= Solution
Use the continuous representative of $\bar u$ furnished by the one-dimensional <Sobolev embedding theorem>, and define $u^*(x)=\bar u(|x|)$ on $A$. The <one-dimensional Sobolev representative> is absolutely continuous; <Holder inequality> applied to its <fundamental theorem of calculus> gives
$$
|\bar u(r)-\bar u(s)|
=\left|\int_s^r\bar u'(t)\,dt\right|
\leq |r-s|^{5/6}\|\bar u'\|_{L^6(1,2)}.
$$
Also $\|\bar u\|_\infty\leq\|\bar u\|_6+\|\bar u'\|_6$: select a point where $|\bar u|$ is no larger than its $L^6$ <norm> on the unit-length interval and apply the same <fundamental theorem of calculus>. Thus the one-dimensional <Hölder norm> is controlled by $\|\bar u\|_{W^{1,6}(1,2)}$.
The two <Hölder seminorms> must be compared using the correct distances. By the <reverse triangle inequality>,
$$
\frac{|u^*(x)-u^*(y)|}{|x-y|^{5/6}}
\leq[\bar u]_{C^{0,5/6}(1,2)}
\frac{\bigl||x|-|y|\bigr|^{5/6}}{|x-y|^{5/6}}
\leq[\bar u]_{C^{0,5/6}(1,2)}.
$$
Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two <Hölder seminorms>. Their <supremum norms> are equal as well. Part (i) now gives
$$
\boxed{\|u^*\|_{C^{0,5/6}(A)}
=\|\bar u\|_{C^{0,5/6}(1,2)}
\leq C\|\bar u\|_{W^{1,6}(1,2)}
\leq C'\|u\|_{W^{1,6}(A)}.}
$$
The representative agrees with $u$ <almost everywhere>, and the assignment is linear and injective on the radial <Sobolev space>. This proves the requested <continuous embedding>. The inner radius being positive makes the <radial profile> <Sobolev norm> comparable to its weighted radial version.
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