}(0)}\) are positive operators. Hence \[\left| \mathrm{e}^{-t\tilde{L}(0)}w \right| \leq \mathrm{e}^{-t\tilde{L}(0)} \left| w \right|\] is true for every \(w \in \mathrm{L}^{2}(\Omega)\). By Theorem 2.21 on page 60 in [3] \[\tilde{a}(0)(|u|, |v|) \;\leq \;\Re \tilde{a}(0)(u, v)\] is implied. Hence we conclude to \[\tilde{a}(\overline{B})(|u|, |v|) \;\leq \;\Re \tilde{a}(B)(u, v)\] which proves the claim by using Theorem 2.21 in [3] once again.

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Proof.

i) We use Theorem 4.3 on page 13 in [4] and conclude that the restriction of \(\mathrm{e}^{-tL(-\beta)}\) to \(C(\overline{\Omega})\) is a strongly-continuous operator semigroup in \(C(\overline{\Omega})\). Furthermore \(-L(-\beta)\) restricted to \(C(\overline{\Omega})\) is its generator.

ii) Using Lemma 2.10 on page 18 in [5] we conclude that \[\left\| \mu ( \mu + L(-\beta) )^{-1} \mathbb{1} - \mathbb{1} \right\|_{\infty} \;\to \;0\] holds for \(\mu \to \infty\) where \(\mathbb{1} \in C(\overline{\Omega})\) is a constant function mapping \(\overline{\Omega}\) to \(1\).

iii) Choose \(\mu_{0} > 0\) sufficiently large such that \[h = \mu_{0} ( \mu_{0} + L(-\beta) )^{-1} \mathbb{1} \;\in \;C(\overline{\Omega}) \cap D \left( L(-\beta) \right)\] satsifies \(\frac{1}{2} \;\leq \;h \;\leq \;2\) everywhere in \(\overline{\Omega}\).

iv) For every \(v \in H^{1}(\Omega)\) we argue that \[\begin{align} \langle L(-\beta)h, v \rangle_{\mathrm{L}^{2}(\Omega)} & = a(-\beta)(h,v) & \\ & = \int_{\Omega} A \nabla h \cdot \overline{\nabla v} \,\mathrm{d}x - \int_{\partial \Omega} \beta h \;\overline{\gamma(v)} \,\mathrm{d}\sigma(x) & \\ & = \int_{\Omega} A \nabla h \cdot \overline{\nabla v} \,\mathrm{d}x + \int_{\partial \Omega} \overline{B} h \;\overline{\gamma(v)} \,\mathrm{d}\sigma(x) & \\ & = a( \overline{B} )(h,v) & \end{align}\] is true. Hence \(h \in D \left( L( \overline{B}) \right)\) with \(L( \overline{B})h = L( -\beta)h\).

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Proof.

i) Define \(w = \left( \mu_{0} + L(\overline{B}) \right)h \in \mathrm{L}^{2}(\Omega)\). Then \(w = \left( \mu_{0} + L(-\beta) \right)h = \mu_{0} \geq 0\) is true almost everywhere in \(\Omega\). Furthermore \[h = \left( \mu_{0} + L(-\beta) \right)^{-1}w = \int_{0}^{\infty} \mathrm{e}^{-\mu_{0}s} \mathrm{e}^{-sL(\overline{B})}w \,\mathrm{d}s\] is implied. Then we use the continuity of \(\mathrm{e}^{-tL(\overline{B})}\) in \(\mathrm{L}^{2}(\Omega)\) as well as its positivity to conclude to \[\begin{align} \mathrm{e}^{-tL(\overline{B})} h & = \mathrm{e}^{-tL(\overline{B})} \int_{0}^{\infty} \mathrm{e}^{-\mu_{0}s} \mathrm{e}^{-sL(\overline{B})}w \,\mathrm{d}s & \\ & = \int_{0}^{\infty} \mathrm{e}^{-\mu_{0}s} \mathrm{e}^{-(s+t)L(\overline{B})}w \,\mathrm{d}s & \\ & = \mathrm{e}^{\mu_{0}t} \int_{t}^{\infty} \mathrm{e}^{-\mu_{0}\tau} \mathrm{e}^{-\tau L(\overline{B})}w \,\mathrm{d}\tau & \\ & \leq \mathrm{e}^{\mu_{0}t} \int_{0}^{\infty} \mathrm{e}^{-\mu_{0}\tau} \mathrm{e}^{-\tau L(\overline{B})}w \,\mathrm{d}\tau & \\ & = \mathrm{e}^{\mu_{0}t} \left( \mu_{0} + L(-\beta) \right)^{-1}w = \mathrm{e}^{\mu_{0}t} h \leq 2 \mathrm{e}^{\mu_{0}t} & \\ \end{align}\] almost everywhere in \(\Omega\) where \[\int_{t}^{\infty} \mathrm{e}^{-\mu_{0}\tau} \mathrm{e}^{-\tau L(\overline{B})}w \,\mathrm{d}\tau \leq \int_{0}^{\infty} \mathrm{e}^{-\mu_{0}\tau} \mathrm{e}^{-\tau L(\overline{B})}w \,\mathrm{d}\tau\] is true since \(\mathrm{e}^{-\tau L(\overline{B})}w \geq 0\) is implied almost everywhere in \(\Omega\) since \(\mathrm{e}^{-tL(\overline{B})}\) are positive operators for any time \(t \geq 0\) and \(w \geq 0\) is a positive function.

