Morrey's inequality has no $p=n$ endpoint

Membership in $W^{1,n}$. Off the origin $f$ is smooth and radial with $\mathrm{\mid }Df(x)\mathrm{\mid }=\frac{1}{\mathrm{\mid }x\mathrm{\mid }(\mathrm{\mid }x\mathrm{\mid }+1)\log (1+1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })}\mathrm{.}$ For $0<r=\mathrm{\mid }x\mathrm{\mid }\le \frac{1}{2}$, this is at most $1\mathrm{/}(r\log (1\mathrm{/}r))$, so polar coordinates [F1] give ${\int }_{B(0,1\mathrm{/}2)}\mathrm{\mid }Df{\mathrm{\mid }}^{n}dx\le \sigma (S^{n-1}){\int }_0^{1\mathrm{/}2}\frac{dr}{r(\log (1\mathrm{/}r){)}^n}=\sigma (S^{n-1}){\int }_{\log 2}^{\mathrm{\infty }}{s}^{-n}ds<\mathrm{\infty },$ where $s=\log (1\mathrm{/}r)$ and $n\ge 2$. On $1\mathrm{/}2\le r<1$, the exact derivative is bounded by $4\mathrm{/}(3\log 2)$, since $r\ge 1\mathrm{/}2$, $r+1\ge 3\mathrm{/}2$, and $\log (1+1\mathrm{/}r)\ge \log 2$; hence its $n$th-power polar integral on this annulus is finite as well. Also ${\int }_B\mathrm{\mid }f{\mathrm{\mid }}^{n}dx<\mathrm{\infty }$: near $0$, $\log \log (1+1\mathrm{/}r)=O(r^{-1\mathrm{/}2})$, so the radial $L^n$ majorant is $O(r^{n\mathrm{/}2-1})$, which is integrable for $n\ge 2$, and $f$ is bounded on $1\mathrm{/}2\le r<1$. To identify the weak gradient, fix a test $\phi \in C_c^{\mathrm{\infty }}(B)$ and a coordinate $i$. For every transverse coordinate other than the zero vector, the coordinate line avoids the origin; its intersection with $B$ is an interval on which $f$ is smooth, and $\phi$ vanishes near the endpoints. The fundamental theorem applied to $f\phi$ gives $\int f{\mathrm{\partial }}_i\phi =-\int ({\mathrm{\partial }}_{i}f)\phi$ on that line. The omitted transverse singleton is null because $n-1\ge 1$; both integrands are globally integrable since $f,Df\in L^n(B)\subseteq L^1(B)$. Fubini [F5] therefore integrates the section identities into the weak derivative identity. Thus $Df$ is the weak gradient and $f\in W^{1,n}(B)$ by [F2].

No bounded representative. For every $M>0$ the set $\{x\in B\setminus \{0\}:f(x)>M\}$ is a punctured neighbourhood of $0$ (since $f(x)\to +\mathrm{\infty }$ as $x\to 0$) and has positive measure; hence if a measurable $g$ equals $f$ almost everywhere on $B$, then for every $\epsilon >0$ the set $\{x\in B(0,\epsilon ):g(x)>M\}$ has positive measure, so $g$ is unbounded on every neighbourhood of $0$. In particular no representative of $f$ is bounded, and a fortiori none is uniformly continuous or Holder on $B$.

Failure of every Holder bound. Suppose there were $\alpha >0$ and $C$ with $[u{]}_{C^{0,\alpha }(B)}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,n}(B)}$ for every $u\in W^{1,n}(B)$. Applying this to the class of $f$ would produce a representative $f^{\ast }$ with $\mathrm{\mid }{f}^{\ast }(x)-f^{\ast }(y)\mathrm{\mid }\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{W^{1,n}}\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }$ for all $x,y\in B$; such an $f^{\ast }$ is continuous and bounded on the bounded set $B$. But $f^{\ast }$ is a representative of $f$, contradicting step 1.2. Hence no such constants exist, Morrey's inequality has no $p=n$ endpoint, and the hypothesis $p>n$ in Morrey's inequality for $p>n$ is essential.