[proofplan] We prove both directions by reducing to the [Hille-Yosida Theorem](/theorems/3139): a densely defined closed operator generates a contraction $C_0$-semigroup if and only if $(0, \infty) \subset \rho(A)$ with $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$. Here $\rho(A) := \{\lambda \in \mathbb{C} : \lambda I - A : D(A) \to X \text{ is bijective with bounded inverse}\}$ is the **resolvent set**, $R(\lambda, A) := (\lambda I - A)^{-1} \in \mathcal{L}(X)$ is the **resolvent operator** for $\lambda \in \rho(A)$, and $\mathcal{L}(X)$ denotes the Banach algebra of bounded linear operators $X \to X$ with the operator norm (notation as in the [Hille-Yosida Theorem](/theorems/3139) and [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144)). The forward direction follows from the elementary Hilbert-space identity $\operatorname{Re}(AT(t)x, T(t)x)_H = \frac{1}{2}\frac{d}{dt}\|T(t)x\|_X^2$ combined with the fact that contractivity of $T(t)$ forces $\|T(t)x\|_X^2$ to be non-increasing. The reverse direction is the substantive content: dissipativity gives the resolvent estimate $\|(\lambda I - A)x\|_X \ge \lambda \|x\|_X$ via a single Cauchy-Schwarz computation, and surjectivity at one $\lambda_0$ propagates to all $\lambda > 0$ by a connectedness argument on $\rho(A)$. Together, these give the Hille-Yosida hypotheses, after first showing $A$ is closed. [/proofplan]

[step:Derive the basic dissipative inequality $\|(\lambda I - A)x\|_X \ge \lambda \|x\|_X$] Suppose $A$ is dissipative. Fix $\lambda > 0$ and $x \in D(A)$. Then \begin{align*} \|(\lambda I - A) x\|_X^2 &= ((\lambda I - A)x, (\lambda I - A)x)_H \\ &= \lambda^2 \|x\|_X^2 - 2\lambda \operatorname{Re}(Ax, x)_H + \|Ax\|_X^2 \\ &\ge \lambda^2 \|x\|_X^2, \end{align*} where the inequality uses $-2\lambda \operatorname{Re}(Ax, x)_H \ge 0$ (dissipativity, $\lambda > 0$) and $\|Ax\|_X^2 \ge 0$. Taking square roots: \begin{align*} \|(\lambda I - A) x\|_X \ge \lambda \|x\|_X \quad \text{for all } x \in D(A), \ \lambda > 0. \end{align*} This has two immediate consequences: - **Injectivity:** $\lambda I - A$ is injective on $D(A)$ for every $\lambda > 0$ (the inequality forces $(\lambda I - A) x = 0 \Rightarrow x = 0$). - **Bounded inverse on the range:** if $y \in \operatorname{Range}(\lambda I - A)$, write $y = (\lambda I - A) x$ for the unique $x \in D(A)$, and then $\|x\|_X \le \frac{1}{\lambda}\|y\|_X$. So $(\lambda I - A)^{-1}: \operatorname{Range}(\lambda I - A) \to D(A)$ is bounded with norm at most $1/\lambda$. [/step]

[step:Show $A$ is closed using the surjectivity hypothesis at $\lambda_0$] Let $\lambda_0 > 0$ be the value provided by hypothesis (2): $\lambda_0 I - A: D(A) \to X$ is surjective. Combined with injectivity (Step 1), $\lambda_0 I - A$ is a bijection from $D(A)$ onto $X$. The inverse $(\lambda_0 I - A)^{-1}: X \to D(A) \subseteq X$ is bounded by $1/\lambda_0$ (Step 1), hence is a closed operator on $X$ (every bounded everywhere-defined operator on a Banach space is closed). The graph of $(\lambda_0 I - A)^{-1}$ is $\{(y, x) : x \in D(A), \ y = (\lambda_0 I - A) x\}$. The graph of $A$ is related: $(x, Ax)$ corresponds to $((\lambda_0 I - A)x, x)$ swapped and shifted. Concretely, the map $(x, Ax) \mapsto ((\lambda_0 I - A)x, x) = (\lambda_0 x - Ax, x)$ is a continuous bijection between the graph of $A$ and the graph of $(\lambda_0 I - A)^{-1}$ (its inverse is also continuous). Since the graph of $(\lambda_0 I - A)^{-1}$ is closed in $X \times X$ and the bijection is a homeomorphism, the graph of $A$ is closed in $X \times X$. Hence $A$ is a closed operator. [/step]

