I was looking at Nicolaescu's Lectures on the Geometry of Manifolds (3rd edition). In Theorem 10.2.29 he presents (without proof) the following inequality:

For $m \geq 1, p \geq 1, 0 < r \leq R$ there exists a $C > 0$ s.t. for every $j < m, \epsilon >0$ and $ u \in W^{m, p}(\mathbb{R}^n)$ supported on $B_R$: \begin{equation} \| u \|_{j, p, \mathbb{R}^n} \leq C \left( \epsilon \|u\|_{m, p, \mathbb{R}^n} + \epsilon^{-j/(m-j)} \|u\|_{0, p, B_r} \right) \end{equation} (Actually the book says "$\epsilon^{-j(m-j)}$" in the second summand but believe this is a mistake. There is also some talk about picking cutoff functions but I don't think that's relevant to the main point.)

The author references Adams and Fournier's book on Sobolev spaces for a proof. Indeed Theorem 5.12 of that book is a very similar result: For a sufficiently nice open set $\Omega \subseteq \mathbb{R}^n$ it gives a bound \begin{equation} \| u \|_{j, p, \Omega} \leq C \left( \epsilon \| u\|_{m, p, \Omega} + \epsilon^{-j/(m-j)} \|u\|_{0, p, \Omega_\epsilon}\right) \end{equation}

for $u \in W^{m, p}(\Omega)$. However here $\Omega_\epsilon$ is only some mystery subset of $\Omega$ whose closure lies in $\Omega$ and which depends on $\epsilon$.

My question is how/if you can get the first inequality from the second given that the second one offers no control on $\Omega_\epsilon$.

I was looking at Nicolaescu's Lectures on the Geometry of Manifolds (3rd edition). In Theorem 10.2.29 he presents (without proof) the following inequality:

For $m \geq 1, p \geq 1, 0 < r \leq R$ there exists a $C > 0$ s.t. for every $j < m, \epsilon >0$ and $ u \in W^{m, p}(\mathbb{R}^n)$ supported on $B_R$: \begin{equation} \| u \|_{j, p, \mathbb{R}^n} \leq C \left( \epsilon \|u\|_{m, p, \mathbb{R}^n} + \epsilon^{-j/(m-j)} \|u\|_{0, p, B_r} \right) \end{equation} Note that the last norm is taken over $B_r$, not $B_R$. (Actually the book says "$\epsilon^{-j(m-j)}$" in the second summand but believe this is a mistake. There is also some talk about picking cutoff functions but I don't think that's relevant to the main point.)

The author references Adams and Fournier's book on Sobolev spaces for a proof. Indeed Theorem 5.12 of that book is a very similar result: For a sufficiently nice open set $\Omega \subseteq \mathbb{R}^n$ it gives a bound \begin{equation} \| u \|_{j, p, \Omega} \leq C \left( \epsilon \| u\|_{m, p, \Omega} + \epsilon^{-j/(m-j)} \|u\|_{0, p, \Omega_\epsilon}\right) \end{equation}

for $u \in W^{m, p}(\Omega)$. However here $\Omega_\epsilon$ is only some mystery subset of $\Omega$ whose closure lies in $\Omega$ and which depends on $\epsilon$.

My question is how/if you can get the first inequality from the second given that the second one offers no control on $\Omega_\epsilon$.

I was looking at Nicolaescu's Lectures on the Geometry of Manifolds (3rd edition). In Theorem 10.2.29 he presents (without proof) the following inequality:

For $m \geq 1, p \geq 1, 0 < r \leq R$ there exists a $C > 0$ s.t. for every $j < m, \epsilon >0$ and $ u \in W^{m, p}(\mathbb{R}^n)$ supported on $B_R$: \begin{equation} \| u \|_{j, p, \mathbb{R}^n} \leq C \left( \epsilon \|u\|_{m, p, \mathbb{R}^n} + \epsilon^{-j/(m-j)} \|u\|_{0, p, B_r} \right) \end{equation} (Actually the book says "$\epsilon^{-j(m-j)}$" in the second summand but believe this is a mistake. There is also some talk about picking cutoff functions but I don't think that's relevant to the main point.)

The author references Adams and Fournier's book on Sobolev spaces for a proof. Indeed Theorem 5.12 of that book is a very similar result: For a sufficiently nice open set $\Omega \subseteq \mathbb{R}^n$ it gives a bound \begin{equation} \| u \|_{j, p, \Omega} \leq C \left( \epsilon \| u\|_{m, p, \Omega} + \epsilon^{-j/(m-j)} \|u\|_{0, p, \Omega_\epsilon}\right) \end{equation}

for $u \in W^{m, p}(\Omega)$. However here $\Omega_\epsilon$ is only some mystery subset of $\Omega$ whose closure lies in $\Omega$ and which depends on $\epsilon$.

My question is how/if you can get the second inequality from the first given that the second one offers no control on $\Omega_\epsilon$.

I was looking at Nicolaescu's Lectures on the Geometry of Manifolds (3rd edition). In Theorem 10.2.29 he presents (without proof) the following inequality:

For $m \geq 1, p \geq 1, 0 < r \leq R$ there exists a $C > 0$ s.t. for every $j < m, \epsilon >0$ and $ u \in W^{m, p}(\mathbb{R}^n)$ supported on $B_R$: \begin{equation} \| u \|_{j, p, \mathbb{R}^n} \leq C \left( \epsilon \|u\|_{m, p, \mathbb{R}^n} + \epsilon^{-j/(m-j)} \|u\|_{0, p, B_r} \right) \end{equation} (Actually the book says "$\epsilon^{-j(m-j)}$" in the second summand but believe this is a mistake. There is also some talk about picking cutoff functions but I don't think that's relevant to the main point.)

The author references Adams and Fournier's book on Sobolev spaces for a proof. Indeed Theorem 5.12 of that book is a very similar result: For a sufficiently nice open set $\Omega \subseteq \mathbb{R}^n$ it gives a bound \begin{equation} \| u \|_{j, p, \Omega} \leq C \left( \epsilon \| u\|_{m, p, \Omega} + \epsilon^{-j/(m-j)} \|u\|_{0, p, \Omega_\epsilon}\right) \end{equation}

for $u \in W^{m, p}(\Omega)$. However here $\Omega_\epsilon$ is only some mystery subset of $\Omega$ whose closure lies in $\Omega$ and which depends on $\epsilon$.

My question is how/if you can get the first inequality from the second given that the second one offers no control on $\Omega_\epsilon$.