[proofplan] Set $A = -\Delta_D$, so $A$ is nonnegative and [self-adjoint](/page/Self-Adjoint%20Operators) on $L^2(\Omega)$. The spectral theorem gives a projection-valued spectral measure for $A$ and realizes the semigroup as the spectral functional calculus multiplier $e^{-t\lambda}$. Holomorphicity follows by applying the same functional calculus to $e^{-z\lambda}$ for $\operatorname{Re} z > 0$, while the derivative estimate follows from the scalar bound $\lambda e^{-t\lambda} \leq (et)^{-1}$ on $[0,\infty)$. [/proofplan] [step:Represent the semigroup by the spectral functional calculus] Let $H := L^2(\Omega)$ with respect to $\mathcal{L}^n$, and let $A := -\Delta_D$ with domain $D(A) \subset H$. Let $\mathcal{L}(H)$ denote the [Banach space](/page/Banach%20Space) of bounded linear maps $T: H \to H$, equipped with the operator norm $\|T\|_{\mathcal{L}(H)}$. By hypothesis, $A$ is a nonnegative [self-adjoint](/page/Self-Adjoint%20Operators) operator on $H$. We use the spectral theorem for nonnegative self-adjoint operators and its spectral functional calculus, results not yet represented by wiki theorem entries. Thus there exists a projection-valued measure $E_A$ on the Borel subsets of $[0,\infty)$ such that, for every bounded Borel function $m: [0,\infty) \to \mathbb{C}$, the operator $m(A) \in \mathcal{L}(H)$ is defined by the spectral functional calculus and satisfies \begin{align*} \|m(A)\|_{\mathcal{L}(H)} \leq \sup_{\lambda \in [0,\infty)} |m(\lambda)|. \end{align*} For $t \geq 0$, define the bounded Borel function $m_t: [0,\infty) \to \mathbb{C}$ by $m_t(\lambda) = e^{-t\lambda}$. Since $\Delta_D = -A$ generates the semigroup $(S(t))_{t \geq 0}$, the spectral functional calculus gives \begin{align*} S(t) = e^{-tA} = m_t(A). \end{align*} [/step] [step:Extend the semigroup holomorphically to the right half-plane] Let $\mathbb{C}_+ := \{z \in \mathbb{C} : \operatorname{Re} z > 0\}$. For each $z \in \mathbb{C}_+$, define the bounded Borel function $m_z: [0,\infty) \to \mathbb{C}$ by $m_z(\lambda)=e^{-z\lambda}$. Since $\operatorname{Re} z > 0$ and $\lambda \geq 0$, one has \begin{align*} |m_z(\lambda)| = e^{-(\operatorname{Re} z)\lambda} \leq 1. \end{align*} Therefore $m_z(A) \in \mathcal{L}(H)$ and \begin{align*} \|m_z(A)\|_{\mathcal{L}(H)} \leq 1. \end{align*} Define $S: \mathbb{C}_+ \to \mathcal{L}(H)$ by $S(z)=m_z(A)$. We prove that this map is holomorphic in the operator norm. Fix $z_0 \in \mathbb{C}_+$, and choose $r > 0$ such that $r < \operatorname{Re} z_0$. If $h \in \mathbb{C}$ satisfies $0 < |h| < r/2$, then $z_0 + h \in \mathbb{C}_+$ and the line segment from $z_0$ to $z_0+h$ lies in the half-plane $\{z \in \mathbb{C} : \operatorname{Re} z \geq \operatorname{Re} z_0 - r/2\}$. Define $b_{z_0}: [0,\infty) \to \mathbb{C}$ by $b_{z_0}(\lambda) = \lambda e^{-z_0\lambda}$. Since $\operatorname{Re} z_0 > 0$, the function $b_{z_0}$ is bounded on $[0,\infty)$; indeed its absolute value is bounded by $(e\operatorname{Re} z_0)^{-1}$. Hence $A S(z_0)=b_{z_0}(A)$ is a bounded operator on $H$. Define the difference-quotient multiplier $d_h: [0,\infty) \to \mathbb{C}$ by \begin{align*} d_h(\lambda) = \frac{e^{-(z_0+h)\lambda} - e^{-z_0\lambda}}{h} + \lambda e^{-z_0\lambda}. \end{align*} The spectral multiplier estimate gives \begin{align*} \left\|\frac{S(z_0+h)-S(z_0)}{h} + A