# Precompactness of a set vs. compactness of some operator

I found a remark in Adams' book Sobolev spaces (Acad. Press) which I cannot understand completely. It is on page 34 and says:

We remark that Theorem 2.22 is just a setting, suitable for our purposes, of a well-known theorem stating that the operator-norm limit of a sequence of compact operators is compact.

Theorem 2.22 is stated as follows:

Theorem Let $1\le p <+\infty$ and let $K\subset L^p(\Omega)$. Suppose there exists a sequence $(\Omega_j)$ of subdomains of $\Omega$ having the following properties:

a. For each $j$, $\Omega_j \subset \Omega_{j+1}$;

b. For each $j$ the set of restrictions to $\Omega_j$ of the functions in $K$ is precompact in $L^p(\Omega_j)$;

c. For every $\varepsilon > 0$ there exists $j$ such that

$$\int_{\Omega \setminus \Omega_j} \lvert u(x)\rvert^p\, dx <\varepsilon \qquad \text{for every}\ u\in K.$$

Then $K$ is precompact in $L^p(\Omega)$.

What "sequence of operators" is he talking about?

Thank you.

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The sequence of restriction operators to $\Omega_j$. Restriction is a linear operator, right?
Yes, of course. But it seems to me that it is not compact. Am I wrong? Also, what is the relationship between restriction operators and precompactness of $K$? – Giuseppe Negro Jun 24 '12 at 23:18
Restriction operators $R_j$ are not compact. However, $R_j(K)$ is assumed to be precompact for every $j$. Property (c) says that $R_j$ converge to the identity $I$ uniformly in norm when restricted to $K$. (The restriction of a restriction operator! :) This implies that $I(K)=K$ is precompact. Details: to get an $\epsilon$-net for $K$, use an $\epsilon/2$-net for $R_j(K)$ where $j$ is sufficiently large. – user31373 Jun 24 '12 at 23:29