Point evaluation is unbounded below the Sobolev continuity threshold

The declared choice principle is exactly Countable Choice. [F1, given] Its uses are confined to the supplier hypotheses listed in the Choice use note [F2, F3, F10, F12, F19, F20]; the integer cutoffs below use the Archimedean property and the displayed limit, not additional choices.

Openness supplies a positive closed ball, and the common cutoff bound is finite. [F5, F7, F9, construct] Choose $R>0$ so that ${\bar{B}}_R(0)\subseteq \mathrm{\Omega }$, after taking an open ball about $0$ and reducing its radius. Take $\varphi$ from [F5] and let $M=1+{\sup }_{{\bar{B}}_1(0)}\mathrm{\mid }\varphi \mathrm{\mid }+{\max }_{1\le j\le n}{\sup }_{{\bar{B}}_1(0)}\mathrm{\mid }{\mathrm{\partial }}_j\varphi \mathrm{\mid }+{\sup }_{0\le t\le 1}\mathrm{\mid }{\theta }^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{.}$ This is finite by [F9]. Also $\varphi (0)=1$ and $\mathrm{supp}\varphi \subset B_{3\mathrm{/}4}(0)$.

For $1\le p<n$, the scaled smooth bumps have supports shrinking to zero and values diverging there. [F4, F5, F8, F15, F17, F18, step 1.2, construct] By [F18] choose an integer $q\ge 1$ with $q(n-p)>p$, then choose an integer $k_0>\max (1,R^{-1})$. For each $m\ge 1$, put $k=m+k_0$ and $u_m(x)=k\varphi (k^{q}x)\mathrm{.}$ Since $k\ge k_0>1$ and $q\ge 1$, $k^{-q}\le k^{-1}<R$. The support of $u_m$, and of each of its first derivatives, lies in $B_{k^{-q}}(0)\subset B_R(0)$; thus [F4] gives $u_m\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})\subset C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$. The chain rule gives ${\mathrm{\partial }}_j{u}_m(x)=k^{q+1}({\mathrm{\partial }}_j\varphi )(k^{q}x),u_m(0)=k⟶\mathrm{\infty }\mathrm{.}$ The derivative support assertion follows because a smooth function is zero with all derivatives on the open complement of its support.

If $p=n$, the logarithmic radial cutoffs are smooth, supported, diverge at zero, and have the displayed gradient bound. [F4, F6, F8, F13, F14, F15, F16, F22, step 1.2, construct] By [F22], $h^n\exp (-n{h}^2)\to 0$ as integers $h\to \mathrm{\infty }$: apply its limit assertion with $x=h^2$, polynomial degree $n$, and $a=n$, and use $h^n\le h^{2n}$ for $h\ge 1$ and [F16]. Choose an integer $h_0\ge 1$ so that $h^n\exp (-n{h}^2)\le 1$ for $h\ge h_0$. For each $m\ge 1$, set $h=m+h_0$, $L=h^2$, and $r_h=R\exp (-L)$. Since $L>0$, [F16] gives $0<\exp (-L)<\exp (0)=1$, so $0<r_h<R$. With $Q(x)={\sum }_{j=1}^n{x}_j^2=\mathrm{\mid }x{\mathrm{\mid }}^2$ by [F6], define $w_h(x)=\{\begin{array}{ll}1,& Q(x)\le r_h^2,\\ {\displaystyle \theta \text{\hspace{0.17em}⁣}\left(\frac{\log (R^2\mathrm{/}Q(x))}{2L}\right),}& r_h^2<Q(x)<R^2,\\ 0,& Q(x)\ge R^2,\end{array}{u}_m(x)=h{w}_h(x)\mathrm{.}$ For $r_h<\mathrm{\mid }x\mathrm{\mid }<R$, the scalar argument equals $t=\log (R\mathrm{/}\mathrm{\mid }x\mathrm{\mid })\mathrm{/}L\in (0,1)$, by [F16]. The polynomial $Q$ and the composition with $\log Q$ are smooth where $Q>0$, by [F15]; $w_h$ is constant near $0$. Since $\theta$ is smooth and constant on both half-lines beyond $[0,1]$, the pieces join smoothly at both radii. Its support lies in ${\bar{B}}_R(0)\subseteq \mathrm{\Omega }$, so [F4] gives $u_m\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$, and $u_m(0)=h\to \mathrm{\infty }$. On the annulus, differentiation gives $\mathrm{\mid }{\mathrm{\partial }}_j{u}_m(x)\mathrm{\mid }=\frac{h\mathrm{\mid }{\theta }^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{\mid }{x}_j\mathrm{\mid }}{L\mathrm{\mid }x{\mathrm{\mid }}^2}\le \frac{Mh}{L\mathrm{\mid }x\mathrm{\mid }}\mathrm{.}$ Outside the annulus the derivatives vanish, including at the joining radii because the step is smooth and constant on the adjacent half-lines.

The subcritical construction has a uniform $W^{1,p}$ bound. [F2, F3, F5, F10, F11, F12, F17, step 1.2, step 2.1] The ball $B_{k^{-q}}(0)$ lies in a cube of measure $(2k^{-q}{)}^n$, by [F10]. The pointwise bounds from [F5] and step 1.2, measurability from [F12], and integral monotonicity and homogeneity [F11] give ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{u}_m{\mathrm{\mid }}^{p}dx\le 2^n{k}^{p-qn},{\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}_j{u}_m{\mathrm{\mid }}^{p}dx\le 2^n{M}^p{k}^{p(q+1)-qn}=2^n{M}^p{k}^{p-q(n-p)}\mathrm{.}$ Both exponents are negative: $p-q(n-p)<0$ by the choice of $q$, and $p-qn<0$ follows since $qn>q(n-p)>p$. Thus [F17] makes both quantities uniformly bounded for $k\ge 1$. Since the classical derivatives are weak derivatives by [F3], [F2] now gives ${\sup }_m\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_{W^{1,p}}<\mathrm{\infty }$.

Polar integration gives a uniform $L^n$ bound for each critical gradient term. [F16, F19, F20, step 2.2] For each $j$, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}_j{u}_m{\mathrm{\mid }}^{n}dx\le \sigma (S^{n-1}){\left(\frac{Mh}{L}\right)}^n{\int }_{r_h}^R\frac{dr}{r}\mathrm{.}$ Since $\log (R\mathrm{/}{r}_h)=L$, [F16] and the fundamental theorem in [F20] give ${\int }_{r_h}^{R}dr\mathrm{/}r=L$. Therefore ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}_j{u}_m{\mathrm{\mid }}^{n}dx\le M^n\sigma (S^{n-1})\frac{h^n}{L^{n-1}}=M^n\sigma (S^{n-1})h^{2-n},$ which is uniformly bounded because $n\ge 2$ and $h\ge 1$.

The core and annulus estimates give a uniform critical $L^n$ bound for the function term. [F10, F11, F13, F14, F16, F17, F19, F20, F21, step 1.2, step 2.2] On the core ${\bar{B}}_{r_h}(0)$, [F10] bounds the measure by $(2r_h{)}^n$; hence ${\int }_{{\bar{B}}_{r_h}}\mathrm{\mid }{u}_m{\mathrm{\mid }}^{n}dx\le (2R{)}^n{h}^n\exp (-n{h}^2)\le (2R{)}^n$ by the choice of $h_0$. For the annulus, the mean value theorem and the derivative bound in step 1.2 imply $0\le \theta (t)\le Mt$ for $0\le t\le 1$. The substitution $r=R\exp (-s)$ is decreasing: its oriented endpoints are $L$ and $0$, and reversing them gives the positive integral on $[0,L]$ with Jacobian $R\exp (-s)$. Thus [F20] gives $\begin{array}{rl}{\displaystyle {\int }_{r_h}^R{\mid \theta \text{\hspace{0.17em}⁣}\left(\frac{\log (R\mathrm{/}r)}{L}\right)\mid }^n{r}^{n-1}dr}& {\displaystyle \le \frac{M^n{R}^n}{L^n}{\int }_0^L{s}^n\exp (-ns)ds}\\ {\displaystyle }& {\displaystyle \le \frac{M^n{R}^{n}n!}{L^n}{\int }_0^L\exp (-s)ds}\\ {\displaystyle }& {\displaystyle \le \frac{M^n{R}^{n}n!}{L^n}\mathrm{.}}\end{array}$ Here [F21] gives $s^n\le n!\exp (s)$; since $n\ge 2$, monotonicity of the exponential in [F16] gives $\exp (-(n-1)s)\le \exp (-s)$, and [F20] evaluates ${\int }_0^L\exp (-s)ds=1-\exp (-L)\le 1$. Applying [F19] to this radial annulus integral and multiplying by $h^n$ yields ${\int }_{B_R\setminus {\bar{B}}_{r_h}}\mathrm{\mid }{u}_m{\mathrm{\mid }}^{n}dx\le \sigma (S^{n-1})M^n{R}^{n}n!\frac{h^n}{L^n}=\sigma (S^{n-1})M^n{R}^{n}n!h^{-n},$ also uniformly bounded.

These estimates bound the critical $W^{1,n}$ norm while the origin values diverge. [F2, F3, step 3.2, step 3.3] Steps 3.2 and 3.3 bound the function term and each of the $n$ first derivative terms in [F2] uniformly in $L^n$. By [F3], the classical derivatives are the weak derivatives, so ${\sup }_m\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_{W^{1,n}}<\mathrm{\infty }$, while $u_m(0)=h\to \mathrm{\infty }$.

Any bounded extension contradicts divergence of the corresponding test sequence at zero. [F23, step 3.1, step 4.1, assume-contra, discharge-contradiction] Suppose a bounded linear functional $T:W^{1,p}(\mathrm{\Omega };\mathbb{K})\to \mathbb{K}$ agreed with ordinary point evaluation on all test functions. Boundedness would give a constant $C$ such that, for the corresponding sequence, $\mathrm{\mid }{u}_m(0)\mathrm{\mid }=\mathrm{\mid }T([u_m])\mathrm{\mid }\le C\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_{W^{1,p}}\mathrm{.}$ The right side is uniformly bounded by step 3.1 or step 4.1, while the left side tends to infinity. This contradiction proves unboundedness and rules out the asserted bounded extension. ∎