[proofplan] The core of the proof is the Laplace-transform representation of the resolvent: for $\operatorname{Re}(\lambda) > \omega_0$, the integral $R(\lambda) := \int_0^\infty e^{-\lambda t} T(t) x \, d\mathcal{L}^1(t)$ converges absolutely and equals $(\lambda I - A)^{-1} x$. This shows the right half-plane $\{\operatorname{Re}(\lambda) > \omega_0\}$ lies in $\rho(A)$, hence $s(A) \le \omega_0$ (part 1). Part 3 (forward direction) follows from the same construction with the resolvent bound $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le M / (\operatorname{Re}(\lambda) - \omega)$. Part 3 (reverse direction) is the [Hille-Yosida Theorem](/theorems/3139) applied to the rescaled semigroup $S(t) := e^{-\omega t} T(t)$. Part 2 uses the spectral mapping theorem for analytic semigroups: $\sigma(T(t)) \setminus \{0\} = e^{t \sigma(A)}$, which translates the spectral radius $r(T(t)) = e^{t\, s(A)}$ into the equality $\omega_0 = s(A)$. We give part 1 in full detail and state parts 2 and 3 concisely. [/proofplan] [step:Fix the standing exponential bound and define the Laplace transform of the orbit] By the boundedness theorem for $C_0$-semigroups, there exist $M \ge 1$ and $\omega_1 \in \mathbb{R}$ with $\|T(t)\|_{\mathcal{L}(X)} \le M e^{\omega_1 t}$ for all $t \ge 0$. By definition of the growth bound, for every $\omega > \omega_0$ there exists $M_\omega \ge 1$ such that $\|T(t)\|_{\mathcal{L}(X)} \le M_\omega e^{\omega t}$ for all $t \ge 0$. Fix $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) > \omega_0$. Choose $\omega \in (\omega_0, \operatorname{Re}(\lambda))$ and let $M_\omega$ be the corresponding constant. For $x \in X$, define \begin{align*} R(\lambda) x &:= \int_0^\infty e^{-\lambda t} T(t) x \, d\mathcal{L}^1(t). \end{align*} The integrand $t \mapsto e^{-\lambda t} T(t) x$ is continuous (strong continuity of the semigroup) with norm bounded by $|e^{-\lambda t}|\, \|T(t)\|_{\mathcal{L}(X)}\, \|x\|_X \le M_\omega\, e^{-(\operatorname{Re}(\lambda) - \omega) t}\, \|x\|_X$, an integrable function on $[0, \infty)$ since $\operatorname{Re}(\lambda) - \omega > 0$. Hence the Bochner integral converges absolutely and \begin{align*} \|R(\lambda) x\|_X &\le \frac{M_\omega}{\operatorname{Re}(\lambda) - \omega}\, \|x\|_X. \end{align*} The map $R(\lambda): X \to X$ is bounded and linear. [/step] [step:Show $R(\lambda) = (\lambda I - A)^{-1}$ for $\operatorname{Re}(\lambda) > \omega_0$] [claim:For every $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) > \omega_0$, the operator $R(\lambda) \in \mathcal{L}(X)$ satisfies $R(\lambda) X \subset D(A)$, $(\lambda I - A) R(\lambda) = I$, and $R(\lambda)(\lambda I - A) = I$ on $D(A)$.] [/claim] [proof] Fix $\lambda$ with $\operatorname{Re}(\lambda) > \omega_0$ and $x \in X$. We compute the right derivative of $T(h) R(\lambda) x$ at $h = 0$. For $h > 0$, using the semigroup law and the change of variable $s = t + h$ in the integral: \begin{align*} T(h) R(\lambda) x &= T(h) \int_0^\infty e^{-\lambda t} T(t) x\, d\mathcal{L}^1(t) = \int_0^\infty e^{-\lambda t} T(t + h) x\, d\mathcal{L}^1(t) \\ &= \int_h^\infty e^{-\lambda(s - h)} T(s) x\, d\mathcal{L}^1(s) = e^{\lambda h}\, \int_h^\infty e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s). \end{align*} Subtracting $R(\lambda) x = \int_0^\infty e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s)$: \begin{align*} T(h) R(\lambda) x - R(\lambda) x &= (e^{\lambda h} - 1) \int_h^\infty e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s) - \int_0^h e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s). \end{align*} Dividing by $h$ and letting $h \to 0^+$: For the first term, $(e^{\lambda h} - 1)/h \to \lambda$ in $\mathbb{C}$ and $\int_h^\infty e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s) \to \int_0^\infty e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s) = R(\lambda) x$ (dominated convergence with majorant $M_\omega e^{-(\operatorname{Re}(\lambda) - \omega) s}\|x\|_X$). So the first term tends to $\lambda\, R(\lambda) x$. For the second term, $\frac{1}{h}\int_0^h e^{-\lambda s} T(s) x\, d\mathcal{L}^1(s) \to e^{0} T(0) x = x$ by the Lebesgue Differentiation Theorem for vector-valued continuous integrands. Combining: \begin{align*} \lim_{h \to 0^+} \frac{T(h) R(\lambda) x - R(\lambda) x}{h} &= \lambda R(\lambda) x - x. \end{align*} By definition of the generator, this means $R(\lambda) x \in D(A)$ and \begin{align*} A R(\lambda) x &= \lambda R(\lambda) x - x, \quad \text{i.e.