Weak Derivatives and Sobolev Spaces — Examples

The set $Q_{+}=\{x_n>0\}\cap Q=Y\times (0,1)$ is a box, so [F3] makes it measurable with finite measure $2^{n-1}$, and [F2] makes $u={\mathbf{1}}_{Q_{+}}$ measurable. For finite $p$, ${\int }_Q\mathrm{\mid }u{\mathrm{\mid }}^p={\lambda }_n(Q_{+})=2^{n-1}<\mathrm{\infty }$; for $p=\mathrm{\infty }$, $\mathrm{\mid }u\mathrm{\mid }\le 1$, so $u\in L^p(Q)$ for every $1\le p\le \mathrm{\infty }$, and in particular $u\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(Q)$ by [F4].

Let $\phi \in C_c^{\mathrm{\infty }}(Q)$. Since $\mathrm{supp}\phi$ is compact in $Q$, [F5] and [F6] apply to the integrable function $u{\mathrm{\partial }}_n\phi$ and reduce the integral to the iterated integral over $Y\times I$. For fixed $y\in Y$ the function $t\mapsto \phi (y,t)$ is $C^1$ on the interval $[0,1]$ and vanishes at $t=1$, so [F7] gives ${\int }_0^1{\mathrm{\partial }}_t\phi (y,t)dt=\phi (y,1)-\phi (y,0)=-\phi (y,0)$. Hence ${\int }_{Q}u{\mathrm{\partial }}_n\phi dx={\int }_Y({\int }_0^1{\mathrm{\partial }}_t\phi (y,t)dt)dy=-{\int }_Y\phi (y,0)dy,$ and by the distributional sign convention the normal derivative of $T_u$ is the surface functional $⟨{\mathrm{\partial }}_n{T}_u,\phi ⟩=-{\int }_{Q}u{\mathrm{\partial }}_n\phi dx={\int }_Y\phi (y,0)dy\mathrm{.}$

Suppose $v\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(Q)$ represented that normal derivative, that is ${\int }_{Q}u{\mathrm{\partial }}_n\phi dx=-{\int }_{Q}v\phi dx$ for every test $\phi$ by [F8]. Combined with step 1.2 this means ${\int }_{Q}v\phi dx={\int }_Y\phi (y,0)dy\text{for every }\phi \in C_c^{\mathrm{\infty }}(Q)\mathrm{.}$

Fix $\psi$ as in [F9] and let $c={\int }_Y\psi >0$. Let $\eta :\mathbb{R}\to [0,1]$ be smooth, equal to $1$ on $[-1\mathrm{/}2,1\mathrm{/}2]$ and supported in $I$; for $0<\delta <1\mathrm{/}2$ put ${\eta }_{\delta }(t)=\eta (t\mathrm{/}\delta )$ and ${\phi }_{\delta }=\psi \otimes {\eta }_{\delta }$, which is a test by [F10]. Step 2.1 evaluated at ${\phi }_{\delta }$ gives ${\int }_{Q}v{\phi }_{\delta }dx={\int }_Y\psi (y){\eta }_{\delta }(0)dy=c$ for every $\delta$. On the other hand $v\psi \in L^1(K)$ for the compact product $K=\mathrm{supp}\psi \times [-1\mathrm{/}2,1\mathrm{/}2]$ by [F4], and the supports of the ${\phi }_{\delta }$ are contained in the slabs $A_{\delta }=\{-\delta <t<\delta \}\cap K$ with ${\lambda }_n(A_{\delta })\le 2\delta \cdot {\lambda }_{n-1}(\mathrm{supp}\psi )\to 0$ by [F3]. Since $\mathrm{\mid }{\phi }_{\delta }\mathrm{\mid }\le 1$, [F11] applied to $v\psi$ on $K$ yields $\mid {\int }_{Q}v{\phi }_{\delta }dx\mid \le {\int }_{A_{\delta }}\mathrm{\mid }v\psi \mathrm{\mid }dx⟶0,$ contradicting the constant value $c>0$. Hence no locally integrable $v$ represents ${\mathrm{\partial }}_n{T}_u$.

It remains to note the tangential directions. For $j<n$ and any test $\phi$, the same Fubini reduction gives ${\int }_{Q}u{\mathrm{\partial }}_j\phi ={\int }_{I}H(t)({\int }_Y{\mathrm{\partial }}_j\phi (y,t)dy)dt=0$, because [F7] applied along the $j$-th coordinate of the compactly supported function $y\mapsto \phi (y,t)$ makes the inner integral vanish; so the tangential distributional derivatives are represented by the zero function. That does not remove the obstruction of step 3.1: by [F8] membership of $u$ in $W^{1,p}(Q)$ for any $1\le p\le \mathrm{\infty }$ would require an $L^p$ class with a locally integrable representative for the multi-index $\alpha =e_n$, which step 3.1 rules out. Therefore $u\mathrm{\notin }{W}^{1,p}(Q)$ for every $1\le p\le \mathrm{\infty }$, including both endpoints, although $u\in L^p(Q)$ for every $p$ by step 1.1. The assumption used is Countable Choice [F1], spent through the box-measure, product-completion, Fubini and one-dimensional fundamental-theorem interfaces; no full Axiom of Choice occurs. $\mathrm{\square }$

Let $(X_n{)}_{n\in \mathbb{N}}$ be any sequence of nonempty sets. By AC, [F1], the range $\{X_n:n\in \mathbb{N}\}$ has a choice function $s$; then $n\mapsto s(X_n)$ selects from every $X_n$, proving Countable Choice. this discharges precisely the CC assumptions in [F5] (Cantor nullity), [F6] (the Cantor measure and its properties), [F10] (weak-derivative locality/uniqueness), and [F13] (Riemann–Stieltjes/Lebesgue–Stieltjes agreement). The Cantor staircase is a real continuous nondecreasing function on $[0,1]$, with endpoint values $0$ and $1$, and is constant on every closed gap $[a,b]$ of $C$ by [F2]–[F4]. Apart from invoking AC for this reduction, no further family of choices is made; each later CC use is through a named interface.

Put $U=(0,1)\setminus C$. It is open because $C$ is closed by [F22] and $(0,1)$ is open, so [F8] writes it as at most countably many disjoint intervals $J=(a,b)$. Every endpoint lies in $[0,1]$. The endpoints $0$ and $1$ belong to $C$ by [F23]; if an endpoint in $(0,1)$ were outside $C$, closedness would put it in the open set $U$, and a neighborhood would enlarge the component, contradicting maximality. Thus $a,b\in C$ and $(a,b)\cap C=\mathrm{\varnothing }$. By [F3], $c$ is constant on $[a,b]$; by [F17] its restriction is absolutely continuous, and its classical derivative is $0$ throughout $J$. Meanwhile $C$ is null by [F5].

Let $\phi \in C_c^{\mathrm{\infty }}(0,1)$ be real-valued and extend it by zero to $[0,1]$. Then $\phi (0)=\phi (1)=0$. For a partition $0=t_0<\cdots <t_N=1$, the telescoping product identity is $\phi (1)c(1)-\phi (0)c(0)={\sum }_{i=1}^N\phi (t_i)(c(t_i)-c(t_{i-1}))+{\sum }_{i=1}^{N}c(t_{i-1})(\phi (t_i)-\phi (t_{i-1}))\mathrm{.}$ The first sum uses right-endpoint tags for ${\int }_0^1\phi dc$, which exists by [F12]; the second uses left-endpoint tags for ${\int }_0^{1}cd\phi$, which exists by [F11]. As the mesh tends to zero, each sum converges to its Riemann–Stieltjes integral, so the zero boundary term gives ${\int }_0^1\phi dc=-{\int }_0^{1}cd\phi$. By [F11], the right side is $-{\int }_0^{1}c(x){\phi }^{\mathrm{\prime }}(x)dx$. The agreement in [F13] identifies the left side with ${\int }_{(0,1]}\phi d{\mu }_c$; since $\phi$ is supported inside $(0,1)$, this equals ${\int }_{\mathbb{R}}\phi d{\mu }_c$. Thus ${\int }_{\mathbb{R}}\phi d{\mu }_c=-{\int }_0^{1}c(x){\phi }^{\mathrm{\prime }}(x)dx=⟨\mathrm{\partial }{T}_c,\phi ⟩\mathrm{.}$ For a complex test, apply the real identity separately to its real and imaginary parts; complex linearity of the distribution pairing and the measure integral then gives the same identity. The order-zero estimate in [F19] makes $\phi \mapsto \int \phi d{\mu }_c$ a distribution on $(0,1)$. Hence $\mathrm{\partial }{T}_c$ is exactly the distribution induced by ${\mu }_c{\mathrm{\mid }}_{(0,1)}$. Since [F6] makes ${\mu }_c$ singular and concentrated on the null set $C$, and [F7] gives ${\mu }_c((0,1))=1$, this restricted measure is singular and nonzero.