ii) Now let \(u \in \mathrm{L}^{\infty}(\Omega)\) be arbitrary but fixed. Then \[\begin{align} \left| \mathrm{e}^{-tL(\overline{B})} u \right| & \leq \mathrm{e}^{-tL(\overline{B})} |u| \leq \| u \|_{\infty} \mathrm{e}^{-tL(\overline{B})} \mathbb{1} = 2 \| u \|_{\infty} \mathrm{e}^{-tL(\overline{B})} \frac{1}{2} & \\ & \leq 2 \| u \|_{\infty} \mathrm{e}^{-tL(\overline{B})} h \leq 4 \mathrm{e}^{\mu_{0}t} \|u\|_{\infty} & \end{align}\] is implied by using the positivty of \(\mathrm{e}^{-tL(\overline{B})}\) once again.

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Proof.

i) Please notice that we have focus on the auxiliary form \(\tilde{a}(B)^{\ast}\) from Section 2 to be able to use \[\Re \tilde{a}(B)^{\ast}(v,v) \geq \frac{\alpha}{2} \| \nabla v \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2}\] as described above. Therefore we include \(\lambda_{0}\) sufficiently large such that the auxiliary form \(\tilde{a}(B)^{\ast}\) defined as \[\tilde{a}(B)^{\ast}(u,v) \;= \;\int_{\Omega} A^{T} \nabla u \cdot \overline{\nabla v} \,\mathrm{d}x + \int_{\partial \Omega} B^{\ast} \gamma(u) \;\overline{\gamma(v)} \,\mathrm{d}\sigma(x) + \lambda_{0} \int_{\Omega} u\overline{v} \,\mathrm{d}x\] for \(u,v \in D \left( \tilde{a}(B)^{\ast} \right) = D \left( \tilde{a}(B) \right) = H^{1} \left( \Omega \right)\) is accretive. We argue that \[\Re \tilde{a}(B)^{\ast}(u,u) = \Re \tilde{a}(B)(u,u)\] is true. Hence we conclude to \[\label{Ellipticity95A} \Re \tilde{a}(B)^{\ast}(u,u) \;= \;\Re \tilde{a}(B)(u,u) \;\geq \;\frac{\alpha}{2} \| \nabla u \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2}\tag{8}\] for any \(u \in D \left( \tilde{a}(B)^{\ast} \right)\) by Remark (9).

