[proofplan] We construct the infinitesimal generator $K$ by taking the strong derivative of $U(t)$ at $t=0$. Smooth averaging shows that the domain of $K$ is dense, and the group law plus unitarity show that $K$ is closed, skew-symmetric, and has both resolvent ranges $\operatorname{Range}(I-K)=H$ and $\operatorname{Range}(I+K)=H$. These range identities upgrade skew-symmetry to skew-adjointness, so $A:=iK$ is self-adjoint. The [spectral theorem for self-adjoint operators](/theorems/6911) then defines $e^{-itA}$, and comparison of infinitesimal generators proves that this spectral group is exactly the original group $U(t)$; uniqueness follows because the derivative at $t=0$ recovers the self-adjoint generator. [/proofplan] [step:Define the strong infinitesimal generator] Define the domain \begin{align*} D(K):=\left\{\psi\in H:\lim_{t\to 0}\frac{U(t)\psi-\psi}{t}\text{ exists in }H\right\}. \end{align*} Define the operator $K:D(K)\subset H\to H$ by \begin{align*} K\psi:=\lim_{t\to 0}\frac{U(t)\psi-\psi}{t}. \end{align*} For $\psi\in D(K)$ and $s\in\mathbb{R}$, the group law gives \begin{align*} \frac{U(t)U(s)\psi-U(s)\psi}{t}=U(s)\frac{U(t)\psi-\psi}{t}. \end{align*} Since $U(s)\in\mathcal{L}(H)$ is bounded, passing to the limit as $t\to 0$ gives $U(s)\psi\in D(K)$ and \begin{align*} K U(s)\psi=U(s)K\psi. \end{align*} Thus $D(K)$ is invariant under every $U(s)$, and $K$ commutes with the group on its domain. [/step] [step:Show that smooth averages lie in the generator domain and are dense] Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $f\in C_c^\infty(\mathbb{R})$ and $\psi\in H$. Define the Bochner integral \begin{align*} \psi_f:=\int_{\mathbb{R}} f(s)U(s)\psi\,d\mathcal{L}^1(s)\in H. \end{align*} The map $s\mapsto f(s)U(s)\psi$ is continuous with compact support, hence Bochner integrable. For $t\neq 0$, using the substitution $r=s+t$, with $d\mathcal{L}^1(r)=d\mathcal{L}^1(s)$ and integration domain still $\mathbb{R}$, we obtain \begin{align*} \frac{U(t)\psi_f-\psi_f}{t}=\int_{\mathbb{R}}\frac{f(r-t)-f(r)}{t}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Because $f\in C_c^\infty(\mathbb{R})$, the difference quotients converge uniformly to $-f'$ and are supported in a fixed compact set for $|t|$ small. Therefore the [Dominated Convergence Theorem](/theorems/4), applied in its Bochner-integral form to the uniformly dominated $H$-valued integrands, gives $\psi_f\in D(K)$ and \begin{align*} K\psi_f=-\int_{\mathbb{R}} f'(r)U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Now choose a standard approximate identity $(\rho_n)_{n\in\mathbb{N}}\subset C_c^\infty(\mathbb{R})$ with $\operatorname{supp}\rho_n\subset(-1/n,1/n)$, $\rho_n\geq 0$, and \begin{align*} \int_{\mathbb{R}}\rho_n(s)\,d\mathcal{L}^1(s)=1. \end{align*} For each $\psi\in H$, define \begin{align*} \psi_n:=\int_{\mathbb{R}}\rho_n(s)U(s)\psi\,d\mathcal{L}^1(s). \end{align*} Each $\psi_n$ lies in $D(K)$ by the preceding paragraph. Strong continuity of $U$ at $0$ implies \begin{align*} \|\psi_n-\psi\|_H\leq \int_{\mathbb{R}}\rho_n(s)\|U(s)\psi-\psi\|_H\,d\mathcal{L}^1(s)\to 0. \end{align*} Hence $D(K)$ is dense in $H$. [guided] The generator is defined by differentiating $U(t)$ at $t=0$, so the first serious issue is domain: why should there be many vectors for which that derivative exists? The standard device is to smooth a vector by averaging