How exactly does the constant $C$ in the Sobolev inequality depend on the domain? The Sobolev inequality theorem -as stated here- says

Let $U$ be a bounded open subset of $\mathbb{R}^N$, with a $C^1$ boundary. Assume $u \in W^{k,p}(U)$. If $k<n/p$ then $u \in L^q(U)$, where $$ \frac{1}{q}=\frac{1}{p}-\frac{k}{n}. $$ We have in addition the estimate $$ \Vert u \Vert_{L^q(U)} \leq C \Vert u \Vert_{W^{k,p}(U)} $$ the constant C depending only on $k,p,n$ and $U$.


My question is: How does $C$ depend on $U$ exactly? Does it depend only on the measure of $U$? Is it independent of translation of $U$?

The reason I'm asking this is that I am able to estimate solutions to some PDE as $$ \Vert u \Vert_{L^q(B(x,1))} \leq C \Vert u \Vert_{W^{k,p}(B(x,1))} $$ for some $q,k,p$ and any $x \in \mathbb{R}^N$. However, I need a uniform bound, i.e. a bound not depending on the center $x$ of the balls $B(x,1)$. So if the Sobolev inequality depended only on the measure of $U$, I'd be fine.

Here is a similar question, but I really don't understand the answer.
 A: Let $T:\mathbb R^n\to\mathbb R^n$ be a rigid motion: translation, rotation, reflection, or composition of such things. Then $|D^k u\circ T|=|D^k u|\circ T$ for derivatives of all orders.  Using this and the change of variables formula (in which the absolutely value of Jacobian is $1$), you will find that the neither side of $\Vert u \Vert_{L^q(U)} \leq C \Vert u \Vert_{W^{k,p}(U)}$ is affected by the transformation $T$. Thus, the best Sobolev constant is preserved under rigid motions. 
The scaling is trickier, because it affects the derivatives of different orders differently. I  think the behavior of optimal $C$ in $\Vert u \Vert_{L^q(U)} \leq C \Vert u \Vert_{W^{k,p}(U)}$ under scaling can be pretty complicated. What is true is that one can give an upper bound on $C$ that depends (besides the indices $m,p,q$) only on the parameters of the cone condition satisfied by $U$. See Sobolev spaces by Adams for details. 
Scaling is more tractable for a related inequality $\Vert u \Vert_{L^q(U)} \leq C \Vert D^ku \Vert_{L^{p}(U)}$, which holds for $u\in W^{k,p}_0(U)$. Indeed, here replacing $u$ with $u(\lambda^{-1}  x)$ contributes the factor of $\lambda^{n/q}$ on the left and $\lambda^{-k+n/p}$ on the right. So, if the domain is scaled by $\lambda$, the constant $C$ gets multiplied by $\lambda^{k-n/p+n/q}$. Observe that $k-n/p+n/q\ge 0$ here, with equality for the borderline Sobolev exponent. 
That larger domains have larger constants for $W^{k,p}_0(U)$ inequalities is obvious, because $W^{k,p}_0(U)\subset W^{k,p}_0(V)$ when $U\subset V$. Such monotonicity does not hold for $W^{k,p}(U)$ inequality, which is sensitive to the shape of the domain in a subtle and mostly intractable way.
A: In addition to the answer by user90090, it can be told that in Sobolev spaces by Adams, as well as in its version by Adams–Fournier, no appropriate details are to be found. The details of dependence of the best Sobolev constant on parameters of the cone condition,  usually referred to by experts in the field as well-known, belong to what is now called Mathematical folklore, i.e, something like check-it-youself-if-you-need (http://en.wikipedia.org/wiki/Mathematical_folklore). Cited from Mathematical folklore, the details look like this. For a domain $U\subset\mathbb{R}^n$ of diameter $D$, star-shaped wrt some ball $B\subset U$ of diameter $d$, the best Sobolev constant depends only on the ratio $d/D$, which is established by scaling argument.  For a bounded domain $U\subset\mathbb{R}^n$, represented by the Gagliardo lemma as a union of $N\geqslant 2$ domains $U_i\,$, each star-shaped wrt some ball $B_i\subset U_i\,$, the best Sobolev constant depends only on $N$ and the ratios $d_i/D_i\,$. 
