[proofplan] We reduce compactness of the interior restriction to the standard compact Sobolev embedding on the bounded smooth domain $W$. The restriction from $W$ to $V$ is bounded at order $t$, while the Rellich-Kondrachov theorem gives compactness from $H^s(W)$ to $H^t(W)$ because $s>t$. The bounded-domain statement is the same compact embedding applied directly on $\Omega$. [/proofplan] [step:Factor the interior restriction through the compact embedding on $W$] Define the Sobolev restriction operator \begin{align*} R_{W,V}: H^t(W) &\to H^t(V) \end{align*} \begin{align*} v &\mapsto v|_V. \end{align*} By the standard restriction property for Sobolev spaces on open subsets, $R_{W,V}$ is a [bounded linear operator](/page/Bounded%20Linear%20Operator). Since $W \subset \mathbb{R}^n$ is bounded with smooth boundary and $s>t$, the [Rellich-Kondrachov Theorem](/theorems/64) gives that the inclusion operator \begin{align*} J_{s,t,W}: H^s(W) &\to H^t(W) \end{align*} \begin{align*} u &\mapsto u \end{align*} is compact. Therefore the desired map is the composition \begin{align*} H^s(W) \xrightarrow{J_{s,t,W}} H^t(W) \xrightarrow{R_{W,V}} H^t(V). \end{align*} The composition of a compact linear operator with a bounded linear operator is compact, so $R_{W,V} \circ J_{s,t,W}:H^s(W)\to H^t(V)$ is compact. This composition is exactly $u \mapsto u|_V$. [guided] We want compactness of the map that takes $u \in H^s(W)$ and restricts it to $V$. Rather than proving compactness directly on the smaller set, we factor the map through the larger [Sobolev space](/page/Sobolev%20Space) $H^t(W)$. First define the restriction operator at the lower regularity level: \begin{align*} R_{W,V}: H^t(W) &\to H^t(V) \end{align*} \begin{align*} v &\mapsto v|_V. \end{align*} This map is bounded by the standard restriction property for Sobolev spaces on open subsets: restricting a distribution or Sobolev function from $W$ to the open subset $V$ cannot increase the Sobolev norm by more than a fixed constant depending only on $V$, $W$, and $t$. Next define the identity-on-functions inclusion \begin{align*} J_{s,t,W}: H^s(W) &\to H^t(W) \end{align*} \begin{align*} u &\mapsto u. \end{align*} The hypotheses needed for the [Rellich-Kondrachov Theorem](/theorems/64) are satisfied: $W$ is bounded, $W$ has smooth boundary, and the Sobolev orders satisfy $s>t$. Hence $J_{s,t,W}$ is compact. Now the interior restriction is exactly the composition \begin{align*} R_{W,V} \circ J_{s,t,W}: H^s(W) &\to H^t(V) \end{align*} \begin{align*} u &\mapsto u|_V. \end{align*} A bounded operator applied after a [compact operator](/page/Compact%20Operator) preserves compactness: if $(u_k)_{k=1}^{\infty}$ is bounded in $H^s(W)$, compactness of $J_{s,t,W}$ gives a subsequence $(u_{k_j})_{j=1}^{\infty}$ converging in $H^t(W)$, and boundedness of $R_{W,V}$ gives convergence of $(u_{k_j}|_V)_{j=1}^{\infty}$ in $H^t(V)$. Therefore $u \mapsto u|_V$ is compact from $H^s(W)$ to $H^t(V)$. [/guided] [/step] [step:Apply the same compact embedding on the bounded smooth domain $\Omega$] Assume now that $\Omega \subset \mathbb{R}^n$ is bounded with smooth boundary. Since $s>t$, the [Rellich-Kondrachov Theorem](/theorems/64) applies directly to $\Omega$ and gives that \begin{align*} I_{\Omega}: H^s(\Omega) &\to H^t(\Omega) \end{align*} \begin{align*} u &\mapsto u \end{align*} is compact. This is precisely the asserted compact inclusion $H^s(\Omega)\hookrightarrow H^t(\Omega)$. [/step]