[proofplan] We conjugate the momentum operator by the unitary [Fourier transform](/page/Fourier%20Transform). On test functions, [integration by parts](/theorems/210) shows that $-i\,d/dx$ becomes multiplication by the frequency variable $\xi$. Multiplication by $\xi$ on its natural maximal domain is self-adjoint, and the inverse Fourier transform of that domain is exactly $H^1(\mathbb R)$. Finally, the density of $C_c^\infty(\mathbb R)$ in $H^1(\mathbb R)$ identifies the graph closure of $P_0$ with this self-adjoint operator, proving essential self-adjointness. [/proofplan] [step:Conjugate the test function operator to multiplication by frequency] Let $\mathcal F:L^2(\mathbb R)\to L^2(\mathbb R)$ denote the unitary Fourier transform obtained by extending the map \begin{align*} \mathcal F:C_c^\infty(\mathbb R)\to L^2(\mathbb R), \quad (\mathcal Fu)(\xi)=(2\pi)^{-1/2}\int_{\mathbb R}u(x)e^{-ix\xi}\,d\mathcal L^1(x) \end{align*} from test functions. Define the multiplication operator \begin{align*} M_\xi:D(M_\xi)\subset L^2(\mathbb R)\to L^2(\mathbb R), \quad (M_\xi f)(\xi)=\xi f(\xi) \end{align*} with domain \begin{align*} D(M_\xi)=\{f\in L^2(\mathbb R): \xi f\in L^2(\mathbb R)\}. \end{align*} For $u\in C_c^\infty(\mathbb R)$ and $\xi\in\mathbb R$, [integration by parts](/theorems/2098) is valid because $u$ has compact support. The boundary term vanishes, and hence \begin{align*} (\mathcal F(P_0u))(\xi)=(2\pi)^{-1/2}\int_{\mathbb R}(-iu'(x))e^{-ix\xi}\,d\mathcal L^1(x)=\xi(\mathcal Fu)(\xi). \end{align*} Thus \begin{align*} \mathcal F P_0u=M_\xi\mathcal Fu \end{align*} for every $u\in C_c^\infty(\mathbb R)$. [guided] The reason for introducing the Fourier transform is that differentiation becomes multiplication. We use the unitary Fourier transform \begin{align*} \mathcal F:L^2(\mathbb R)\to L^2(\mathbb R) \end{align*} whose action on test functions is \begin{align*} (\mathcal Fu)(\xi)=(2\pi)^{-1/2}\int_{\mathbb R}u(x)e^{-ix\xi}\,d\mathcal L^1(x). \end{align*} We also define the frequency multiplication operator \begin{align*} M_\xi:D(M_\xi)\subset L^2(\mathbb R)\to L^2(\mathbb R), \quad (M_\xi f)(\xi)=\xi f(\xi), \end{align*} where \begin{align*} D(M_\xi)=\{f\in L^2(\mathbb R): \xi f\in L^2(\mathbb R)\}. \end{align*} Now fix $u\in C_c^\infty(\mathbb R)$. Since $u$ is compactly supported and smooth, the classical integration by parts formula applies on any interval containing $\operatorname{supp}u$, and the boundary term is zero because $u$ vanishes near the endpoints. Therefore \begin{align*} (\mathcal F(P_0u))(\xi)=(2\pi)^{-1/2}\int_{\mathbb R}(-iu'(x))e^{-ix\xi}\,d\mathcal L^1(x). \end{align*} Integrating by parts transfers the derivative from $u$ to the exponential: \begin{align*} (2\pi)^{-1/2}\int_{\mathbb R}(-iu'(x))e^{-ix\xi}\,d\mathcal L^1(x)=(2\pi)^{-1/2}\int_{\mathbb R}u(x)(-i)(i\xi)e^{-ix\xi}\,d\mathcal L^1(x). \end{align*} Since $(-i)(i\xi)=\xi$, this becomes \begin{align*} (\mathcal F(P_0u))(\xi)=\xi(\mathcal Fu)(\xi). \end{align*} Equivalently, \begin{align*} \mathcal F P_0u=M_\xi\mathcal Fu. \end{align*} This is the core algebraic fact of the proof: the momentum operator is Fourier-conjugate to multiplication by the real variable $\xi$. [/guided] [/step] [step:Show that multiplication by $\xi$ is self-adjoint on its maximal domain] We first verify symmetry. If $f,g\in D(M_\xi)$, then $\xi f\in L^2(\mathbb R)$ and $\xi g\in L^2(\mathbb R)$, so the products $(\xi f)\overline g$ and $f\overline{\xi g}$ lie in $L^1(\mathbb R)$ by Cauchy-Schwarz. Since $\xi$ is real-valued, \begin{align*} (M_\xi f,g)_{L^2}=\int_{\mathbb R}\xi f(\xi)\overline{g(\xi)}\,d\mathcal L^1(\xi)=\int_{\mathbb R}f(\xi)\overline{\xi g(\xi)}\,d\mathcal L^1(\xi)=(f,M_\xi g)_{L^2}. \end{align*} It remains to identify the adjoint domain. Let $g\in D(M_\xi^*)$. Then there exists $h\in L^2(\mathbb R)$ such that \begin{align*} \int_{\mathbb R}\xi f(\xi)\overline{g(\xi)}\,d\mathcal L^1(\xi)=\int_{\mathbb R}f(\xi)\overline{h(\xi)}\,d\mathcal L^1(\xi) \end{align*} for every $f\in D(M_\xi)$. Every $f\in C_c^\infty(\mathbb R)$ belongs to $D(M_\xi)$, so the preceding identity implies $\xi g=h$ as distributions on $\mathbb R$. Since $h\in L^2(\mathbb R)$, we obtain $\xi g\in L^2(\mathbb R)$, hence $g\in D(M_\xi)$ and $M_\xi^*g=M_\xi g$. Therefore $D(M_\xi^*)\subset D(M_\xi)$. Symmetry gives $M_\xi\subset M_\xi^*$, so $D(M_\xi)\subset D(M_\xi^*)$. Thus $M_\xi=M_\xi^*$. [/step] [step:Identify the Fourier-side domain with $H^1(\mathbb R)$] We claim that \begin{align*} \mathcal F(H^1(\mathbb R))=D(M_\xi) \end{align*} and, for $u\in H^1(\mathbb R)$, \begin{align*} \mathcal F(Pu)=M_\xi\mathcal Fu. \end{align*} Indeed, suppose $u\in H^1(\mathbb R)$. Let $u'\in L^2(\mathbb R)$ denote its [weak derivative](/page/Weak%20Derivative). The [distributional derivative](/page/Distributional%20Derivative) identity gives \begin{align*} \mathcal F(u')=i\xi\mathcal Fu \end{align*} in $L^2(\mathbb R)$. Since $\mathcal F(u')\in L^2(\mathbb R)$, we have $\xi\mathcal Fu\in L^2(\mathbb R)$, so $\mathcal Fu\in D(M_\xi)$, and \begin{align*} \mathcal F(Pu)=\mathcal F(-iu')=\xi\mathcal Fu=M_\xi\mathcal Fu. \end{align*} Conversely, suppose $f\in D(M_\xi)$ and set \begin{align*} u=\mathcal F^{-1}f\in L^2(\mathbb R). \end{align*} Define \begin{align*} v=\mathcal F^{-1}(i\xi f)\in L^2(\mathbb R). \end{align*} For every $\varphi\in C_c^\infty(\mathbb R)$, the Fourier derivative rule on test functions and unitarity of $\mathcal F$ give \begin{align*} \int_{\mathbb R}u(x)\varphi'(x)\,d\mathcal L^1(x)=-\int_{\mathbb R}v(x)\varphi(x)\,d\mathcal L^1(x). \end{align*} Thus $v$ is the weak derivative of $u$, so $u\in H^1(\mathbb R)$. Therefore $\mathcal F(H^1(\mathbb R))=D(M_\xi)$ and $\mathcal F P\mathcal F^{-1}=M_\xi$. [guided] The domain of the self-adjoint multiplication operator is not arbitrary; it is exactly the Fourier image of $H^1(\mathbb R)$. We prove both inclusions. First let $u\in H^1(\mathbb R)$. By definition, $u\in L^2(\mathbb R)$ and there exists a weak derivative $u'\in L^2(\mathbb R)$. The Fourier transform converts weak differentiation into multiplication by $i\xi$, so \begin{align*} \mathcal F(u')=i\xi\mathcal Fu \end{align*} as an identity in $L^2(\mathbb R)$. Since $\mathcal F$ is unitary and $u'\in L^2(\mathbb R)$, the function $\mathcal F(u')$ lies in $L^2(\mathbb R)$. Hence $i\xi\mathcal Fu\in L^2(\mathbb R)$, equivalently $\xi\mathcal Fu\in L^2(\mathbb R)$. This says precisely that $\mathcal Fu\in D(M_\xi)$. Moreover \begin{align*} \mathcal F(Pu)=\mathcal F(-iu')=-i\mathcal F(u')=-i(i\xi\mathcal Fu)=\xi\mathcal Fu=M_\xi\mathcal Fu. \end{align*} Conversely, let $f\in D(M_\xi)$. We must show that its inverse Fourier transform has one weak derivative in $L^2$. Define \begin{align*} u=\mathcal F^{-1}f\in L^2(\mathbb R). \end{align*} Because $f\in D(M_\xi)$, the product $\xi f$ belongs to $L^2(\mathbb R)$, so the function \begin{align*} v=\mathcal F^{-1}(i\xi f) \end{align*} also lies in $L^2(\mathbb R)$. We now verify that $v$ is the weak derivative of $u$. Let $\varphi\in C_c^\infty(\mathbb R)$. The weak derivative condition requires \begin{align*} \int_{\mathbb R}u(x)\varphi'(x)\,d\mathcal L^1(x)=-\int_{\mathbb R}v(x)\varphi(x)\,d\mathcal L^1(x). \end{align*} Using unitarity of $\mathcal F$ and the test-function identity $\mathcal F(\varphi')=i\xi\mathcal F\varphi$, this equality follows from the definitions of $u$ and $v$. Therefore $u$ has weak derivative $v\in L^2(\mathbb R)$, so $u\in H^1(\mathbb R)$. This proves $\mathcal F(H^1(\mathbb R))=D(M_\xi)$ and the operator identity $\mathcal F P\mathcal F^{-1}=M_\xi$. [/guided] [/step] [step:Conclude that $P$ is self-adjoint] Since $\mathcal F:L^2(\mathbb R)\to L^2(\mathbb R)$ is unitary and $\mathcal FP\mathcal F^{-1}=M_\xi$ with $\mathcal F(D(P))=D(M_\xi)$, the operator $P$ is unitarily equivalent to $M_\xi$. Self-adjointness is preserved under unitary equivalence. Since $M_\xi$ is self-adjoint, $P$ is self-adjoint. [/step] [step:Identify the closure of $P_0$ with $P$] The operator $P_0$ is the restriction of $P$ to $C_c^\infty(\mathbb R)$, because the weak derivative of a [test function](/page/Test%20Function) equals its classical derivative. Hence $\overline{P_0}\subset P$. We prove the reverse inclusion by graph-norm density. The graph norm of $P$ is \begin{align*} \|u\|_{\operatorname{graph}(P)}^2=\|u\|_{L^2(\mathbb R)}^2+\|Pu\|_{L^2(\mathbb R)}^2=\|u\|_{L^2(\mathbb R)}^2+\|u'\|_{L^2(\mathbb R)}^2=\|u\|_{H^1(\mathbb R)}^2. \end{align*} The space $C_c^\infty(\mathbb R)$ is dense in $H^1(\mathbb R)$. Therefore, for every $u\in H^1(\mathbb R)$, there exists a sequence $(u_k)_{k\in\mathbb N}$ in $C_c^\infty(\mathbb R)$ such that \begin{align*} \|u_k-u\|_{H^1(\mathbb R)}\to 0. \end{align*} By the graph-norm identity, this is exactly \begin{align*} \|u_k-u\|_{L^2(\mathbb R)}+\|P_0u_k-Pu\|_{L^2(\mathbb R)}\to 0. \end{align*} Thus every $u\in D(P)$ belongs to $D(\overline{P_0})$ and $\overline{P_0}u=Pu$. Hence $\overline{P_0}=P$. [/step] [step:Deduce essential self-adjointness] We have shown that $P$ is self-adjoint and that $\overline{P_0}=P$. Therefore the closure of $P_0$ is self-adjoint. By definition, $P_0$ is essentially self-adjoint. This also identifies its closure as \begin{align*} P:H^1(\mathbb R)\subset L^2(\mathbb R)\to L^2(\mathbb R), \quad Pu=-iu'. \end{align*} This is the momentum operator on the real line. [/step]