7.2 Properties and characterizations of compact operators

Compact operators are crucial in functional analysis, bridging finite and infinite-dimensional spaces. They generalize finite-rank operators, preserving many of their properties. Understanding compact operators is key to grasping advanced concepts in operator theory.

This section delves into the properties of compact operators, including their behavior under limits and adjoints. We'll explore how compactness interacts with operator products and examine the relationship between compact operators and precompact sets.

Properties of Compact Operators

Compactness of sequence limits

• Consider Banach spaces $X$ and $Y$ and a sequence of compact operators $(T_n)$ from $X$ to $Y$ converging to an operator $T$ in the operator norm
• To prove the compactness of $T$, take a bounded sequence $(x_n)$ in $X$
  • The compactness of $T_1$ guarantees a subsequence $(x_{n_k})$ such that $(T_1 x_{n_k})$ converges in $Y$
  • Repeatedly apply this argument to $(x_{n_k})$ and subsequent operators $T_2, T_3, \ldots$ to extract further subsequences
  • This process yields a diagonal subsequence $(x_{n_k})$ for which $(T_n x_{n_k})$ converges for each $n$
• Demonstrate that $(T x_{n_k})$ forms a Cauchy sequence in $Y$
  • Given $\varepsilon > 0$, select $N$ such that $\|T_n - T\| < \varepsilon/3$ for all $n \geq N$
  • For sufficiently large indices $k$ and $l$, ensure $\|T_N x_{n_k} - T_N x_{n_l}\| < \varepsilon/3$
  • Apply the triangle inequality to show $\|T x_{n_k} - T x_{n_l}\| < \varepsilon$
• The convergence of $(T x_{n_k})$ in $Y$ establishes the compactness of $T$

Properties of compact operators

• An operator $T: X \to Y$ is compact if and only if $T(B_X)$ is precompact in $Y$, where $B_X$ denotes the unit ball in $X$
  • Precompact sets have compact closures
• For a compact operator $T$ and a sequence $(x_n)$ in $B_X$, the sequence $(T x_n)$ admits a convergent subsequence in $Y$
  • This property implies the precompactness of $T(B_X)$
• Conversely, if $T(B_X)$ is precompact and $(x_n)$ is a bounded sequence in $X$, then $(x_n)$ lies within a multiple of $B_X$
  • Consequently, $(T x_n)$ is contained in a multiple of the precompact set $T(B_X)$
  • The existence of a convergent subsequence of $(T x_n)$ confirms the compactness of $T$

Compactness of adjoint operators

• Consider a compact operator $T: X \to Y$ between Banach spaces $X$ and $Y$
• The adjoint operator $T^*: Y^* \to X^*$ is defined by $(T^* f)(x) = f(Tx)$ for all $f \in Y^*$ and $x \in X$
• To establish the compactness of $T^*$, consider a bounded sequence $(f_n)$ in $Y^*$
  • The sequence $(T^* f_n)$ is bounded in $X^*$ due to the inequality $\|T^* f_n\| \leq \|T\| \|f_n\|$
  • For each $x \in X$, the sequence $((T^* f_n)(x))$ is bounded in the scalar field $\mathbb{K}$
    • The compactness of $T$ ensures that $(Tx)$ is precompact in $Y$
    • The boundedness of $(f_n)$ implies the boundedness of $((T^* f_n)(x))$
  • The Banach-Alaoglu theorem guarantees the existence of a weak-* convergent subsequence of $(T^* f_n)$ in $X^*$
• The compactness of $T^*$ follows from the above arguments

Compactness in operator products

• Let $S: X \to Y$ be a compact operator and $T: Y \to Z$ be a bounded linear operator between Banach spaces $X$, $Y$, and $Z$
• To verify the compactness of the product $TS: X \to Z$, consider a bounded sequence $(x_n)$ in $X$
  • The compactness of $S$ implies the existence of a convergent subsequence $(S x_{n_k})$ in $Y$
  • The boundedness of $T$ ensures the convergence of $(TS x_{n_k})$ in $Z$
    • Specifically, $\|TS x_{n_k} - TS x_{n_l}\| \leq \|T\| \|S x_{n_k} - S x_{n_l}\|$
• The convergence of $(TS x_{n_k})$ establishes the compactness of the product operator $TS$