Suppose $c\in W^{1,1}(0,1)$ and let $g\in L^1(0,1)$ be its weak derivative, as supplied by [F9]. For each component interval $J$ of $U$, the constant function $0$ is a weak derivative of $c{\mathrm{\mid }}_J$: for every $\psi \in C_c^{\mathrm{\infty }}(J)$ choose $[a,b]\subset J$ containing its support. By [F21], ${\int }_J{\psi }^{\mathrm{\prime }}={\int }_a^b{\psi }^{\mathrm{\prime }}=\psi (b)-\psi (a)=0$, and constancy of $c$ on $J$ gives ${\int }_{J}c{\psi }^{\mathrm{\prime }}=0$. By [F10], $g=0$ almost everywhere on each $J$. The exceptional sets on these countably many intervals have null union by [F15]; [F5] makes the remaining Cantor set null too. Therefore $g=0$ almost everywhere on $(0,1)$.

The compact intervals $K_m=[1\mathrm{/}m,1-1\mathrm{/}m]$, $m\ge 3$, cover $(0,1)$: for each $x\in (0,1)$, [F20] gives an $m$ with $1\mathrm{/}m<\min \{x,1-x\}$. Since ${\mu }_c((0,1))=1$ by [F7], [F15] implies ${\mu }_c(K_m)>0$ for at least one $m$. Fix one such $m$. The cutoff fact [F16] gives $\chi \in C_c^{\mathrm{\infty }}(0,1)$ with $0\le \chi \le 1$ and $\chi =1$ on a neighborhood of $K_m$. By [F14], $\int \chi d{\mu }_c\ge \int {\mathbf{1}}_{K_m}d{\mu }_c={\mu }_c(K_m)>0.$ But the weak-derivative identity and step 2.1 give $-{\int }_0^{1}c{\chi }^{\mathrm{\prime }}={\int }_0^{1}g\chi =0$, contradicting step 1.3, which identifies the left side with $\int \chi d{\mu }_c>0$. So $c\mathrm{\notin }{W}^{1,1}(0,1)$.

Steps 1.1–1.2 show absolute continuity on every complementary gap away from the null Cantor set, while steps 1.3, 2.1, and 3.1 show that the distributional derivative is the nonzero singular Cantor measure and that no $L^1$ weak derivative exists. This is the claimed counterexample.

Put $E=\{x_0\}$. By [F1], $E$ is measurable and $\mathrm{\mid }E\mathrm{\mid }=0$, so [F2] makes $u_1={\mathbf{1}}_E$ measurable. Both functions are locally integrable; for every compact $K\subseteq \mathrm{\Omega }$, ${\int }_K\mathrm{\mid }{u}_1\mathrm{\mid }dx={\int }_{K\cap E}1dx=0$ by [F6]. They agree at every $x\mathrm{\ne }{x}_0$, hence almost everywhere. For $1\le p<\mathrm{\infty }$, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{u}_1{\mathrm{\mid }}^{p}dx={\int }_{E}1dx=0$ again by [F6]; for $p=\mathrm{\infty }$, every positive superlevel set $\{\mathrm{\mid }{u}_1\mathrm{\mid }>\epsilon \}$ is empty or $E$, so its measure is zero and $\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}=0$. Thus $[u_0]=[u_1]$ in every $L^p$ by [F3], including both endpoints.

For every multi-index $\alpha$, zero has weak derivative zero because both sides of its test identity vanish. The functions $u_0,u_1,0,0$ are locally integrable and $u_0=u_1$ almost everywhere, so [F5] transfers the identity to $u_1$: zero is a weak derivative of both representatives at every order. At $\alpha =0$ the derivative class is their common zero $L^p$ class by [F7]; for $\mathrm{\mid }\alpha \mathrm{\mid }>0$ it is the zero $L^p$ class. By [F4], both belong to $W^{k,p}$ with identical derivative classes through order $k$, so every term in the Sobolev norm is zero. This includes $k=0$, $p=1$, and $p=\mathrm{\infty }$.

The representatives have different point values, $u_0(x_0)=0$ and $u_1(x_0)=1$, although [F3] and [F4] identify them as the same $L^p$ and $W^{k,p}$ elements. A value at $x_0$ therefore cannot be assigned from either class alone. After fixing $x_0$, the construction makes no choices; the stated Countable Choice hypothesis is exactly ${\mathrm{A}\mathrm{C}}_{\omega }$ by [F8] and is carried only through the null-set, representative-independence, and Sobolev-class interfaces. No full Axiom of Choice is used.

