[proofplan] We prove both implications. If $A$ generates a contraction semigroup, the Laplace resolvent formula gives the resolvent on every positive real $\lambda$ and the norm bound $\|\lambda(\lambda I-A)^{-1}\|_{\mathcal{L}(X)}\leq 1$, which immediately implies dissipativity and the range condition. Conversely, dissipativity gives injectivity and the resolvent estimate wherever $\lambda I-A$ is onto. The range condition at one positive parameter is then propagated to every $\lambda>0$ by a Neumann-series perturbation argument and closedness of $A$. With the full positive resolvent set and the contraction resolvent bounds in hand, the contraction form of the Hille-Yosida Generation Theorem yields the desired strongly continuous contraction semigroup. [/proofplan]

[step:Derive the resolvent bound from a contraction semigroup]Assume that $A$ generates a strongly continuous semigroup $T:[0,\infty)\to\mathcal{L}(X)$ with $\|T(t)\|_{\mathcal{L}(X)}\leq 1$ for every $t\geq 0$. For each $\lambda>0$, the Laplace resolvent formula for generators of strongly continuous semigroups gives \begin{align*} R_\lambda:X\to D(A)\subset X,\qquad R_\lambda g=\int_0^\infty e^{-\lambda t}T(t)g\,d\mathcal{L}^1(t), \end{align*} where the integral is the Bochner integral in $X$, and \begin{align*} R_\lambda=(\lambda I-A)^{-1}. \end{align*} The contraction bound on $T(t)$ gives, for every $g\in X$, \begin{align*} \|R_\lambda g\|_X\leq \int_0^\infty e^{-\lambda t}\|T(t)g\|_X\,d\mathcal{L}^1(t)\leq \|g\|_X\int_0^\infty e^{-\lambda t}\,d\mathcal{L}^1(t)=\frac{1}{\lambda}\|g\|_X. \end{align*} Thus $R_\lambda\in\mathcal{L}(X)$ and \begin{align*} \|\lambda R_\lambda\|_{\mathcal{L}(X)}\leq 1. \end{align*} Since $R_\lambda=(\lambda I-A)^{-1}$ is defined on all of $X$, we have $\operatorname{Range}(\lambda I-A)=X$ for every $\lambda>0$.[/step]

[guided]Assume first that $A$ is already known to be the generator of a strongly continuous contraction semigroup $T:[0,\infty)\to\mathcal{L}(X)$. The central tool in this direction is the Laplace resolvent formula for semigroup generators. It says that for each $\lambda>0$, the operator \begin{align*} R_\lambda:X\to D(A)\subset X,\qquad R_\lambda g=\int_0^\infty e^{-\lambda t}T(t)g\,d\mathcal{L}^1(t) \end{align*} is well-defined as a Bochner integral and satisfies \begin{align*} R_\lambda=(\lambda I-A)^{-1}. \end{align*} The hypotheses needed for this formula are exactly the strong continuity of $T$ and the exponential boundedness of the semigroup. Here exponential boundedness holds with growth bound $0$ because $\|T(t)\|_{\mathcal{L}(X)}\leq 1$ for all $t\geq 0$. Now fix $g\in X$. We estimate the Bochner integral using the triangle inequality for the Bochner norm and the contraction property: \begin{align*} \|R_\lambda g\|_X\leq \int_0^\infty e^{-\lambda t}\|T(t)g\|_X\,d\mathcal{L}^1(t)\leq \int_0^\infty e^{-\lambda t}\|g\|_X\,d\mathcal{L}^1(t). \end{align*} The remaining scalar integral is \begin{align*} \int_0^\infty e^{-\lambda t}\,d\mathcal{L}^1(t)=\frac{1}{\lambda}. \end{align*} Therefore \begin{align*} \|R_\lambda g\|_X\leq \frac{1}{\lambda}\|g\|_X. \end{align*} Taking the supremum over all $g\neq 0$ gives \begin{align*} \|\lambda R_\lambda\|_{\mathcal{L}(X)}\leq 1. \end{align*} Since $R_\lambda$ is the inverse of $\lambda I-A$ on all of $X$, the equation $(\lambda I-A)u=g$ has a solution $u=R_\lambda g\in D(A)$ for every $g\in X$. Hence $\operatorname{Range}(\lambda I-A)=X$ for every $\lambda>0$.[/guided]