its orbit under the group. Fix $f\in C_c^\infty(\mathbb{R})$ and $\psi\in H$, and define \begin{align*} \psi_f:=\int_{\mathbb{R}} f(s)U(s)\psi\,d\mathcal{L}^1(s). \end{align*} This is a Bochner integral in $H$. It is valid because $s\mapsto U(s)\psi$ is continuous by strong continuity, $f$ is continuous with compact support, and therefore $s\mapsto f(s)U(s)\psi$ is continuous with compact support as an $H$-valued function. We now compute the difference quotient. For $t\neq 0$, \begin{align*} U(t)\psi_f=\int_{\mathbb{R}} f(s)U(t+s)\psi\,d\mathcal{L}^1(s). \end{align*} Use the substitution $r=s+t$. [Translation invariance](/theorems/4911) of one-dimensional Lebesgue measure gives $d\mathcal{L}^1(r)=d\mathcal{L}^1(s)$, and the domain $\mathbb{R}$ is unchanged. Thus \begin{align*} U(t)\psi_f=\int_{\mathbb{R}} f(r-t)U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Subtracting $\psi_f$ and dividing by $t$ gives \begin{align*} \frac{U(t)\psi_f-\psi_f}{t}=\int_{\mathbb{R}}\frac{f(r-t)-f(r)}{t}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Since $f$ is smooth and compactly supported, the scalar difference quotient $\frac{f(r-t)-f(r)}{t}$ converges uniformly to $-f'(r)$ as $t\to 0$, and for all sufficiently small $|t|$ these difference quotients vanish outside one fixed compact subset of $\mathbb{R}$. The integrand is therefore dominated by an integrable scalar function times the fixed vector norm $\|\psi\|_H$. The [Dominated Convergence Theorem](/theorems/4), applied in its Bochner-integral form, gives \begin{align*} K\psi_f=-\int_{\mathbb{R}} f'(r)U(r)\psi\,d\mathcal{L}^1(r). \end{align*} So every smoothed vector $\psi_f$ belongs to $D(K)$. To prove density, we choose a standard approximate identity $(\rho_n)_{n\in\mathbb{N}}\subset C_c^\infty(\mathbb{R})$ satisfying $\rho_n\geq 0$, $\operatorname{supp}\rho_n\subset(-1/n,1/n)$, and \begin{align*} \int_{\mathbb{R}}\rho_n(s)\,d\mathcal{L}^1(s)=1. \end{align*} For $\psi\in H$, define \begin{align*} \psi_n:=\int_{\mathbb{R}}\rho_n(s)U(s)\psi\,d\mathcal{L}^1(s). \end{align*} Each $\psi_n$ is in $D(K)$ by the smoothing argument. Moreover, \begin{align*} \psi_n-\psi=\int_{\mathbb{R}}\rho_n(s)(U(s)\psi-\psi)\,d\mathcal{L}^1(s), \end{align*} because the integral of $\rho_n$ is $1$. Taking norms and using positivity of $\rho_n$ gives \begin{align*} \|\psi_n-\psi\|_H\leq \int_{\mathbb{R}}\rho_n(s)\|U(s)\psi-\psi\|_H\,d\mathcal{L}^1(s). \end{align*} The support of $\rho_n$ shrinks to $0$, and strong continuity says $\|U(s)\psi-\psi\|_H\to 0$ as $s\to 0$. Therefore the right-hand side tends to $0$. Thus every $\psi\in H$ is a norm limit of vectors in $D(K)$, so $D(K)$ is dense. [/guided] [/step] [step:Prove that the generator is closed and skew-symmetric] First let $\psi\in D(K)$. Since $U(s)\psi\in D(K)$ and $K U(s)\psi=U(s)K\psi$, the map $F_\psi:\mathbb{R}\to H$ defined by \begin{align*} F_\psi(s):=U(s)\psi \end{align*} is strongly differentiable and satisfies $F_\psi'(s)=U(s)K\psi$. Hence, for every $t\in\mathbb{R}$, \begin{align*} U(t)\psi-\psi=\int_0^t U(s)K\psi\,d\mathcal{L}^1(s). \end{align*} Now suppose $(\psi_n)_{n\in\mathbb{N}}\subset D(K)$, $\psi_n\to\psi$ in $H$, and $K\psi_n\to\eta$ in $H$. The preceding identity gives \begin{align*} U(t)\psi_n-\psi_n=\int_0^t U(s)K\psi_n\,d\mathcal{L}^1(s). \end{align*} Passing to the limit in $H$ yields \begin{align*} U(t)\psi-\psi=\int_0^t