= Solution
{c}
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>, gives
$$
|v(x)|\leq\|v\|_{L^p(\mathbb R^n)}+C\|Dv\|_{L^p(\mathbb R^n)}
$$
for every smooth $v$. Hence $u_j$ is a <Cauchy sequence> in the supremum norm and has a bounded continuous limit $u^*$ by <uniform convergence>. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^*$ <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, gives
$$
|u^*(y)-u^*(x)|\leq C|y-x|^\alpha\|Du\|_{L^p(\mathbb R^n)}.
$$
Thus $u^*$ is a <Hölder continuous function> and
$$
\boxed{u^*\in C^{0,1-n/p}(\mathbb R^n),\qquad
\|u^*\|_\infty+[u^*]_{C^{0,1-n/p}}\leq C\|u\|_{W^{1,p}}.}
$$
It is the unique continuous representative: two continuous functions agreeing <almost everywhere> agree everywhere.
For $1\leq p<n$, take a <smooth cutoff function> $\chi$ equal to one near zero and set
$$
\boxed{u(x)=\chi(x)|x|^{-a},\qquad 0<a<\frac np-1.}
$$
Its <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.
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