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.
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.