> [!definition] > > Let $\cx$, $\cy$ be [[Normed Vector Space|normed spaces]] and $T \in L(\cx, \cy)$ be a [[Bounded Linear Map|bounded linear map]]. $T$ is **compact** if $T(B)$ is [[Precompact|precompact]] for all bounded sets $B \subset \cx$. > > The set $K(\cx, \cy)$ is the space of all compact maps from $\cx$ to $\cy$. > [!theorem] > > 1. For any [[Spectrum|eigenvalue]] $\lambda \in \sigma_p(T)$ with $\lambda \ne 0$, the geometric multiplicity of $\lambda$ is finite. > 2. For any enumeration $\seq{\lambda_n} \subset \sigma_p(T)$, $\lambda_n \to 0$ as $n \to \infty$. Therefore $\sigma_p(T)$ is countable. > > *Proof*. Let $\seq{\lambda_n} \subset \sigma_p(T)$ such that $\abs{\lambda_n} \ge \eps > 0$, and $\seq{x_n}$ be their corresponding eigenvectors, then they are all linearly independent. Let $H_n = \span(\seqf{x_j})$, then $T|_{\bigoplus_n H}$ is an isomorphism to an infinite dimensional space, which is impossible. > [!definition] > > Let $\cx$ be a normed space, $T \in L(\cx, \cx)$ be a bounded linear operator, and $\cm \subset \cx$ be a closed subspace. $\cm$ is $T$-**invariant** if $T(\cm) \subset \cm$. If $\cx$ is a [[Hilbert Space|Hilbert space]], then $\cm$ [[Split Subspace|splits]] $\ch$ as $\ch = \cm \oplus \cm^\perp$. > [!theorem] > > Let $\cx$ be a normed space and $\cy$ be a [[Banach Space|Banach space]], then $K(\cx, \cy)$ is a closed subspace of $L(\cx, \cy)$. > > *Proof*. Sums of compact sets are images of a continuous function on a product of two compact sets, thus they are compact. Therefore $K(\cx, \cy)$ is a subspace. > > Let $u \in \ol{K(\cx, \cy)}$, and $B \subset B(0, 1)$. Let $v \in K(\cx, \cy)$ such that $\norm{u - v} < \eps/2$, and $\seqf{B(y_j, \eps/2)}$ be a finite covering of $v(B)$ by open balls of radius $\eps/2$. From here, since for any $x \in B$, > $ > \norm{u(x) - v(x)} < \eps/2 \quad \norm{u(x) - y_j} < \eps/2 > $ > so $\seqf{B(y_j, \eps)}$ is a finite covering of $u(B)$. > > Thus $K(\cx, \cy)$ is closed. > [!theorem] > > Let $\bracs{E_j: 1 \le j \le 4}$ be normed spaces, with the sequence of bounded linear maps > $ > \begin{CD} > E_1 @>{S}>> E_2 @>{T}>> E_3 @>{U}>> E_4 > \end{CD} > $ > If $T$ is compact, then so is $U \circ T \circ S$. Thus $K(\cx, \cx)$ is a two-sided ideal of $L(\cx, \cx)$. > [!theorem] > > Let $E, F$ be Banach spaces and $u \in L(E, F)$, then $u \in K(E, F)$ if and only if it maps [[Weak Topology|weakly convergent sequences]] to strongly convergent sequences. > > *Proof*. Suppose that $u \in K(E, F)$ and let $\seq{x_n}$ be a weakly convergent sequence. By the [[Uniform Boundedness Principle]], since $\seq{\angles{\phi, x_n}}$ is bounded for all $\phi \in F^*$, $\seq{x_n}$ is bounded in norm. By compactness, it admits a convergent subsequence, converging to the weak limit. Since such a subsequence can be extracted from *all* subsequences, $x_n$ converges strongly to the weak limit. > > Suppose that $u$ maps weakly convergent sequences to strongly convergent sequences. Let $\seq{x_n}$ such that $\norm{x_n} \le C$ for all $n \in \nat$. By [[Tychonoff's Theorem]], it admits a weakly convergent subsequence.