[step:Show $\lambda_0 \in \rho(A)$ and propagate to all $\lambda > 0$ by connectedness]From Steps 1-2: $\lambda_0 I - A$ is a bijection from $D(A)$ onto $X$ with bounded inverse $(\lambda_0 I - A)^{-1} \in \mathcal{L}(X)$ of norm at most $1/\lambda_0$. By definition, this means $\lambda_0 \in \rho(A)$ and $\|R(\lambda_0, A)\|_{\mathcal{L}(X)} \le 1/\lambda_0$. [claim:If $\lambda_1 \in \rho(A) \cap (0, \infty)$, then a neighbourhood $(\lambda_1 - \varepsilon, \lambda_1 + \varepsilon) \cap (0, \infty)$ of $\lambda_1$ is contained in $\rho(A)$, and $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$ on this neighbourhood] [/claim] [proof] For $\lambda > 0$ near $\lambda_1$, write \begin{align*} \lambda I - A = (\lambda_1 I - A) + (\lambda - \lambda_1) I = (\lambda_1 I - A)\bigl(I + (\lambda - \lambda_1) R(\lambda_1, A)\bigr). \end{align*} By a standard Neumann-series argument, whenever $\|(\lambda - \lambda_1) R(\lambda_1, A)\|_{\mathcal{L}(X)} < 1$ — i.e., $|\lambda - \lambda_1| < 1/\|R(\lambda_1, A)\|_{\mathcal{L}(X)}$ — the operator $I + (\lambda - \lambda_1) R(\lambda_1, A) \in \mathcal{L}(X)$ is invertible with inverse given by the convergent geometric series $\sum_{k=0}^{\infty} (-(\lambda - \lambda_1) R(\lambda_1, A))^k \in \mathcal{L}(X)$. Hence $\lambda I - A$ is invertible (as a composition of an invertible bounded operator with the bijection $\lambda_1 I - A : D(A) \to X$), so $\lambda \in \rho(A)$ for $\lambda$ in this neighbourhood. For the bound: we proved $\|(\lambda I - A) x\|_X \ge \lambda \|x\|_X$ for all $\lambda > 0$ and $x \in D(A)$ in Step 1 using *only* dissipativity. So whenever $\lambda \in \rho(A) \cap (0, \infty)$, taking $x = R(\lambda, A) y$ gives $\|y\|_X \ge \lambda \|R(\lambda, A) y\|_X$, i.e., $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$. [/proof] The set $S := \rho(A) \cap (0, \infty)$ is non-empty (contains $\lambda_0$) and open in $(0, \infty)$ by the Claim. We show it is also closed in $(0, \infty)$. Suppose $\lambda_n \in S$, $\lambda_n \to \lambda \in (0, \infty)$. By the Claim, $\|R(\lambda_n, A)\|_{\mathcal{L}(X)} \le 1/\lambda_n$, so the resolvents are uniformly bounded for $n$ large. For $y \in X$, set $x_n := R(\lambda_n, A) y \in D(A)$. By the resolvent equation \begin{align*} R(\lambda_n, A) - R(\lambda_m, A) = (\lambda_m - \lambda_n) R(\lambda_n, A) R(\lambda_m, A), \end{align*} applied to $y$, \begin{align*} \|x_n - x_m\|_X = \|(R(\lambda_n, A) - R(\lambda_m, A))y\|_X \le |\lambda_m - \lambda_n|\cdot \|R(\lambda_n, A)\|_{\mathcal{L}(X)} \|R(\lambda_m, A)\|_{\mathcal{L}(X)} \|y\|_X. \end{align*} Since $\|R(\lambda_n, A)\|_{\mathcal{L}(X)} \le 1/\lambda_n$ is uniformly bounded and $|\lambda_m - \lambda_n| \to 0$, the sequence $(x_n)$ is Cauchy in $X$. Let $x = \lim x_n$. From $(\lambda_n I - A) x_n = y$ we get $A x_n = \lambda_n x_n - y \to \lambda x - y$ in $X$. Since $A$ is closed (Step 2) and $x_n \to x$, $A x_n \to \lambda x - y$, we conclude $x \in D(A)$ and $A x = \lambda x - y$, i.e., $(\lambda I - A) x = y$. Combined with the injectivity (Step 1), $\lambda I - A: D(A) \to X$ is a bijection, hence $\lambda \in \rho(A)$. So $S$ is non-empty, open, and closed in the connected set $(0, \infty)$, hence $S = (0, \infty)$. Thus every $\lambda > 0$ is in $\rho(A)$ with $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$.[/step]

[guided]The strategy is to show that the set $S := \rho(A) \cap (0, \infty)$ of positive real $\lambda$ at which the resolvent exists is **non-empty, open, and closed** in $(0, \infty)$ — hence equals all of $(0, \infty)$ by connectedness. Each property comes from a different ingredient. Let us assemble them. **Non-emptiness.** From Steps 1-2, $\lambda_0 I - A : D(A) \to X$ is a bijection with bounded inverse, the bound $\|(\lambda_0 I - A)^{-1}\|_{\mathcal{L}(X)} \le 1/\lambda_0$ coming from the dissipative inequality of Step 1. By the definition of the resolvent set, this means $\lambda_0 \in \rho(A)$ and $\|R(\lambda_0, A)\|_{\mathcal{L}(X)} \le 1/\lambda_0$. So $S \ni \lambda_0$. **Openness.