LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Kernel of identity minus compact is finite dimensional

Statement

Let $X$ be a normed space over $R$ or $C$ and let $K : X \to X$ be a compact operator (Compact linear operator). Then the kernel

$\ker (I - K) = {x \in X : K x = x}$

is a finite-dimensional subspace of $X$ : it admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Facts & Assumptions

[A1]

$K$ is a bounded linear operator, and a bounded linear operator is continuous (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent); $K$ is compact, that is, $\overline{K ({\overline{B}}_{X})}$ is compact (Compact linear operator).

[A2]

The normed space $X$ is a metric space with $d (x, y) = ∥ x - y ∥$ , convergence is metric convergence and limits of sequences are unique (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , A sequence in a metric space has at most one limit, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Complex norm and complex normed space, Realification preserves a complex norm and its metric); the closed unit ball is ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ and is closed (Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[A3]

If the closed unit ball of a normed space is compact, then that space admits an ordered basis of finite length (The closed unit ball is compact if and only if the normed space is finite-dimensional); a closed subset of a compact metric space is a compact metric subspace (A closed subset of a compact metric space is compact). Relative openness is the trace of ambient openness, and compactness is the compactness of the restricted metric (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Proof

technique · direct

Given: A normed space $X$ over $R$ or $C$ , a compact operator $K : X \to X$ , and $N : = \ker (I - K) = {x \in X : K x = x}$ .

1.1A1A2

The set $N$ is a linear subspace of $X$ , since $I - K$ is linear. It is closed without using a sequential-closure criterion: take a bound $C \geq 0$ for $K$ . If $x \notin N$ , put $a = ∥ x - K x ∥ > 0$ . For $∥ y - x ∥ < a / (2 (1 + C))$ , the triangle inequality gives $∥ y - K y ∥ \geq a - ∥ (I - K) (y - x) ∥ \geq a - (1 + C) ∥ y - x ∥ > a / 2 > 0$ . Thus a ball around every $x \notin N$ lies outside $N$ , proving its complement open.

2.1step 1.1A2

Every $x$ in the closed unit ball $N \cap {\overline{B}}_{X}$ of $N$ satisfies $x = K x$ and $∥ x ∥ \leq 1$ , hence $x \in K ({\overline{B}}_{X})$ ; therefore $N \cap {\overline{B}}_{X} \subseteq K ({\overline{B}}_{X}) \subseteq \overline{K ({\overline{B}}_{X})}$ , and $N \cap {\overline{B}}_{X}$ is a closed subset of $X$ by [step 1.1] and [A2].

3.1step 2.1A1A3

The set $\overline{K ({\overline{B}}_{X})}$ is compact by [A1], so by [step 2.1] and [A3] the set $N \cap {\overline{B}}_{X}$ is compact with its metric restricted from $X$ : it is relatively closed in the compact metric subspace $\overline{K ({\overline{B}}_{X})}$ . Restricting that same metric via $N$ gives exactly the same distance on $N \cap {\overline{B}}_{X}$ , so it is also the compact closed unit ball of the normed space $N$ , so $N$ admits an ordered basis of finite length by [A3].

4.1step 3.1∎

Hence $\ker (I - K)$ is finite dimensional, as claimed.

Kernel of identity minus compact is finite dimensional

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources