Absolute value has a Dirac second derivative

The function $\mathrm{\mid }x\mathrm{\mid }$ is continuous, hence locally integrable, so [F1] defines the regular distribution $T_{\mathrm{\mid }x\mathrm{\mid }}$. Since the second-order multi-index has $\mathrm{\mid }\alpha \mathrm{\mid }=2$, [F2] gives $⟨{\mathrm{\partial }}^2{T}_{\mathrm{\mid }x\mathrm{\mid }},\phi ⟩=(-1{)}^2⟨T_{\mathrm{\mid }x\mathrm{\mid }},{\phi }^{\mathrm{\prime }\mathrm{\prime }}⟩={\int }_{\mathbb{R}}\mathrm{\mid }x\mathrm{\mid }{\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx$, a finite integral because ${\phi }^{\mathrm{\prime }\mathrm{\prime }}$ is bounded with compact support. By [F5] fix $R>0$ with $\mathrm{supp}\phi \subseteq (-R,R)$. [F1, F2, F5, given] 1.2 $\mathrm{\mid }x\mathrm{\mid }\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,\mathrm{\infty }}(\mathbb{R})$: let $I$ be a bounded open interval. If $0\mathrm{\notin }I$, then $\mathrm{\mid }x\mathrm{\mid }$ is $C^1$ on $I$ with classical derivative $\pm 1$, so [F7] makes that constant the weak derivative, and both $\mathrm{\mid }x\mathrm{\mid }$ and its derivative lie in $L^{\mathrm{\infty }}(I)$, giving $\mathrm{\mid }x\mathrm{\mid }\in W^{1,\mathrm{\infty }}(I)$. If $0\in I$, [F6] gives $\mathrm{\mid }x\mathrm{\mid }\in W^{1,\mathrm{\infty }}(I)$ directly with bounded weak derivative represented by the sign function. As $I$ was arbitrary, $\mathrm{\mid }x\mathrm{\mid }\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,\mathrm{\infty }}(\mathbb{R})$. [F6, F7, given] 2.1 On the interval $[0,R]$ apply [F4] with $u(x)=x$ and $v={\phi }^{\mathrm{\prime }}$: ${\int }_0^{R}x{\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx=[R{\phi }^{\mathrm{\prime }}(R)-0\cdot {\phi }^{\mathrm{\prime }}(0)]-(\phi (R)-\phi (0))\mathrm{.}$ On $[-R,0]$ apply [F4] with $u(x)=-x$: ${\int }_{-R}^0(-x){\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx=[(-x){\phi }^{\mathrm{\prime }}(x){]}_{-R}^0+{\int }_{-R}^0{\phi }^{\mathrm{\prime }}(x)dx=-R{\phi }^{\mathrm{\prime }}(-R)+\phi (0)-\phi (-R)\mathrm{.}$ By the choice of $R$ in step 1.1 we have $\phi (\pm R)=0$ and ${\phi }^{\mathrm{\prime }}(\pm R)=0$, so both right-hand sides equal $\phi (0)$, and adding the two half-line integrals gives ${\int }_{\mathbb{R}}\mathrm{\mid }x\mathrm{\mid }{\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx=2\phi (0)$. Hence $⟨{\mathrm{\partial }}^2{T}_{\mathrm{\mid }x\mathrm{\mid }},\phi ⟩=2\phi (0)=2{\delta }_0(\phi )$ by [F3], and since $\phi$ was arbitrary, ${\mathrm{\partial }}^2{T}_{\mathrm{\mid }x\mathrm{\mid }}=2{\delta }_0$. [F3, F4, step 1.1, given] 3.1 $\mathrm{\mid }x\mathrm{\mid }\mathrm{\notin }{W}_{\mathrm{l}\mathrm{o}\mathrm{c}}^{2,1}(\mathbb{R})$: suppose otherwise and take $I=(-1,1)$. Then $\mathrm{\mid }x\mathrm{\mid }\in W^{2,1}(I)$, so by [F8] there is $v\in L^1(I)$ representing the second weak derivative: ${\int }_I\mathrm{\mid }x\mathrm{\mid }{\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx={\int }_{I}v(x)\phi (x)dx\text{for every test }\phi \text{ supported in }I\mathrm{.}$ Step 2.1 computes the left side as $2\phi (0)$ for every test function $\phi$ on $\mathbb{R}$, hence for every test function supported in $I$. Let $U_{+}=(0,1)$ and $U_{-}=(-1,0)$. If $\phi$ is a test function supported in $U_{+}$, then $\phi (0)=0$, so ${\int }_{U_{+}}v\phi ={\int }_{I}v\phi =2\phi (0)=0$; as $\phi$ was arbitrary, [F9] applied on the open set $U_{+}$ gives $v=0$ almost everywhere on $U_{+}$. The same computation on $U_{-}$ gives $v=0$ almost everywhere on $U_{-}$, and since $I\setminus (U_{+}\cup U_{-})=\{0\}$ is a singleton, hence null, $v=0$ almost everywhere on $I$. By [F10] applied to the compact set $\{0\}$ inside the open set $I$ there is a test function $\psi \in C_c^{\mathrm{\infty }}(I)$ with $\psi (0)=1$; the weak-derivative identity for this $\psi$ gives ${\int }_{I}v\psi ={\int }_I\mathrm{\mid }x\mathrm{\mid }{\psi }^{\mathrm{\prime }\mathrm{\prime }}=2\psi (0)=2$, while $v=0$ almost everywhere on $I$ gives ${\int }_{I}v\psi =0$, a contradiction. Hence $\mathrm{\mid }x\mathrm{\mid }\mathrm{\notin }{W}^{2,1}(I)$ for this $I$, and therefore $\mathrm{\mid }x\mathrm{\mid }\mathrm{\notin }{W}_{\mathrm{l}\mathrm{o}\mathrm{c}}^{2,1}(\mathbb{R})$. [F8, F9, F10, step 2.1] 4.1 The example is complete: the second distributional derivative of $\mathrm{\mid }x\mathrm{\mid }$ is the point mass $2{\delta }_0$, which has no locally integrable representative, while the first weak derivative exists and is bounded. The conclusion uses only Countable Choice [F11], used by the regular-distribution injectivity interfaces [F1] and [F9], the Lebesgue integration-by-parts interface [F4], the first-derivative interfaces [F6] and [F7], and the Sobolev interface [F8]. The bump construction [F10] and distributional differentiation [F2] are choice-free. The endpoint value of the sign representative at $0$ is irrelevant, since a point is null. $\mathrm{\square }$