$$\int_{\Omega} |\nabla u(x)|^2 dx < \infty$$
could be interpreted as the finite energy. Although the researchers use the embedding theorem everyday, sometimes when we would some more information from the constants, it's not easy, especially when we talk about the fractional order one $W^{s,p}(\Omega)$. The most direct way to illustrate its delicate usage may be a note. However, here we concentrate on the inequality of Soblev and Poincare type instead of the one of Holder type.
$W^{k,p}(\Omega), W_0^{k,p}(\Omega), W^{k,p}(\mathbb{R}^d)$
The definition of $W^{k,p}(\Omega)$ is just say that its all kind of weak derivative belongs to space $L^p(\Omega)$. However, since the function $C^{\infty}_0(\Omega)$ isn't always dense in this space, we have to define another guy as the closure under the norm of the former one. This case has no problem when we talk about the space $W^{k,p}(\mathbb{R}^d)$. The more tricky problem is that $W^{k,p}(\Omega)$ isn't the subspace of $W^{k,p}(\mathbb{R}^d)$ ! Luckily, we have one extension theorem to define $Ext(u)$ such that$$\Vert Ext(u) \Vert_{W^{k,p}(\mathbb{R}^d)} \leq C(d,p,k, \Omega)\Vert u \Vert_{W^{k,p}(\Omega)}$$
But this theorem will use something about the property of $\Omega$. So this gives a very intuitive remark that the estimates about $W_0^{k,p}(\Omega), W^{k,p}(\mathbb{R}^d)$ don't use the property of $\Omega$ while the estimate of $W^{k,p}(\Omega)$ will use.
$W^{k,p}(\mathbb{R^d})$ and inequality of Soblev-Poincare
For $W_0^{k,p}(\Omega), W^{k,p}(\mathbb{R}^d)$, two useful inequalities are the Soblev inequality that$$\Vert u\Vert_{L^{p*}(\mathbb{R^d})} \leq C(d,p,k) \Vert \nabla^{k}u\Vert_{L^{p}(\mathbb{R^d})}$$
where $p*$ is some critical exponent that $p* = \frac{dp}{d - pk}$. Another important inequality is the one of Poincare that
$$\Vert u\Vert_{L^{p}(\mathbb{R^d})} \leq C(d,p,k) d(\Omega) \Vert \nabla u\Vert_{L^{p}(\mathbb{R^d})}$$
while this time it will gives one time of factor of $d(\Omega)$. The proof starts from the dense subspace $C^{\infty}_0(\Omega)$ and use the expression of integral. So it also works for the type of inequality of $u - (u)_{\Omega}$.
The interest of Poincare is that when we define the weighted norm
$$\Vert u\Vert_{\underline{W}^{k,p}(\Omega)} = \sum_{\beta \leq k}|\Omega|^{\frac{|\beta| - |k|}{d}}\Vert \partial^{\beta} u \Vert_{L^p(\Omega)}$$
we could just consider the last term that
$$\Vert u\Vert_{\underline{W}^{k,p}(\Omega)} \simeq \Vert \nabla^{k}u \Vert_{\underline{L}^{p}(\Omega)}$$
However, when we use the Soblev inequality, we get some supplementary power of the size.
$W^{s,p}(\Omega), W_0^{s,p}(\Omega), W^{s,p}(\mathbb{R}^d)$
Secondary, we think about another norm that$$[u]^p_{W^{s,p}(\Omega)} = \int\int_{\Omega \times \Omega} \frac{|u(x) - u(y)|^p}{|x-y|^{d + sp}}dx dy$$
We draw an analogue when we define it in different context. However, this one isn't a norm since we observe that after a difference of constant, this semi-norm doesn't change. So when we talk about its norm, we have to add the one of $L^p$. One remark is that this semi-norm only take in account of the small size fluctuation. Two trivial inequalities are that
$$0 < s \leq s' < 1, [u]_{W^{s,p}(\Omega)} \leq (d(\Omega))^{s'-s}[u]_{W^{s',p}(\Omega)}$$
$$0 < s < 1, [u]_{W^{s,p}(\Omega)} \leq C(d,p,s)(d(\Omega))^{1-s}\Vert \nabla u \Vert_{L^{p}(\Omega)}$$
They are correct for $u \in W_0^{s,p}(\Omega), W^{s,p}(\mathbb{R}^d)$, otherwise we have to add the constant of $C(\Omega)$.These two inequality share the same spirit of Poincare inequality that we use the diameter to replace the derivative. However, to get better description about the weighted norm, we have to think about other type of inequality. After all, we see that transform between the integer order and fractional order is always a little messy, although that we believe that they share the similar property. (As how we define them properly.)
$W^{s,p}(\mathbb{R}^d) \hookrightarrow L^{p*}(\mathbb{R}^d)$
This result looks like the one of Soblev inequality : for $p* = \frac{dp}{d - ps}$ and $u \in W^{s,p}(\mathbb{R}^d)$, we have$$\Vert u\Vert_{L^{p*}(\mathbb{R^d})} \leq C(d,p,k) [u]_{W^{s,p}(\mathbb{R^d})}$$
however, its proof is very tricky which uses many technique like decomposition dyadic. Once it's correct, we could also well take the highest derivative as the main part of the weighted norm that
$$\Vert u\Vert_{\underline{W}^{s,p}(\Omega)} \simeq [u]_{\underline{W}^{s,p}(\Omega)}$$
$W^{1,p}(\mathbb{R}^d) \hookrightarrow W^{s,p*}(\mathbb{R}^d)$
The final theorem, where we neglect the context since it's always same, is that for $0 < s < 1, p* = \frac{dp}{d + (s-1)p} \Leftrightarrow s - \frac{d}{p*} = 1 - \frac{d}{p}$, we have that
$$ [u]_{W^{s,p*}(\mathbb{R^d})} \leq C(d,p,s)\Vert \nabla u\Vert_{L^{p}(\mathbb{R^d})}$$
To prove it, one important tool is Hardy-Littlewood-Soblev inequality. So this completes the relation between fractional and integer order Soblev space.
Conclusion
We got a lot of inequality. How to remember it ? Ok, we only have to take one ingredient in heart. The weighted norm eliminate the constant in the Poincare and Holder inequality, so we could use the highest derivative and any $p$ that we like. However, if we use the weighted norm of different order, the effect of scale always exists.
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