Morrey's inequality for $p>n$

Oscillation on a ball. Fix a ball $B(x,\rho )\subseteq \mathrm{\Omega }$. By [F1], for almost every $z\in B(x,\rho )$, $\mathrm{\mid }u(z)-u_{B(x,\rho )}\mathrm{\mid }\le C_0(n){\int }_{B(x,\rho )}\mathrm{\mid }Du(y)\mathrm{\mid }\mathrm{\mid }z-y{\mathrm{\mid }}^{1-n}dy$. Holder [F2] with exponents $p$ and $p^{\mathrm{\prime }}$ bounds this by $C_0(n)\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,\rho ))}({\int }_{B(z,2\rho )}\mathrm{\mid }z-y{\mathrm{\mid }}^{(1-n)p^{\mathrm{\prime }}}dy{)}^{1\mathrm{/}{p}^{\mathrm{\prime }}}$, where the integration region has been enlarged from $B(x,\rho )\subseteq B(z,2\rho )$ to $B(z,2\rho )$. By [F2] the kernel integral equals $c(n,p){\rho }^{n-(n-1)p^{\mathrm{\prime }}}$ with $n-(n-1)p^{\mathrm{\prime }}>0$ and exponent $(n-(n-1)p^{\mathrm{\prime }})\mathrm{/}{p}^{\mathrm{\prime }}=1-n\mathrm{/}p$, so $\mathrm{\mid }u(z)-u_{B(x,\rho )}\mathrm{\mid }\le C_1(n,p){\rho }^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,\rho ))}$ for almost every $z\in B(x,\rho )$.

Convergence of the ball means and the representative. Fix a ball $B(x,r)$ with $B(x,2r)⋐\mathrm{\Omega }$ and let $0<\rho \le \sigma \le r$. For almost every $z\in B(x,\rho )$ step 1.1 applied on $B(x,\sigma )$ gives $\mathrm{\mid }u(z)-u_{B(x,\sigma )}\mathrm{\mid }\le C_1{\sigma }^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,\sigma ))}\le C_1{\sigma }^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,r))}$; averaging in $z$ over $B(x,\rho )$ yields $\mathrm{\mid }{u}_{B(x,\rho )}-u_{B(x,\sigma )}\mathrm{\mid }\le C_1{\sigma }^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,r))}$. Hence the net $(u_{B(x,\rho )}{)}_{\rho \downarrow 0}$ is Cauchy and we may define $u^{\ast }(x):={\lim }_{\rho \downarrow 0}{u}_{B(x,\rho )}$ for every $x$ such that some $B(x,2r)⋐\mathrm{\Omega }$, with $\mathrm{\mid }{u}^{\ast }(x)-u_{B(x,\rho )}\mathrm{\mid }\le C_1{\rho }^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,r))}$ for $0<\rho \le r$. The zero extension of $u$ to ${\mathbb{R}}^n$ lies in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$ by Holder [F2] on bounded balls, since $u\in L^p(\mathrm{\Omega })$; its sufficiently small ball means at each interior point are those of $u$. Thus [F3] and [F4] give $u_{B(x,\rho )}\to u(x)$ for almost every $x$, so $u^{\ast }=u$ almost everywhere: $u^{\ast }$ is a representative of the class. The Countable-Choice interface used here is supplied by [F5].

The Holder bound and continuity. Let $z,y\in B(x,r)$ and put $\mathrm{\ell }:=\mathrm{\mid }z-y\mathrm{\mid }$; the case $\mathrm{\ell }=0$ is trivial. If $\mathrm{\ell }\le r\mathrm{/}2$, apply step 1.1 on the balls $B(z,\mathrm{\ell })$ and $B(y,\mathrm{\ell })$ and step 2.1 on their means: $\mathrm{\mid }{u}^{\ast }(z)-u_{B(z,\mathrm{\ell })}\mathrm{\mid }\le C_1{\mathrm{\ell }}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(z,\mathrm{\ell }))}$ and similarly at $y$; both balls lie in $B(x,r+\mathrm{\ell })\subseteq B(x,2r)$ when $\mathrm{\ell }\le r\mathrm{/}2$, so the two norms are at most $\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$. For the difference of the two means, both balls $B(z,\mathrm{\ell })$ and $B(y,\mathrm{\ell })$ lie in $B(z,2\mathrm{\ell })\subseteq B(x,r+2\mathrm{\ell })\subseteq B(x,2r)$, and step 1.1 on $B(z,2\mathrm{\ell })$ bounds $\mathrm{\mid }u(a)-u(b)\mathrm{\mid }\le 2C_1(2\mathrm{\ell }{)}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$ for almost every $a\in B(z,\mathrm{\ell })$, $b\in B(y,\mathrm{\ell })$; averaging gives $\mathrm{\mid }{u}_{B(z,\mathrm{\ell })}-u_{B(y,\mathrm{\ell })}\mathrm{\mid }\le 2C_1(2\mathrm{\ell }{)}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$. Combining the three terms, $\mathrm{\mid }{u}^{\ast }(z)-u^{\ast }(y)\mathrm{\mid }\le C_2(n,p){\mathrm{\ell }}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$. If $\mathrm{\ell }>r\mathrm{/}2$, use the mean over the fixed ball $B(x,2r)$: by step 1.1 on $B(x,2r)$, for almost every $a\in B(x,2r)$ one has $\mathrm{\mid }u(a)-u_{B(x,2r)}\mathrm{\mid }\le C_1(2r{)}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$, and since $B(z,\rho )\subseteq B(x,2r)$ for $0<\rho <r$, averaging over $B(z,\rho )$ and letting $\rho \downarrow 0$ gives $\mathrm{\mid }{u}^{\ast }(z)-u_{B(x,2r)}\mathrm{\mid }\le C_1(2r{)}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$, with the same bound at $y$; hence $\mathrm{\mid }{u}^{\ast }(z)-u^{\ast }(y)\mathrm{\mid }\le 2C_1(2r{)}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}\le C_2^{\mathrm{\prime }}{\mathrm{\ell }}^{1-n\mathrm{/}p}\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(B(x,2r))}$ because $(2r{)}^{1-n\mathrm{/}p}\le 4^{1-n\mathrm{/}p}{\mathrm{\ell }}^{1-n\mathrm{/}p}$ when $\mathrm{\ell }>r\mathrm{/}2$. This proves the displayed estimate with a constant depending only on $n$ and $p$; since the exponent $1-n\mathrm{/}p$ is positive, $u^{\ast }$ is continuous on every ball $B(x,r)$ with $B(x,2r)⋐\mathrm{\Omega }$, hence on all of $\mathrm{\Omega }$, and the equivalent Holder-norm statement follows from [F4].