U(s)\eta\,d\mathcal{L}^1(s). \end{align*} Dividing by $t$ and letting $t\to 0$, strong continuity gives \begin{align*} \lim_{t\to 0}\frac{U(t)\psi-\psi}{t}=\eta. \end{align*} Thus $\psi\in D(K)$ and $K\psi=\eta$, so $K$ is closed. For $\psi,\varphi\in D(K)$, unitarity gives \begin{align*} (U(t)\psi,U(t)\varphi)_H=(\psi,\varphi)_H. \end{align*} Differentiating at $t=0$ gives \begin{align*} (K\psi,\varphi)_H+(\psi,K\varphi)_H=0. \end{align*} Therefore \begin{align*} (K\psi,\varphi)_H=-(\psi,K\varphi)_H, \end{align*} so $K$ is skew-symmetric. [/step] [step:Construct resolvents on the positive and negative half-lines] For $\lambda>0$, define the bounded [linear map](/page/Linear%20Map) $R_\lambda:H\to H$ by \begin{align*} R_\lambda\psi:=\int_0^\infty e^{-\lambda t}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} The integral is Bochner integrable because $\|e^{-\lambda t}U(t)\psi\|_H=e^{-\lambda t}\|\psi\|_H$ and $e^{-\lambda t}\in L^1((0,\infty),\mathcal{L}^1)$. For $h\neq 0$, translation by $h$ in the Bochner integral gives \begin{align*} \frac{U(h)R_\lambda\psi-R_\lambda\psi}{h}=\frac{e^{\lambda h}-1}{h}\int_h^\infty e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)-\frac{1}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Here the substitution is $r=t+h$, the measure is unchanged by translation, and the interval $(0,\infty)$ becomes $(h,\infty)$. The formula is interpreted with oriented Bochner integrals: if $h<0$, then $\int_0^h=-\int_h^0$. We separate the two limiting terms. First, \begin{align*} \left\|\int_h^\infty e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)-R_\lambda\psi\right\|_H\leq \int_0^{|h|} e^{\lambda |h|} \|\psi\|_H\,d\mathcal{L}^1(r)=|h|e^{\lambda |h|}\|\psi\|_H, \end{align*} so the translated tail converges in $H$ to $R_\lambda\psi$. Since $(e^{\lambda h}-1)/h\to\lambda$, the first term converges to $\lambda R_\lambda\psi$. Second, the averaged interval term satisfies \begin{align*} \frac{1}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)\to \psi \end{align*} by strong continuity of $r\mapsto e^{-\lambda r}U(r)\psi$ at $0$. Hence $R_\lambda\psi\in D(K)$ and \begin{align*} K R_\lambda\psi=\lambda R_\lambda\psi-\psi. \end{align*} Therefore \begin{align*} (\lambda I-K)R_\lambda\psi=\psi. \end{align*} So $\operatorname{Range}(\lambda I-K)=H$ for every $\lambda>0$. For $\lambda<0$, define the bounded linear map $R_\lambda:H\to H$ by \begin{align*} R_\lambda\psi:=-\int_{-\infty}^0 e^{-\lambda t}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} Since $\lambda<0$, the scalar function $t\mapsto e^{-\lambda t}$ belongs to $L^1((-\infty,0),\mathcal{L}^1)$. The same translation computation, now with the interval $(-\infty,0)$ translated to $(-\infty,h)$ and with $\int_0^h$ understood as an oriented Bochner integral for $h<0$, yields \begin{align*} \frac{U(h)R_\lambda\psi-R_\lambda\psi}{h}=\frac{e^{\lambda h}-1}{h}R_\lambda\psi-\frac{e^{\lambda h}}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} The first term converges to $\lambda R_\lambda\psi$ by the same dominated-convergence argument on the translated half-line, and the second term converges to $\psi$ because the oriented averages over $[0,h]$ converge to the value at $0$ by strong continuity of $U$. Thus $R_\lambda\psi\in D(K)$ and \begin{align*} (\lambda I-K)R_\lambda\psi=\psi. \end{align*} So $\operatorname{Range}(\lambda I-K)=H$ for every $\lambda<0$. In particular, \begin{align*} \operatorname{Range}(I-K)=H,\qquad \operatorname{Range}(I+K)=H. \end{align*} [/step] [step:Upgrade skew-symmetry to skew-adjointness] Since $K$ is densely defined and skew-symmetric, $K\subset -K^*$. We prove the reverse inclusion. Let $y\in D(K^*)$. Because $\operatorname{Range}(I+K)=H$, choose $x\in D(K)$ such that \begin{align*} (I+K)x=y-K^*y. \end{align*} For every $v\in D(K)$, skew-symmetry gives \begin{align*} ((I-K)v,x)_H=(v,(I+K)x)_H=(v,y-K^*y)_H. \end{align*} By the definition of $K^*$, \begin{align*} (v,y-K^*y)_H=((I-K)v,y)_H. \end{align*} Therefore \begin{align*} ((I-K)v,y-x)_H=0 \end{align*} for every $v\in D(K)$. Since $\operatorname{Range}(I-K)=H$, it follows that $y=x$. Hence $y\in D(K)$, and on $D(K)$ the relation $K^*=-K$ holds. Thus $K^*=-K$, so $K$ is skew-adjoint. Define $A:D(A)\subset H\to H$ by setting $D(A):=D(K)$ and \begin{align*} A\psi:=iK\psi. \end{align*} Because $K^*=-K$, we have \begin{align*} A^*=(iK)^*=-iK^*=iK=A. \end{align*} Thus $A$ is self-adjoint. [/step] [step:Use the spectral theorem to recover the original unitary group] Let $\mathcal{B}(\mathbb{R})$ denote the Borel $\sigma$-algebra on $\mathbb{R}$. By the [Spectral Theorem for Self-Adjoint Operators](/theorems/440), applied to the self-adjoint operator $A$, there is a projection-valued measure \begin{align*} E:\mathcal{B}(\mathbb{R})\to \mathcal{L}(H) \end{align*} such that \begin{align*} A=\int_{\mathbb{R}}\lambda\,dE(\lambda). \end{align*} For each $t\in\mathbb{R}$, define the bounded operator $V(t):H\to H$ by \begin{align*} V(t):=e^{-itA}:=\int_{\mathbb{R}}e^{-it\lambda}\,dE(\lambda). \end{align*} The spectral functional calculus in the [Spectral Theorem for Self-Adjoint Operators](/theorems/440) gives that each $V(t)$ is unitary, $V(t+s)=V(t)V(s)$, and $t\mapsto V(t)\psi$ is continuous for each $\psi\in H$. For $\psi\in D(A)$, define the finite scalar measure $\mu_\psi:\mathcal{B}(\mathbb{R})\to[0,\infty)$ by \begin{align*} \mu_\psi(B):=(E(B)\psi,\psi)_H. \end{align*} The domain condition $\psi\in D(A)$ means \begin{align*} \int_{\mathbb{R}}\lambda^2\,d\mu_\psi(\lambda)<\infty. \end{align*} Using the spectral calculus and the [Dominated Convergence Theorem](/theorems/4), with domination by $|\lambda|$ in $L^2(\mu_\psi)$, we obtain \begin{align*} \lim_{t\to 0}\frac{V(t)\psi-\psi}{t}=-iA\psi. \end{align*} Since $A=iK$, we have $-iA=K$. Therefore $V$ and $U$ have the same infinitesimal generator on $D(K)$. We also need domain invariance for the comparison argument. The spectral functional calculus gives $V(t)D(A)\subset D(A)$ and $AV(t)\psi=V(t)A\psi$ for every $\psi\in D(A)$, because multiplication by $e^{-it\lambda}$ preserves the weighted condition $\int_{\mathbb{R}}\lambda^2\,d\mu_\psi(\lambda)<\infty$ and commutes with multiplication by $\lambda$. Since $D(A)=D(K)$ and $K=-iA$, this gives $V(t)D(K)\subset D(K)$ and $KV(t)\psi=V(t)K\psi$ for every $\psi\in D(K)$. For $\psi\in D(K)$, define $W_\psi:\mathbb{R}\to H$ by \begin{align*} W_\psi(t):=U(-t)V(t)\psi. \end{align*} We first record the derivative of the inverse orbit. If $\phi\in D(K)$, then \begin{align*} \frac{U(-(t+h))\phi-U(-t)\phi}{h}=U(-t)\frac{U(-h)\phi-\phi}{h}, \end{align*} and the quotient on the right converges to $-U(-t)K\phi$ because $\phi\in D(K)$ and $K$ commutes