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For $p>n$, the Hölder continuous function representative $u^{\ast }$ of $u\in W^{1,p}({\mathbb{R}}^n)$ has a classical Frechet derivative at almost every point, equal to its weak derivative. At a point $x$ where the $p$-mean oscillation of $Du$ tends to zero, apply the Morrey inequality on a cube to $u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. On a cube of side comparable to $|y-x|$, its oscillation is at mostThe required points have full measure by the Lebesgue differentiation theorem.

Put $Q=Q_r(x)$ and $v_Q=|Q{|}^{-1}{\int }_{Q}v$. The fundamental theorem of calculus along a line segment, followed by averaging, givesFor fixed $t$, use the change of variables formula $w=y+t(z-y)$. Its Jacobian determinant is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a convex set. Moreover, $w\in Q_t$ implies $|w-y|\le \sqrt{n}rt$. Thus Tonelli theorem givesThe diagonal $w=y$ is a null set. The Holder inequality with $p^{\prime }=p/(p-1)$ now applies: $p>n$ implies $(n-1)p^{\prime }<n$, and integration in spherical coordinates bounds the kernel byThis proves the Morrey inequality on a cube, uniformly even when $y$ approaches its boundary:Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain

Let $u_j\in C_c^{\infty }({\mathbb{R}}^n)$ tend to $u$ in the Sobolev space $W^{1,p}$, using density of smooth functions in a Sobolev space. Write $\alpha =1-n/p>0$. Applying the Morrey inequality on a cube to a unit cube centered at $x$, and bounding its average by the Holder inequality, givesfor every smooth $v$. Hence $u_j$ is a Cauchy sequence in the supremum norm and has a bounded continuous limit $u^{\ast }$ by uniform convergence. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^{\ast }$ almost everywhere.

For $x\ne y$, choose $r=4|x-y|$, so that $y\in Q_r(x)$. The second estimate from part (a), followed by passage to the limit, givesThus $u^{\ast }$ is a Hölder continuous function andIt is the unique continuous representative: two continuous functions agreeing almost everywhere agree everywhere.

For $1\le p<n$, take a smooth cutoff function $\chi$ equal to one near zero and setIts weak derivative has size $O(|x{|}^{-a-1})$ near zero, whose $p$th power is integrable because $p(a+1)<n$. Thus $u\in W^{1,p}$, but it is essentially unbounded in every neighborhood of zero and has no continuous representative. The value assigned at zero is irrelevant.