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In Hilbert spaces, an operator $T$ is compact if every bounded sequence $(x_n)$ has a subsequence $(x_{n_k})$ such that $(Tx_{n_k})$ is convergent.

In Sobolev spaces, the Rellich-Kondrachov theorem asserts that certain embeddings (inclusions) of Sobolev spaces are compact; specifically, the embedding from a higher order Sobolev space to a lower order space is compact under certain conditions related to the domain's boundedness and smoothness.

Fredholm Alternative states that for a compact operator $T$ and scalar $\lambda \neq 0$, either the equation $(T - \lambda I)x = 0$ has a non-trivial solution, or the equation $(T - \lambda I)x = y$ has a unique solution for every $y$ in the space.

A family of functions $\mathcal{F}$ on a compact space $K$ is relatively compact in $C(K)$, the space of continuous functions on $K$, if and only if $\mathcal{F}$ is uniformly bounded and equicontinuous.

In $L^p$ spaces for $1 \leq p < \infty$, an operator $T: L^p(\mu) \to L^p(\mu)$ is compact if it satisfies conditions such as mapping weakly convergent sequences to strongly convergent sequences.

For any sequence $(x_n)$ in $X$ such that $T(x_n)$ has no convergent subsequence, $T$ is not a compact operator.

The spectrum of a compact operator on an infinite-dimensional space consists of zero and an at most countable set of eigenvalues with no accumulation point except possibly at zero.

Compact operators are bounded, linear, and continuous, but not every bounded linear operator is compact.

If $X$ is a Hilbert space, any compact self-adjoint operator on $X$ has an orthonormal basis of eigenvectors.

A linear operator $T: X \to Y$ between two Banach spaces is called compact if it maps bounded sets into precompact (or totally bounded) sets, which means their closure is compact.

If $T: X \to Y$ is a compact operator and $X$ is infinite-dimensional, then the dimension of the closure of $T(X)$ is also infinite.

If $X$ and $Y$ are normed spaces and $X$ is compact, then every continuous linear operator $T: X \to Y$ is a compact operator.

For a compact operator $T$, the norm and weak topologies on $T(B)$ for any bounded set $B$ are equivalent, meaning a sequence in $T(B)$ converges in norm if and only if it converges weakly.

If $T_1: X_1 \to Y_1$ and $T_2: X_2 \to Y_2$ are compact operators, then the operator $T: X_1 \times X_2 \to Y_1 \times Y_2$ defined by $T(x_1, x_2) = (T_1 x_1, T_2 x_2)$ is also compact.

The Volterra operator $V: L^2[0,1] \to L^2[0,1]$ defined by $(Vf)(x) = \int_0^x f(t)dt$ is an example of a compact operator in function space.

If $T$ is a compact operator on an infinite-dimensional Hilbert space, any non-zero eigenvalue $\lambda$ of $T$ has finite multiplicity and the corresponding eigenspace is also compact.

If a Banach space $X$ has a Schauder basis and $T: X \to X$ is a compact operator, then $T$ can be approximated by operators with finite-dimensional ranges.

In a locally convex topological vector space, an operator is termed Hausdorff if its range is not only precompact but also if the closure of its range is actually compact.

A linear operator $T$ is compact if and only if it is the limit (in the operator norm) of a sequence of finite rank operators.