** We claim that if $\lambda_1 \in S$, a whole neighbourhood of $\lambda_1$ in $(0, \infty)$ lies in $S$, and the resolvent bound $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$ holds throughout. Why does this follow? Algebraically factor \begin{align*} \lambda I - A = (\lambda_1 I - A) + (\lambda - \lambda_1) I = (\lambda_1 I - A)\bigl(I + (\lambda - \lambda_1) R(\lambda_1, A)\bigr). \end{align*} The first factor $\lambda_1 I - A : D(A) \to X$ is a bijection by hypothesis. The second factor is the bounded operator $I + (\lambda - \lambda_1) R(\lambda_1, A) \in \mathcal{L}(X)$. When is it invertible? By a standard Neumann-series argument: whenever the perturbation has operator norm $\|(\lambda - \lambda_1) R(\lambda_1, A)\|_{\mathcal{L}(X)} < 1$, the geometric series $\sum_{k=0}^\infty (-(\lambda - \lambda_1) R(\lambda_1, A))^k$ converges absolutely in $\mathcal{L}(X)$ (since $\mathcal{L}(X)$ is a Banach algebra) and is the inverse of $I + (\lambda - \lambda_1) R(\lambda_1, A)$. The convergence radius is $|\lambda - \lambda_1| < 1/\|R(\lambda_1, A)\|_{\mathcal{L}(X)}$. For $\lambda$ in this neighbourhood, $\lambda I - A$ is therefore the composition of two bijections (the bounded operator $I + (\lambda - \lambda_1) R(\lambda_1, A)$ on $X$ followed by $\lambda_1 I - A : D(A) \to X$), so $\lambda I - A : D(A) \to X$ is bijective, and $\lambda \in \rho(A)$. Restricting to $\lambda > 0$ gives a neighbourhood in $S$. For the resolvent bound on this neighbourhood: the dissipative inequality from Step 1, $\|(\lambda I - A) x\|_X \ge \lambda \|x\|_X$, holds for **all** $\lambda > 0$ and $x \in D(A)$ — its proof used only dissipativity, never any spectral information. So for any $\lambda \in \rho(A) \cap (0, \infty)$ and any $y \in X$, applying it to $x := R(\lambda, A) y \in D(A)$ gives $\|y\|_X = \|(\lambda I - A) R(\lambda, A) y\|_X \ge \lambda \|R(\lambda, A) y\|_X$. Taking the supremum over $\|y\|_X \le 1$ yields $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$, as claimed. **Closedness.** This is the technical part. Suppose $\lambda_n \in S$ with $\lambda_n \to \lambda$ and $\lambda \in (0, \infty)$. We must show $\lambda \in \rho(A)$. The key tool is the **resolvent equation** \begin{align*} R(\lambda_n, A) - R(\lambda_m, A) = (\lambda_m - \lambda_n) R(\lambda_n, A) R(\lambda_m, A), \end{align*} a standard identity valid for any $\lambda_n, \lambda_m \in \rho(A)$ (it follows from $(\lambda_m I - A) - (\lambda_n I - A) = (\lambda_m - \lambda_n) I$ after multiplying by $R(\lambda_n, A)$ on the left and $R(\lambda_m, A)$ on the right). By the openness step, $\|R(\lambda_n, A)\|_{\mathcal{L}(X)} \le 1/\lambda_n$, and since $\lambda_n \to \lambda > 0$, these norms are uniformly bounded by $2/\lambda$ for $n$ large. Fix $y \in X$ and define $x_n := R(\lambda_n, A) y \in D(A)$. Applying the resolvent equation to $y$: \begin{align*} \|x_n - x_m\|_X &= \|(R(\lambda_n, A) - R(\lambda_m, A))y\|_X \\ &\le |\lambda_m - \lambda_n|\cdot \|R(\lambda_n, A)\|_{\mathcal{L}(X)}\, \|R(\lambda_m, A)\|_{\mathcal{L}(X)}\, \|y\|_X \\ &\le |\lambda_m - \lambda_n| \cdot \frac{4}{\lambda^2}\, \|y\|_X \end{align*} for $n, m$ large. Since $|\lambda_m - \lambda_n| \to 0$, the sequence $(x_n)$ is Cauchy in $X$, hence converges to some $x \in X$. Now we extract the candidate for $R(\lambda, A) y$ and verify it. From the defining relation $(\lambda_n I - A) x_n = y$, that is $A x_n = \lambda_n x_n - y$, we read off \begin{align*} A x_n = \lambda_n x_n - y \to \lambda x - y \quad \text{in } X \text{ as } n \to \infty, \end{align*} since $\lambda_n \to \lambda$ and $x_n \to x$. So we have a sequence $(x_n) \subset D(A)$ with $x_n \to x$ and $A