[step:Convert the resolvent bound into dissipativity] Fix $u\in D(A)$ and $\lambda>0$. Define $g\in X$ by \begin{align*} g=(\lambda I-A)u. \end{align*} Since $R_\lambda=(\lambda I-A)^{-1}$, we have $u=R_\lambda g$. The estimate from the previous step gives \begin{align*} \|u\|_X=\|R_\lambda g\|_X\leq \frac{1}{\lambda}\|g\|_X=\frac{1}{\lambda}\|(\lambda I-A)u\|_X. \end{align*} Multiplying by $\lambda$ yields \begin{align*} \|(\lambda I-A)u\|_X\geq \lambda\|u\|_X. \end{align*} Thus $A$ is dissipative. The range condition at some $\lambda_0>0$ follows because the previous step gives $\operatorname{Range}(\lambda I-A)=X$ for every $\lambda>0$. [/step]

[step:Use dissipativity to obtain uniqueness and resolvent estimates wherever the range is all of $X$] Assume conversely that $A$ is dissipative and that $\operatorname{Range}(\lambda_0 I-A)=X$ for some $\lambda_0>0$. For every $\lambda>0$ and every $u\in D(A)$, dissipativity gives \begin{align*} \|(\lambda I-A)u\|_X\geq \lambda\|u\|_X. \end{align*} If $(\lambda I-A)u=0$, this estimate gives $\lambda\|u\|_X\leq 0$, hence $u=0$. Therefore $\lambda I-A$ is injective for every $\lambda>0$. Whenever $\operatorname{Range}(\lambda I-A)=X$, define \begin{align*} R(\lambda,A):X\to D(A)\subset X,\qquad R(\lambda,A)g=(\lambda I-A)^{-1}g. \end{align*} For $g\in X$ and $u=R(\lambda,A)g$, the dissipativity estimate gives \begin{align*} \lambda\|R(\lambda,A)g\|_X=\lambda\|u\|_X\leq \|(\lambda I-A)u\|_X=\|g\|_X. \end{align*} Hence \begin{align*} \|R(\lambda,A)\|_{\mathcal{L}(X)}\leq \frac{1}{\lambda}. \end{align*} [/step]

[step:Prove that the operator is closed] We first show that $A$ is closed. Since $\operatorname{Range}(\lambda_0 I-A)=X$ and $\lambda_0 I-A$ is injective by dissipativity, the inverse \begin{align*} R_0:X\to D(A)\subset X,\qquad R_0g=(\lambda_0 I-A)^{-1}g \end{align*} is well-defined and satisfies \begin{align*} \|R_0\|_{\mathcal{L}(X)}\leq \frac{1}{\lambda_0}. \end{align*} Let $(u_n)_{n=1}^{\infty}$ be a sequence in $D(A)$ such that $u_n\to u$ in $X$ and $Au_n\to v$ in $X$. Define $g\in X$ by \begin{align*} g=\lambda_0 u-v. \end{align*} Then \begin{align*} (\lambda_0 I-A)u_n\to g \end{align*} in $X$. Since $u_n=R_0((\lambda_0 I-A)u_n)$ and $R_0$ is bounded, we obtain \begin{align*} u_n\to R_0g \end{align*} in $X$. Limits in the [Banach space](/page/Banach%20Space) $X$ are unique, so $u=R_0g$. Thus $u\in D(A)$ and \begin{align*} (\lambda_0 I-A)u=g=\lambda_0 u-v. \end{align*} Therefore $Au=v$, proving that $A$ is closed. [/step]