By [F3], $\lambda (E)=1$, $\lambda (I)=2$, and each compact $K\subset I$ has finite measure; hence [F2] makes $H$ measurable. For finite $p$, $\mathrm{\mid }H{\mathrm{\mid }}^p={\mathbf{1}}_E$, so [F6] gives ${\int }_I\mathrm{\mid }H{\mathrm{\mid }}^{p}dx=\lambda (E)=1$; for $p=\mathrm{\infty }$, $\mathrm{\mid }H\mathrm{\mid }\le 1$ gives a finite essential bound by [F5]. Also ${\int }_K\mathrm{\mid }H\mathrm{\mid }dx\le {\int }_{K}1dx=\lambda (K)<\mathrm{\infty }$ by [F4, F7, F8], so $H\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(I)$ by [F22] and its regular distribution is defined by [F11].

For $\phi \in \mathcal{D}(I)$, [F8, F9, F10, F11, F12] and step 1.1 give $⟨\mathrm{\partial }{T}_H,\phi ⟩=-{\int }_{I}H{\phi }^{\mathrm{\prime }}dx=-{\int }_I{\mathbf{1}}_{(0,1)}{\phi }^{\mathrm{\prime }}dx$. The indicator convention [F8] applies to nonnegative integrands; for signed or complex ${\phi }^{\mathrm{\prime }}$, apply it to the positive and negative parts of each real component and subtract, using the componentwise integral in [F9]. The endpoints $\{0,1\}$ are measurable and null by [F3]; [F21] gives zero integral of $\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$ on them, so the difference between the ${\mathbf{1}}_{(0,1)}{\phi }^{\mathrm{\prime }}$ and ${\mathbf{1}}_{[0,1]}{\phi }^{\mathrm{\prime }}$ integrals is zero by [F20]. Let $\tilde{\phi }$ be the smooth zero extension from [F10]; the interval FTC [F19] gives $-{\int }_{[0,1]}{\tilde{\phi }}^{\mathrm{\prime }}dx=-(\tilde{\phi }(1)-\tilde{\phi }(0))=\phi (0)={\delta }_0(\phi )$ by [F13]. Thus $\mathrm{\partial }{T}_H={\delta }_0$ in ${\mathcal{D}}^{\mathrm{\prime }}(I)$.

Suppose $v\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(I;\mathbb{K})$ were a weak derivative. Then [F14] and step 2.1 give ${\int }_{I}v\phi dx=\phi (0)$ for every test $\phi$. Choose one $\eta$ as in [F17]. For $0<ϵ<1\mathrm{/}2$, set ${\phi }_{ϵ}(x)=\eta (x\mathrm{/}ϵ)$; [F18] makes it smooth and its support is compactly contained in $(-ϵ,ϵ)\subset I$, so it is a test by [F10], with ${\phi }_{ϵ}(0)=1$ and $\mathrm{\mid }{\phi }_{ϵ}\mathrm{\mid }\le 1$. For $K=[-1\mathrm{/}2,1\mathrm{/}2]$, [F22] gives ${\int }_K\mathrm{\mid }v\mathrm{\mid }<\mathrm{\infty }$. The measurable sets $A_{ϵ}=[-ϵ,ϵ]\subset K$ have $\lambda (A_{ϵ})=2ϵ\to 0$ by [F3]; [F16] on the restricted measure space $K$ gives ${\int }_{A_{ϵ}}\mathrm{\mid }v\mathrm{\mid }dx\to 0$. But [F20], [F7], and the support and bound of ${\phi }_{ϵ}$ give $1=\mid {\int }_{I}v{\phi }_{ϵ}dx\mid \le {\int }_I\mathrm{\mid }v{\phi }_{ϵ}\mathrm{\mid }dx\le {\int }_{A_{ϵ}}\mathrm{\mid }v\mathrm{\mid }dx\to 0$, a contradiction. Hence no locally integrable function represents $\mathrm{\partial }{T}_H$.

By [F15], membership of $H$ in any $W^{1,p}(I;\mathbb{K})$ would require a locally integrable representative of its weak first derivative, which step 3.1 rules out for every $1\le p\le \mathrm{\infty }$, including both endpoints. The assumption is exactly Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$ by [F1]; it is used through the interval-measure and compact-measure facts [F3, F4] and the interval FTC [F19]. The regular-distribution injection is not used, and no full Axiom of Choice or sequence of selections occurs.