ii) Now we can use Nash’s inequality. Therefore let \(\mathrm{e}^{-t\tilde{L}(B)^{\ast}}\) be the associated semigroup operators to the auxiliary form \(\tilde{a}(B)^{\ast}\). Furthermore let \(u \in D ( \tilde{L}(B)^{\ast} )\) and \(t > 0\) be arbitrary but fixed. We define \[v(t) = \mathrm{e}^{-t\lambda_{0}} \mathrm{e}^{-tL(B)^{\ast}} u = \mathrm{e}^{-t\tilde{L}(B)^{\ast}} u \in D ( \tilde{L}(B)^{\ast} ) \subset \mathrm{L}^{2} \left( \Omega \right).\] Furthermore \(v(t)\) and \(u\) are both contained in \(\mathrm{L}^{1} \left( \Omega \right)\) since \(\Omega\) is bounded. Hence \[\left\| v(t) \right\|_{\mathrm{L}^{1} \left( \Omega \right)} \;\leq \;\mathrm{e}^{- t \lambda_{0}} \; \left\| \mathrm{e}^{-tL(B)^{\ast}} \right\|_{\mathrm{L}^{1} \to \mathrm{L}^{1}} \; \left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)} \leq \;4 \mathrm{e}^{t(\mu_{0} - \lambda_{0})} \left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)}\] is implied by Corollary 21. We start the argumentation with \[\begin{align} \frac{\partial}{\partial t} \| v(t) \|^{2}_{\mathrm{L}^{2}\left( \Omega \right)} & = 2 \Re \langle v^{\prime}(t), v(t) \rangle_{\mathrm{L}^{2}\left( \Omega \right)} = - 2 \Re \langle \tilde{L}(B)^{\ast} v(t), v(t) \rangle_{\mathrm{L}^{2}\left( \Omega \right)} & \\ & = -2 \Re \tilde{a}(B)^{\ast}(v(t), v(t)) \leq - \alpha \| \nabla u \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2} & \end{align}\] by (8 ). Now we use Nash’s inequality (9 ) to infer that \[\frac{\partial}{\partial t} \| v(t) \|^{2}_{\mathrm{L}^{2}\left( \Omega \right)} \;\leq \; - \alpha \| \nabla u \|_{\mathrm{L}^{2}(\Omega)^{d}}^{2} \;\leq \;- \frac{\alpha}{C_{d}} \frac{ \| v(t) \|_{\mathrm{L}^{2}(\Omega)}^{2+\frac{4}{d}} }{ \| v(t) \|_{\mathrm{L}^{1}(\Omega)}^{\frac{4}{d}} }\] is true. Therefore we use the chain rule to further argue that \[\begin{align} \frac{\partial}{\partial t} \left( \| v(t) \|^{2}_{\mathrm{L}^{2}\left( \Omega \right)} \right)^{-\frac{2}{d}} \;& \geq \;\left( -\frac{2}{d} \right) \left( \| v(t) \|^{2}_{\mathrm{L}^{2}\left( \Omega \right)} \right)^{-\frac{2}{d}-1} \left( - \frac{\alpha}{C_{d}} \right) \frac{ \| v(t) \|_{\mathrm{L}^{2}(\Omega)}^{2+\frac{4}{d}} }{ \| v(t) \|_{\mathrm{L}^{1}(\Omega)}^{\frac{4}{d}} } \\ \;\\ & = \;\frac{2 \alpha}{d C_{d}} \;\| v(t) \|_{\mathrm{L}^{1}(\Omega)}^{-\frac{4}{d}} & \\ \;\\ & \geq \;\frac{2 \alpha }{d C_{d}} \;4^{-4/d} \exp \left( -\frac{4t}{d}(\mu_{0} - \lambda_{0}) \right) \left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)}^{-\frac{4}{d}} & \end{align}\] is implied. Hence we conclude to \[\begin{align} \| v(t) \|_{\mathrm{L}^{2}\left( \Omega \right)}^{-\frac{4}{d}} & = \;\int_{0}^{t} \frac{\partial}{\partial t} \;\| v(t) \|_{\mathrm{L}^{2}\left( \Omega \right)}^{-\frac{4}{d}} \,\mathrm{d}t + \| u \|_{\mathrm{L}^{2}\left( \Omega \right)}^{-\frac{4}{d}} & \\ \;\\ & \geq \; \frac{2 \alpha }{d C_{d}} \;4^{-4/d} \exp \left( -\frac{4t}{d}(\mu_{0} - \lambda_{0}) \right) t \left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)}^{-\frac{4}{d}} & \end{align}\] and end up with \[\| \mathrm{e}^{-t\tilde{L}(B)^{\ast}} u \|_{\mathrm{L}^{2}\left( \Omega \right) } \;\leq \; 4 \left( \frac{d C_{d}}{2 \alpha} \right)^{d/4} \;t^{-d/4} \;\exp \left( t(\mu_{0} - \lambda_{0}) \right) \; \left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)}\] for any function \(u \in D ( \tilde{L}(B)^{\ast} )\).

iii) Note that \(D ( \tilde{L}(B)^{\ast} )\) is a dense subset in \(\mathrm{L}^{2} \left( \Omega \right)\) due to \(\tilde{L}(B)^{\ast}\) being a generator of a \(C_{0}\)-semigroup in \(\mathrm{L}^{2} \left( \Omega \right)\). Furthermore \(\Omega\) is bounded and therefore \(D ( \tilde{L}(B)^{\ast} )\) is even a dense subset in \(\mathrm{L}^{1} \left( \Omega \right)\) with respect to \(\| \cdot \|_{\mathrm{L}^{1} \left( \Omega \right)}\). Finally we argue that there exists an unique extension of \(\mathrm{e}^{-tL_{B}^{\ast}}\) to a continuous operator from \(\mathrm{L}^{1}\left( \Omega \right)\) to \(\mathrm{L}^{2}\left( \Omega \right)\) satisfying \[\exists C > 0 \;\forall u \in \mathrm{L}^{1} \left( \Omega \right) \;: \; \| \mathrm{e}^{-tL(B)^{\ast}} u \|_{\mathrm{L}^{2}\left( \Omega \right) } \;\leq \; C t^{-d/4} \;\mathrm{e}^{t\mu_{0}} \;\left\| u \right\|_{\mathrm{L}^{1} \left( \Omega \right)}\] where we used \(\mathrm{e}^{-t\tilde{L}(B)^{\ast}} = \mathrm{e}^{-t\lambda_{0}} \mathrm{e}^{-tL(B)^{\ast}}\).

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