[step:Propagate the range condition to all positive parameters]Let \begin{align*} S=\{\lambda>0:\operatorname{Range}(\lambda I-A)=X\}. \end{align*} The hypothesis gives $\lambda_0\in S$, so $S$ is nonempty. First we show that $S$ is open in $(0,\infty)$. Fix $\mu\in S$ and define \begin{align*} R_\mu:X\to D(A)\subset X,\qquad R_\mu g=(\mu I-A)^{-1}g. \end{align*} For $u\in D(A)$, \begin{align*} (\lambda I-A)u=(I+(\lambda-\mu)R_\mu)(\mu I-A)u. \end{align*} If $|\lambda-\mu|<\mu$, then \begin{align*} \|(\lambda-\mu)R_\mu\|_{\mathcal{L}(X)}\leq \frac{|\lambda-\mu|}{\mu}<1. \end{align*} The Neumann series theorem gives that $I+(\lambda-\mu)R_\mu$ is invertible on $X$. Hence $\lambda I-A$ maps $D(A)$ onto $X$, so $\lambda\in S$. Next we show that $S$ is closed in $(0,\infty)$. Let $(\lambda_n)_{n=1}^{\infty}$ be a sequence in $S$ with $\lambda_n\to\lambda>0$. Fix $g\in X$ and define \begin{align*} u_n=R(\lambda_n,A)g. \end{align*} For $m,n\in\mathbb{N}$, subtracting the equations $(\lambda_n I-A)u_n=g$ and $(\lambda_m I-A)u_m=g$ gives \begin{align*} (\lambda_n I-A)(u_n-u_m)=(\lambda_m-\lambda_n)u_m. \end{align*} Using the resolvent estimate at $\lambda_n$ and then at $\lambda_m$, \begin{align*} \|u_n-u_m\|_X\leq \frac{|\lambda_m-\lambda_n|}{\lambda_n}\|u_m\|_X\leq \frac{|\lambda_m-\lambda_n|}{\lambda_n\lambda_m}\|g\|_X. \end{align*} Since $\lambda_n\to\lambda>0$, the sequence $(u_n)_{n=1}^{\infty}$ is Cauchy in $X$. Let $u\in X$ be its limit. From \begin{align*} Au_n=\lambda_n u_n-g \end{align*} we get $Au_n\to \lambda u-g$ in $X$. Since $A$ is closed, $u\in D(A)$ and $Au=\lambda u-g$. Thus \begin{align*} (\lambda I-A)u=g. \end{align*} As $g\in X$ was arbitrary, $\operatorname{Range}(\lambda I-A)=X$, so $\lambda\in S$. The set $(0,\infty)$ is connected, and $S\subset(0,\infty)$ is nonempty, open, and closed. Hence \begin{align*} S=(0,\infty). \end{align*}[/step]