with $U(-h)$ on $D(K)$. Since $V(t)\psi\in D(K)$ and $KV(t)\psi=V(t)K\psi$, the product rule for strongly differentiable $H$-valued maps gives \begin{align*} W_\psi'(t)=U(-t)(-K)V(t)\psi+U(-t)KV(t)\psi=0. \end{align*} [guided] We now prove in full that the spectral group $V$ is the original group $U$. Let $\mathcal{B}(\mathbb{R})$ denote the Borel $\sigma$-algebra on $\mathbb{R}$. Since $A$ is self-adjoint, the [Spectral Theorem for Self-Adjoint Operators](/theorems/440) gives a projection-valued measure \begin{align*} E:\mathcal{B}(\mathbb{R})\to\mathcal{L}(H) \end{align*} such that \begin{align*} A=\int_{\mathbb{R}}\lambda\,dE(\lambda). \end{align*} For each $t\in\mathbb{R}$, define \begin{align*} V(t):=e^{-itA}:=\int_{\mathbb{R}}e^{-it\lambda}\,dE(\lambda). \end{align*} The spectral functional calculus gives that $V(t)$ is unitary, $V(t+s)=V(t)V(s)$, and $t\mapsto V(t)\psi$ is continuous for every $\psi\in H$. Fix $\psi\in D(A)$. Define the finite scalar measure $\mu_\psi:\mathcal{B}(\mathbb{R})\to[0,\infty)$ by \begin{align*} \mu_\psi(B):=(E(B)\psi,\psi)_H. \end{align*} The spectral characterization of $D(A)$ says \begin{align*} \int_{\mathbb{R}}\lambda^2\,d\mu_\psi(\lambda)<\infty. \end{align*} For each $\lambda\in\mathbb{R}$, \begin{align*} \frac{e^{-it\lambda}-1}{t}\to -i\lambda \end{align*} as $t\to0$. Also, \begin{align*} \left|\frac{e^{-it\lambda}-1}{t}\right|\leq |\lambda| \end{align*} for all $t\neq0$, by the fundamental theorem of calculus applied to $s\mapsto e^{-is\lambda}$. Since $|\lambda|\in L^2(\mu_\psi)$, the [Dominated Convergence Theorem](/theorems/4) in the spectral $L^2(\mu_\psi)$ norm gives \begin{align*} \lim_{t\to0}\frac{V(t)\psi-\psi}{t}=-iA\psi. \end{align*} Because $A=iK$, we have $-iA=K$. Thus $V$ and $U$ have the same infinitesimal generator on $D(K)=D(A)$. We also need the domain to remain stable under $V(t)$. If $\psi\in D(A)$, multiplication by $e^{-it\lambda}$ does not change the weighted integrability condition \begin{align*} \int_{\mathbb{R}}\lambda^2\,d\mu_\psi(\lambda)<\infty, \end{align*} and multiplication by $e^{-it\lambda}$ commutes with multiplication by $\lambda$. Hence $V(t)D(A)\subset D(A)$ and \begin{align*} AV(t)\psi=V(t)A\psi. \end{align*} Since $D(A)=D(K)$ and $K=-iA$, this is exactly \begin{align*} V(t)D(K)\subset D(K),\qquad KV(t)\psi=V(t)K\psi. \end{align*} Now fix $\psi\in D(K)$ and define $W_\psi:\mathbb{R}\to H$ by \begin{align*} W_\psi(t):=U(-t)V(t)\psi. \end{align*} The point of this definition is that $W_\psi$ measures the difference between the two groups. We compute its derivative. If $\phi\in D(K)$, then the group law gives \begin{align*} \frac{U(-(t+h))\phi-U(-t)\phi}{h}=U(-t)\frac{U(-h)\phi-\phi}{h}. \end{align*} As $h\to0$, the quotient on the right converges to $-K\phi$, so \begin{align*} \frac{d}{dt}U(-t)\phi=-U(-t)K\phi. \end{align*} Since $V(t)\psi\in D(K)$ and $KV(t)\psi=V(t)K\psi$, the product rule for strongly differentiable $H$-valued maps gives \begin{align*} W_\psi'(t)=U(-t)(-K)V(t)\psi+U(-t)KV(t)\psi=0. \end{align*} Therefore $W_\psi$ is constant. Evaluating at $t=0$ gives \begin{align*} W_\psi(t)=W_\psi(0)=\psi. \end{align*} Thus \begin{align*} U(-t)V(t)\psi=\psi, \end{align*} which is equivalent to \begin{align*} V(t)\psi=U(t)\psi. \end{align*} This equality holds for every $\psi\in D(K)$. Since $D(K)$ is dense in $H$ and both $U(t)$ and $V(t)$ are bounded operators, it extends to every $\psi\in H$. Hence \begin{align*} U(t)=V(t)=e^{-itA} \end{align*} for every $t\in\mathbb{R}$. [/guided] Thus $W_\psi(t)=W_\psi(0)=\psi$ for every $t\in\mathbb{R}$. Hence $V(t)\psi=U(t)\psi$ for every $\psi\in D(K)$. Since $D(K)$ is dense in $H$ and both $U(t)$ and $V(t)$ are bounded operators, the equality extends to all $\psi\in H$. Therefore \begin{align*} U(t)=V(t)=e^{-itA} \end{align*} for every $t\in\mathbb{R}$. [/step] [step:Recover the operator from the group and prove uniqueness] Suppose $B:D(B)\subset H\to H$ is another self-adjoint operator such that \begin{align*} U(t)=e^{-itB} \end{align*} for every $t\in\mathbb{R}$. Applying the [Spectral Theorem for Self-Adjoint Operators](/theorems/440) to $B$, the spectral representation identifies $B$ with multiplication by the real variable $\lambda$ on an $L^2$ space and identifies $e^{-itB}$ with multiplication by $e^{-it\lambda}$. For vectors in $D(B)$, the [Dominated Convergence Theorem](/theorems/4), with domination by $|\lambda|$ in the corresponding $L^2$ space, gives the strong derivative $-iB$ exactly as above. Conversely, suppose a vector belongs to the strong generator domain of $t\mapsto e^{-itB}$. In the spectral representation, the difference quotients \begin{align*} \frac{e^{-it\lambda}-1}{t}f(\lambda) \end{align*} converge in $L^2$ as $t\to0$. Taking a sequence $t_n\to0$ along which the convergence is almost everywhere, the pointwise limit is $-i\lambda f(\lambda)$. Since the same sequence converges in $L^2$, its limit must agree with that pointwise limit, so $\lambda f(\lambda)\in L^2$. This is precisely the spectral characterization of $D(B)$, and the generator value is $-iBf$. Thus the infinitesimal generator of $t\mapsto e^{-itB}$ is $-iB$ with domain exactly $D(B)$. [guided] The uniqueness argument uses the fact that the strongly differentiable vectors for the spectral unitary group are exactly the operator domain. In the spectral representation supplied by the [Spectral Theorem for Self-Adjoint Operators](/theorems/440), the operator $B$ is multiplication by $\lambda$, and $e^{-itB}$ is multiplication by $e^{-it\lambda}$. If $f\in D(B)$, then $\lambda f\in L^2$, and the [Dominated Convergence Theorem](/theorems/4) applied with domination by $|\lambda f(\lambda)|$ gives \begin{align*} \frac{e^{-it\lambda}-1}{t}f(\lambda)\to -i\lambda f(\lambda) \end{align*} in $L^2$. Thus every vector in $D(B)$ lies in the strong generator domain, with generator value $-iBf$. Conversely, suppose the difference quotients converge in $L^2$. Choose a sequence $t_n\to0$ along which the $L^2$ convergence has an almost-everywhere subsequence. Pointwise, the scalar quotients satisfy \begin{align*} \frac{e^{-it_n\lambda}-1}{t_n}f(\lambda)\to -i\lambda f(\lambda). \end{align*} The $L^2$ limit of the quotients must therefore be represented by $-i\lambda f(\lambda)$, so $\lambda f\in L^2$. This is exactly the spectral domain condition for $B$. Hence the generator of $t\mapsto e^{-itB}$ is $-iB$ on the exact domain $D(B)$. [/guided] The infinitesimal generator of $U$ was defined uniquely by the strong derivative at $0$, so \begin{align*} -iB=K=-iA \end{align*} as operators, including equality of domains. Multiplying by $i$ gives $B=A$. Therefore the self-adjoint operator $A$ is unique. This completes the proof. [/step]