S(z_0)\right\|_{\mathcal{L}(H)} \leq \sup_{\lambda \in [0,\infty)} |d_h(\lambda)|. \end{align*} We show that the supremum tends to $0$ as $h \to 0$. For each fixed $\lambda \geq 0$, the one-variable [holomorphic function](/page/Holomorphic%20Function) $\zeta \mapsto e^{-\zeta\lambda}$ gives $d_h(\lambda) \to 0$. To make this convergence uniform in $\lambda$, use Taylor's formula with integral remainder along the segment from $z_0$ to $z_0+h$: \begin{align*} e^{-(z_0+h)\lambda} = e^{-z_0\lambda} - h\lambda e^{-z_0\lambda} + h^2 \lambda^2 \int_0^1 (1-s)e^{-(z_0+sh)\lambda}\,d\mathcal L^1(s). \end{align*} Hence \begin{align*} |d_h(\lambda)| \leq |h|\lambda^2 e^{-(\operatorname{Re} z_0-r/2)\lambda}\int_0^1 (1-s)\,d\mathcal L^1(s). \end{align*} Since $\int_0^1 (1-s)\,d\mathcal L^1(s)=1/2$, and since the function $\lambda \mapsto \lambda^2 e^{-\beta\lambda}$ on $[0,\infty)$ has maximum $4e^{-2}\beta^{-2}$ when $\beta:=\operatorname{Re} z_0-r/2>0$, we obtain \begin{align*} \sup_{\lambda \in [0,\infty)} |d_h(\lambda)| \leq 2e^{-2}\beta^{-2}|h|. \end{align*} Therefore \begin{align*} \lim_{h\to 0}\left\|\frac{S(z_0+h)-S(z_0)}{h} + A S(z_0)\right\|_{\mathcal{L}(H)} = 0. \end{align*} Thus $S: \mathbb{C}_+ \to \mathcal{L}(H)$ is complex differentiable in operator norm at the arbitrary point $z_0$, with derivative $S'(z_0)=-A S(z_0)$. Therefore $S$ is a bounded holomorphic extension of the heat semigroup to $\mathbb{C}_+$. [guided] The goal is to extend the real-time formula $S(t) = e^{-tA}$ from $t > 0$ to complex time $z$ with positive real part. The spectral calculus reduces this operator-norm question to scalar estimates for the functions $\lambda \mapsto e^{-z\lambda}$ on the spectrum of $A$, which is contained in $[0,\infty)$ because $A$ is nonnegative. Let \begin{align*} \mathbb{C}_+ := \{z \in \mathbb{C} : \operatorname{Re} z > 0\}. \end{align*} For each $z \in \mathbb{C}_+$, define $m_z: [0,\infty) \to \mathbb{C}$ by $m_z(\lambda)=e^{-z\lambda}$. This is the correct scalar multiplier because $S(t)$ is $e^{-tA}$ for real $t > 0$. The condition $\operatorname{Re} z > 0$ is exactly what keeps this multiplier bounded on the nonnegative spectrum: \begin{align*} |m_z(\lambda)| = |e^{-z\lambda}| = e^{-(\operatorname{Re} z)\lambda} \leq 1 \end{align*} for every $\lambda \geq 0$. The spectral calculus therefore defines a bounded operator $m_z(A) \in \mathcal{L}(H)$, and its norm satisfies \begin{align*} \|m_z(A)\|_{\mathcal{L}(H)} \leq \sup_{\lambda \in [0,\infty)} |m_z(\lambda)| \leq 1. \end{align*} We set $S: \mathbb{C}_+ \to \mathcal{L}(H)$ by $S(z)=m_z(A)$. It remains to prove holomorphicity in the operator norm, not merely in scalar matrix coefficients. Fix $z_0 \in \mathbb{C}_+$, and choose $r > 0$ with $r < \operatorname{Re} z_0$. If $h \in \mathbb{C}$ and $0 < |h| < r/2$, then $z_0+h \in \mathbb{C}_+$, and every point $z_0 + sh$ with $s \in [0,1]$ satisfies \begin{align*} \operatorname{Re}(z_0+sh) \geq \operatorname{Re} z_0-r/2. \end{align*} Define $b_{z_0}: [0,\infty) \to \mathbb{C}$ by $b_{z_0}(\lambda)=\lambda e^{-z_0\lambda}$. Since $\operatorname{Re} z_0 > 0$, the scalar function $b_{z_0}$ is bounded on $[0,\infty)$, with \begin{align*} |b_{z_0}(\lambda)| \leq \lambda e^{-(\operatorname{Re} z_0)\lambda} \leq (e\operatorname{Re} z_0)^{-1}. \end{align*} Thus the spectral calculus defines $A