}\quad (\lambda I - A) R(\lambda) x = x. \end{align*} Hence $(\lambda I - A) R(\lambda) = I$ on $X$. For the reverse identity, let $x \in D(A)$. Since $A$ is closed (part 2 of the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144)) and commutes with bounded Bochner integrals against the orbit (because $A$ commutes with the strongly continuous semigroup on $D(A)$, by part 3 of the same theorem): \begin{align*} A R(\lambda) x &= A \int_0^\infty e^{-\lambda t} T(t) x\, d\mathcal{L}^1(t) = \int_0^\infty e^{-\lambda t} T(t) A x\, d\mathcal{L}^1(t) = R(\lambda) Ax. \end{align*} Therefore $R(\lambda) (\lambda I - A) x = \lambda R(\lambda) x - R(\lambda) Ax = \lambda R(\lambda) x - A R(\lambda) x = x$, where the last equality uses $(\lambda I - A) R(\lambda) x = x$. [/proof] [/step] [step:Conclude $s(A) \le \omega_0$ (part 1)] For every $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) > \omega_0$, the previous step shows $\lambda \in \rho(A)$ with $R(\lambda, A) = R(\lambda)$. Therefore $\sigma(A) \subset \{\lambda \in \mathbb{C} : \operatorname{Re}(\lambda) \le \omega_0\}$, which by definition of the spectral bound gives \begin{align*} s(A) &= \sup\{\operatorname{Re}(\lambda) : \lambda \in \sigma(A)\} \le \omega_0. \end{align*} [/step] [step:Prove the forward direction of part 3 (growth bound implies resolvent bound)] Suppose $\|T(t)\|_{\mathcal{L}(X)} \le M e^{\omega t}$ for all $t \ge 0$. For any $\lambda$ with $\operatorname{Re}(\lambda) > \omega$, the Laplace transform construction in Step 1 (with $\omega$ in place of $\omega_0$ and constant $M$ in place of $M_\omega$) gives $\lambda \in \rho(A)$ with \begin{align*} \|R(\lambda, A)\|_{\mathcal{L}(X)} &\le \frac{M}{\operatorname{Re}(\lambda) - \omega}. \end{align*} This is the uniform resolvent bound on the half-plane $\{\operatorname{Re}(\lambda) > \omega\}$. [/step] [step:State the reverse direction of part 3 via the Hille-Yosida theorem] Conversely, suppose every $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda) > \omega$ lies in $\rho(A)$ and satisfies the iterated bound \begin{align*} \|R(\lambda, A)^n\|_{\mathcal{L}(X)} &\le \frac{M}{(\operatorname{Re}(\lambda) - \omega)^n} \quad \text{for all } n \ge 1. \end{align*} Define $B := A - \omega I$ on $D(B) = D(A)$. Then $\rho(B) \supset \{\mu : \operatorname{Re}(\mu) > 0\}$ with \begin{align*} \|R(\mu, B)^n\|_{\mathcal{L}(X)} &= \|R(\mu + \omega, A)^n\|_{\mathcal{L}(X)} \le \frac{M}{(\operatorname{Re}(\mu))^n}, \quad n \ge 1. \end{align*} The general form of the [Hille-Yosida Theorem](/theorems/3139) states that this resolvent estimate is equivalent to $B$ generating a $C_0$-semigroup $\{S(t)\}_{t \ge 0}$ with $\|S(t)\|_{\mathcal{L}(X)} \le M$ for all $t \ge 0$. The shifted semigroup $T(t) := e^{\omega t} S(t)$ then satisfies $\|T(t)\|_{\mathcal{L}(X)} \le M e^{\omega t}$, and its generator is $B + \omega I = A$. By uniqueness of the semigroup associated to a generator, this $T(t)$ coincides with the original semigroup. (Hence the standing assumption that $A$ already generates the semigroup makes this direction a verification rather than a construction.) This proves part 3. [/step] [step:Prove part 2 via the spectral mapping theorem for analytic semigroups] Suppose $\{T(t)\}_{t \ge 0}$ is an analytic semigroup. Combined with part 1 ($s(A) \le \omega_0$), it suffices to show $\omega_0 \le s(A)$. For analytic semigroups, the spectral mapping theorem for the point and continuous spectrum extending to the whole spectrum gives \begin{align*} \sigma(T(t)) \setminus \{0\} &= e^{t \sigma(A)} \quad \text{for all } t > 0, \end{align*} where $e^{t \sigma(A)} := \{e^{t \lambda} : \lambda \in \sigma(A)\}$. Taking the supremum of moduli on each side, \begin{align*} r(T(t)) &= \sup_{\mu \in \sigma(T(t))} |\mu| = \sup_{\lambda \in \sigma(A)} |e^{t \lambda}| = e^{t\, s(A)}, \end{align*} where $r(T(t))$ denotes the spectral radius of $T(t)$ (using $0 \in \sigma(T(t))$ does not affect the supremum since $|0| = 0 \le e^{t s(A)}$ when $\sigma(A)$ is non-empty, and the empty case gives both sides as $0$). By the Gelfand spectral radius formula, $r(T(t)) = \lim_{n \to \infty} \|T(t)^n\|_{\mathcal{L}(X)}^{1/n} = \lim_{n \to \infty} \|T(nt)\|_{\mathcal{L}(X)}^{1/n}$, and a standard argument for $C_0$-semigroups identifies \begin{align*} \omega_0 &= \lim_{t \to \infty} \frac{\log \|T(t)\|_{\mathcal{L}(X)}}{t} = \frac{1}{t} \log r(T(t)) \quad \text{for any } t > 0. \end{align*} (The first equality is the well-known formula for the growth bound; the second uses Gelfand.) Combining with $r(T(t)) = e^{t\, s(A)}$: \begin{align*} \omega_0 &= \frac{1}{t} \log e^{t\, s(A)} = s(A). \end{align*} With part 1 ($s(A) \le \omega_0$), we obtain $s(A) = \omega_0$, completing part 2 and the proof. [/step]