x_n \to \lambda x - y$. **The closedness of $A$ (Step 2) is essential here:** by definition of a closed operator, this implies $x \in D(A)$ and $A x = \lambda x - y$, i.e., $(\lambda I - A) x = y$. Without closedness, we could not conclude that $x$ lies in the domain of $A$. So $\lambda I - A : D(A) \to X$ is surjective at $\lambda$ (we found a preimage for arbitrary $y$). It is also injective by the dissipative inequality from Step 1 (which holds for all $\lambda > 0$). Therefore $\lambda I - A$ is bijective. Its inverse is bounded by $1/\lambda$ by the same dissipative inequality (Step 1, second consequence). By definition, $\lambda \in \rho(A)$, so $\lambda \in S$, proving $S$ is closed in $(0, \infty)$. **Conclusion.** $S$ is a non-empty, open, and closed subset of the connected set $(0, \infty)$, so $S = (0, \infty)$. Hence every $\lambda > 0$ belongs to $\rho(A)$, with $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$ throughout, by the openness argument applied at any point. Why is dissipativity *not* enough on its own for surjectivity? A general dissipative operator may have $\lambda I - A$ injective with closed range strictly smaller than $X$. The surjectivity hypothesis (sometimes called "maximal dissipativity") rules this out — and once $\lambda I - A$ is surjective at one $\lambda$, the propagation argument above gives surjectivity for all $\lambda > 0$.[/guided]

[step:Apply Hille-Yosida to conclude $A$ generates a contraction semigroup] By Step 3, $A$ is closed (Step 2), densely defined (hypothesis), and satisfies $(0, \infty) \subset \rho(A)$ with $\|R(\lambda, A)\|_{\mathcal{L}(X)} \le 1/\lambda$. The [Hille-Yosida Theorem](/theorems/3139) applies and yields a $C_0$-semigroup of contractions $\{T(t)\}_{t \ge 0}$ generated by $A$. This proves the implication: dissipativity + surjectivity at one $\lambda_0 > 0$ $\Longrightarrow$ generation of a contraction semigroup. [/step]

[step:Prove the converse: if $A$ generates a contraction semigroup, then $A$ is dissipative and $\lambda I - A$ is surjective for every $\lambda > 0$] Suppose $A$ generates a $C_0$-semigroup $\{T(t)\}_{t \ge 0}$ with $\|T(t)\|_{\mathcal{L}(X)} \le 1$. **Dissipativity.** For $x \in D(A)$, the function $\varphi: [0, \infty) \to \mathbb{R}$, $t \mapsto \|T(t)x\|_X^2 = (T(t)x, T(t)x)_H$ is differentiable at $t = 0$ by the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), with \begin{align*} \varphi'(0) = \lim_{t \to 0^+} \frac{\|T(t)x\|_X^2 - \|x\|_X^2}{t}. \end{align*} Computing via the inner product: \begin{align*} \frac{\|T(t)x\|_X^2 - \|x\|_X^2}{t} &= \left(\frac{T(t)x - x}{t}, T(t)x\right)_H + \left(x, \frac{T(t)x - x}{t}\right)_H \\ &\to (Ax, x)_H + (x, Ax)_H = 2\operatorname{Re}(Ax, x)_H \quad \text{as } t \to 0^+, \end{align*} using $T(t)x \to x$ and $\frac{T(t)x - x}{t} \to Ax$ as $t \to 0^+$. On the other hand, contractivity gives $\|T(t)x\|_X^2 \le \|x\|_X^2$, so $\varphi(t) - \varphi(0) \le 0$ for all $t \ge 0$, hence $\varphi'(0) = \lim_{t \to 0^+}(\varphi(t) - \varphi(0))/t \le 0$. Combining: $2\operatorname{Re}(Ax, x)_H \le 0$, i.e., $\operatorname{Re}(Ax, x)_H \le 0$. So $A$ is dissipative. **Surjectivity.** We invoke the [Hille-Yosida Theorem](/theorems/3139) (forward direction). Its hypotheses require $A$ to be closed and densely defined: by [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), the generator of any $C_0$-semigroup is automatically closed and densely defined, so both conditions hold here. The conclusion of Hille-Yosida gives $(0, \infty) \subset \rho(A)$, meaning $\lambda I - A: D(A) \to X$ is bijective and in particular surjective for every $\lambda > 0$. So hypothesis (2) holds for every $\lambda > 0$ (not just one). This proves the reverse implication, completing the equivalence. [/step]