The trace estimate fails on an outward cusp above the critical sharpness

The inside norm. The function $u_{\delta }$ is the restriction to ${\mathrm{\Omega }}_{\alpha }$ of the $C^{\mathrm{\infty }}({\mathbb{R}}^2)$ function $(x,y)\mapsto \theta (y\mathrm{/}\delta )$, whose classical derivatives are its weak derivatives on ${\mathrm{\Omega }}_{\alpha }$ by [F1], with ${\mathrm{\partial }}_x{u}_{\delta }=0$ and ${\mathrm{\partial }}_y{u}_{\delta }={\delta }^{-1}{\theta }^{\mathrm{\prime }}(y\mathrm{/}\delta )$ by [F1]. Its support in ${\mathrm{\Omega }}_{\alpha }$ lies in the strip $0<y<2\delta$, whose area is ${\int }_0^{2\delta }2y^{\alpha }dy=2^{\alpha +2}{\delta }^{\alpha +1}\mathrm{/}(\alpha +1)$ by [F2]. Since $\mathrm{\mid }{u}_{\delta }\mathrm{\mid }\le 1$ and $\mathrm{\mid }{\mathrm{\partial }}_y{u}_{\delta }\mathrm{\mid }\le {\delta }^{-1}\mathrm{\parallel }{\theta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, this gives $\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{L^p}^p\le C{\delta }^{\alpha +1}$ and $\mathrm{\parallel }{\mathrm{\partial }}_y{u}_{\delta }{\mathrm{\parallel }}_{L^p}^p\le C{\delta }^{\alpha +1-p}$, hence $u_{\delta }\in W^{1,p}({\mathrm{\Omega }}_{\alpha })$ and $\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{W^{1,p}({\mathrm{\Omega }}_{\alpha })}^p\le C^{\mathrm{\prime }}{\delta }^{\alpha +1-p}$ for $0<\delta <\frac{1}{2}$ and a constant independent of $\delta$.

The boundary mass. Fix $0<\epsilon <\delta$. On each compact regular subarc $\{(\pm y^{\alpha },y):\epsilon \le y\le \delta \}$ one has $u_{\delta }\equiv 1$. The chart formula [F3] applies away from the cusp and gives length ${\int }_{\epsilon }^{\delta }\sqrt{1+{\alpha }^2{y}^{2\alpha -2}}dy\ge \delta -\epsilon$. Positivity of the boundary integral therefore gives its $p$-th power at least $2(\delta -\epsilon )$ for every $\epsilon >0$; letting $\epsilon \downarrow 0$ yields $2\delta$, without assigning a regular hypersurface chart at the cusp itself. Hence $\mathrm{\parallel }{u}_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}{\mathrm{\parallel }}_{L^p(\mathrm{\partial }{\mathrm{\Omega }}_{\alpha })}^p\ge 2\delta$; the top segment contributes nothing because $u_{\delta }$ vanishes there for $\delta <\frac{1}{2}$.

Matching bounds. On $0<y<2\delta <1$, the arclength density is at most $\sqrt{1+{\alpha }^2}$, and $\mathrm{\mid }{u}_{\delta }\mathrm{\mid }\le 1$; outside these two arc portions the restriction vanishes. Thus $\mathrm{\parallel }{u}_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}{\mathrm{\parallel }}_{L^p}^p\le 4\sqrt{1+{\alpha }^2}\delta$. Since $\theta (1)=1$ and $\theta (2)=0$, its derivative is nonzero at some point of $(1,2)$; continuity supplies $1<a<b<2$ and $c_0>0$ with $\mathrm{\mid }{\theta }^{\mathrm{\prime }}(s)\mathrm{\mid }\ge c_0$ on $[a,b]$. Integrating over the strip $a\delta \le y\le b\delta$ gives $\mathrm{\parallel }{\mathrm{\partial }}_y{u}_{\delta }{\mathrm{\parallel }}_{L^p}^p\ge 2c_0^p{\delta }^{-p}{\int }_{a\delta }^{b\delta }{y}^{\alpha }dy=c_1{\delta }^{\alpha +1-p}$, where $c_1=2c_0^p(b^{\alpha +1}-a^{\alpha +1})\mathrm{/}(\alpha +1)>0$. Together with steps 1.1 and 1.2 these estimates give $c{\delta }^{(p-\alpha )\mathrm{/}p}\le \mathrm{\parallel }{u}_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}{\mathrm{\parallel }}_{L^p}\mathrm{/}\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{W^{1,p}}\le C{\delta }^{(p-\alpha )\mathrm{/}p}$ for positive constants independent of $\delta$.

The ratio diverges and no bounded extension exists. Combining steps 1.1 and 1.2, the ratio of norms satisfies $\mathrm{\parallel }{u}_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}{\mathrm{\parallel }}_{L^p}\mathrm{/}\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{W^{1,p}}\ge c{\delta }^{1\mathrm{/}p}{\delta }^{-(\alpha +1-p)\mathrm{/}p}=c{\delta }^{(p-\alpha )\mathrm{/}p}$, which diverges as $\delta \downarrow 0$ because $\alpha >p$. If a bounded linear $S$ agreeing with classical restriction on $C^1({\bar{\mathrm{\Omega }}}_{\alpha })$ existed, then $S{u}_{\delta }=u_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}$ for each $\delta$ and [F4] would give the uniform bound $\mathrm{\parallel }{u}_{\delta }{\mathrm{\mid }}_{\mathrm{\partial }{\mathrm{\Omega }}_{\alpha }}\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{W^{1,p}}$ for all $\delta$, contradicting the divergence. Therefore no such operator exists, and the uniform-graph hypothesis of The $L^p$ trace operator on a bounded $C^1$ domain cannot be replaced by mere $C^1$ regularity of the boundary arcs.