[guided]The goal is to upgrade one solvable equation, \begin{align*} (\lambda_0 I-A)u=g, \end{align*} to solvability of \begin{align*} (\lambda I-A)u=g \end{align*} for every positive $\lambda$. We encode the set of good parameters by \begin{align*} S=\{\lambda>0:\operatorname{Range}(\lambda I-A)=X\}. \end{align*} The assumption says $\lambda_0\in S$, so this set is nonempty. We first prove openness. Fix $\mu\in S$. Since $\mu I-A$ is onto and injective, its inverse is the bounded operator \begin{align*} R_\mu:X\to D(A)\subset X,\qquad R_\mu g=(\mu I-A)^{-1}g. \end{align*} Dissipativity gives the bound \begin{align*} \|R_\mu\|_{\mathcal{L}(X)}\leq \frac{1}{\mu}. \end{align*} For any $u\in D(A)$, we factor $\lambda I-A$ through $\mu I-A$: \begin{align*} (\lambda I-A)u=(\mu I-A)u+(\lambda-\mu)u. \end{align*} Because $u=R_\mu(\mu I-A)u$, this becomes \begin{align*} (\lambda I-A)u=(I+(\lambda-\mu)R_\mu)(\mu I-A)u. \end{align*} Thus solving $(\lambda I-A)u=g$ is equivalent to first solving \begin{align*} (I+(\lambda-\mu)R_\mu)h=g \end{align*} in $X$ and then solving $(\mu I-A)u=h$. The second equation is solvable because $\mu\in S$. The first equation is solvable whenever the bounded perturbation $(\lambda-\mu)R_\mu$ has operator norm less than $1$. Indeed, \begin{align*} \|(\lambda-\mu)R_\mu\|_{\mathcal{L}(X)}\leq \frac{|\lambda-\mu|}{\mu}. \end{align*} Therefore, if $|\lambda-\mu|<\mu$, the Neumann series theorem applies to $I+(\lambda-\mu)R_\mu$, and this operator is invertible on $X$. Hence every $\lambda$ sufficiently close to $\mu$ belongs to $S$. This proves that $S$ is open in $(0,\infty)$. We next prove closedness. Let $(\lambda_n)_{n=1}^{\infty}$ be a sequence in $S$ converging to some $\lambda>0$. Fix $g\in X$ and define \begin{align*} u_n=R(\lambda_n,A)g. \end{align*} We must show that the equation $(\lambda I-A)u=g$ has a solution. The natural candidate is the limit of the $u_n$, so we prove that $(u_n)$ is Cauchy. Since \begin{align*} (\lambda_n I-A)u_n=g \end{align*} and \begin{align*} (\lambda_m I-A)u_m=g, \end{align*} subtracting the two identities and rewriting relative to $\lambda_n I-A$ gives \begin{align*} (\lambda_n I-A)(u_n-u_m)=(\lambda_m-\lambda_n)u_m. \end{align*} Because $\lambda_n\in S$, the inverse $R(\lambda_n,A)$ exists and has norm at most $1/\lambda_n$. Applying it gives \begin{align*} \|u_n-u_m\|_X\leq \frac{|\lambda_m-\lambda_n|}{\lambda_n}\|u_m\|_X. \end{align*} Using the resolvent estimate again, now for $u_m=R(\lambda_m,A)g$, gives \begin{align*} \|u_m\|_X\leq \frac{1}{\lambda_m}\|g\|_X. \end{align*} Thus \begin{align*} \|u_n-u_m\|_X\leq \frac{|\lambda_m-\lambda_n|}{\lambda_n\lambda_m}\|g\|_X. \end{align*} Since $\lambda_n\to\lambda>0$, the denominators stay bounded away from $0$, while $|\lambda_m-\lambda_n|\to 0$ as $m,n\to\infty$. Hence $(u_n)$ is Cauchy and converges to some $u\in X$. Now we use closedness of $A$, which was proved in the previous step. From the equation defining $u_n$, \begin{align*} Au_n=\lambda_n u_n-g. \end{align*} Since $\lambda_n\to\lambda$ and $u_n\to u$, the right-hand side converges to $\lambda u-g$. Therefore $u_n\to u$ and $Au_n\to \lambda u-g$. Closedness of $A$ implies $u\in D(A)$ and \begin{align*} Au=\lambda u-g. \end{align*} Equivalently, \begin{align*} (\lambda I-A)u=g. \end{align*} Because $g\in X$ was arbitrary, $\lambda I-A$ is onto, so $\lambda\in S$. Thus $S$ is closed in $(0,\infty)$. We have shown that $S$ is nonempty, open, and closed in the connected interval $(0,\infty)$. Therefore \begin{align*} S=(0,\infty). \end{align*} This is the step where the single range hypothesis at $\lambda_0$ becomes the full positive resolvent condition needed for the generation theorem.[/guided]

[step:Apply the contraction Hille-Yosida theorem] From the previous step, for every $\lambda>0$ the operator $\lambda I-A$ is bijective from $D(A)$ onto $X$. Define \begin{align*} R(\lambda,A):X\to D(A)\subset X,\qquad R(\lambda,A)g=(\lambda I-A)^{-1}g. \end{align*} The dissipativity estimate gives \begin{align*} \|\lambda R(\lambda,A)\|_{\mathcal{L}(X)}\leq 1 \end{align*} for every $\lambda>0$. We have also proved that $A$ is closed, and density of $D(A)$ is part of the theorem hypothesis. Thus $A$ satisfies the contraction case of the Hille-Yosida Generation Theorem (citing a result not yet in the wiki: Hille-Yosida Generation Theorem): a densely defined closed operator $A$ on a Banach space generates a strongly continuous contraction semigroup if and only if $(0,\infty)$ is contained in the resolvent set of $A$ and \begin{align*} \|\lambda(\lambda I-A)^{-1}\|_{\mathcal{L}(X)}\leq 1 \end{align*} for every $\lambda>0$. Hence $A$ generates a strongly continuous semigroup \begin{align*} T:[0,\infty)\to\mathcal{L}(X) \end{align*} with \begin{align*} \|T(t)\|_{\mathcal{L}(X)}\leq 1 \end{align*} for every $t\geq 0$. This is precisely the desired contraction semigroup generation conclusion. [/step]