S(z_0)=b_{z_0}(A)$ as a bounded operator on $H$. To compute the derivative, define the difference-quotient error multiplier $d_h: [0,\infty) \to \mathbb{C}$ by \begin{align*} d_h(\lambda)=\frac{e^{-(z_0+h)\lambda}-e^{-z_0\lambda}}{h}+\lambda e^{-z_0\lambda}. \end{align*} The spectral multiplier estimate gives \begin{align*} \left\|\frac{S(z_0+h)-S(z_0)}{h}+A S(z_0)\right\|_{\mathcal{L}(H)} \leq \sup_{\lambda \in [0,\infty)} |d_h(\lambda)|. \end{align*} Therefore the operator-norm derivative will follow once this scalar supremum tends to $0$. We use Taylor's formula with integral remainder for the one-variable holomorphic function $\zeta \mapsto e^{-\zeta\lambda}$ along the segment $\zeta = z_0+sh$, where $s \in [0,1]$. For each $\lambda \geq 0$, \begin{align*} e^{-(z_0+h)\lambda} = e^{-z_0\lambda}-h\lambda e^{-z_0\lambda}+h^2\lambda^2\int_0^1 (1-s)e^{-(z_0+sh)\lambda}\,d\mathcal{L}^1(s). \end{align*} Substituting this identity into the definition of $d_h$ gives \begin{align*} |d_h(\lambda)| \leq |h|\lambda^2 e^{-(\operatorname{Re} z_0-r/2)\lambda}\int_0^1 (1-s)\,d\mathcal{L}^1(s). \end{align*} The one-dimensional integral equals $1/2$. If \begin{align*} \beta := \operatorname{Re} z_0-r/2, \end{align*} then $\beta > 0$, and the function $\lambda \mapsto \lambda^2 e^{-\beta\lambda}$ on $[0,\infty)$ has maximum $4e^{-2}\beta^{-2}$. Consequently, \begin{align*} \sup_{\lambda \in [0,\infty)} |d_h(\lambda)| \leq 2e^{-2}\beta^{-2}|h|. \end{align*} The right-hand side tends to $0$ as $h \to 0$, so \begin{align*} \lim_{h\to 0}\left\|\frac{S(z_0+h)-S(z_0)}{h}+A S(z_0)\right\|_{\mathcal{L}(H)} = 0. \end{align*} Thus $S$ is complex differentiable in operator norm at $z_0$, with derivative $S'(z_0)=-A S(z_0)$. Since $z_0$ was arbitrary, $S: \mathbb{C}_+ \to \mathcal{L}(H)$ is holomorphic and extends the real heat semigroup. [/guided] [/step] [step:Estimate the generator applied to the semigroup by a scalar spectral bound] Fix $t > 0$. Define the scalar function $q_t: [0,\infty) \to \mathbb{C}$ by $q_t(\lambda)=\lambda e^{-t\lambda}$. The function $q_t$ is bounded. Indeed, for $\lambda > 0$ the derivative of $\lambda \mapsto \lambda e^{-t\lambda}$ is $(1-t\lambda)e^{-t\lambda}$, so its maximum on $[0,\infty)$ occurs at $\lambda = t^{-1}$ and equals $(et)^{-1}$. Hence \begin{align*} \sup_{\lambda \in [0,\infty)} \lambda e^{-t\lambda} = \frac{1}{e t}. \end{align*} The spectral theorem also identifies domains of functions of $A$: a vector $w \in H$ belongs to $D(A)$ exactly when $\lambda \mapsto \lambda$ is square-integrable with respect to the spectral measure associated with $w$. Since $q_t(\lambda)=\lambda e^{-t\lambda}$ is bounded on $[0,\infty)$, the functional calculus implies $e^{-tA}H \subset D(A)$ and $A e^{-tA}=q_t(A)$ as a bounded operator on $H$. Therefore \begin{align*} \|A S(t)\|_{\mathcal{L}(H)} = \|A e^{-tA}\|_{\mathcal{L}(H)} \leq \frac{1}{e t}. \end{align*} Consequently, for every $u_0 \in H$, \begin{align*} \|A S(t)u_0\|_{L^2(\Omega)} \leq \frac{1}{e t}\|u_0\|_{L^2(\Omega)}. \end{align*} Since $A = -\Delta_D$, this also gives $S(t)u_0 \in D(A) = D(\Delta_D)$ and \begin{align*} \|\Delta_D S(t)u_0\|_{L^2(\Omega)} = \|A S(t)u_0\|_{L^2(\Omega)} \leq \frac{1}{e t}\|u_0\|_{L^2(\Omega)}. \end{align*} This proves the claimed estimate with $C = e^{-1}$, and hence with any $C \geq e^{-1}$. [/step]