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L^2 kernel operators are bounded and compact

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $L^{2} = L^{2} (R^{3}; C)$ and let $k \in L^{2} (R^{3} \times R^{3}; C)$ , with $T_{k}$ the $L^{2}$ -kernel operator of L^2 kernel operators and Hilbert-Schmidt operators on R^3. Then:

(a) $D (T_{k}) = L^{2}$ ; $T_{k}$ is linear and bounded with $∥ T_{k} f ∥_{2} \leq ∥ k ∥_{L^{2} (R^{3} \times R^{3})} ∥ f ∥_{2}$ for every $f \in L^{2}$ , so $∥ T_{k} ∥ \leq ∥ k ∥_{L^{2} (R^{3} \times R^{3})}$ ;

(b) if $n \geq 1$ , $f_{1}, \dots, f_{n}, g_{1}, \dots, g_{n} \in L^{2}$ and $k (x, y) = \sum_{i = 1}^{n} f_{i} (x) \overline{g_{i} (y)}$ a.e., then $k$ is a class in $L^{2} (R^{3} \times R^{3}; C)$ , the range of $T_{k}$ is contained in $span {f_{1}, \dots, f_{n}}$ , and $T_{k}$ is compact;

(d) for every sequence $(k_{m})$ in $L^{2} (R^{3} \times R^{3}; C)$ with $∥ k_{m} - k ∥_{L^{2} (R^{3} \times R^{3})} \to 0$ one has $T_{k_{m}} - T_{k} = T_{k_{m} - k}$ and $∥ T_{k_{m}} - T_{k} ∥ \leq ∥ k_{m} - k ∥_{L^{2} (R^{3} \times R^{3})} \to 0$ ; in particular $T_{k}$ is the operator-norm limit of finite-rank kernel operators.

No statement is made about the spectrum of $T_{k}$ , about its singular values, or about $L^{2}$ kernels of a given compact operator.

Facts & Assumptions

Given: the Hilbert space $L^{2}$ and its product $L^{2} (R^{3} \times R^{3}; C)$ ; a kernel $k$ of the stated class; the operators $T_{k}$ of L^2 kernel operators and Hilbert-Schmidt operators on R^3.

[F1]

Cauchy-Schwarz for $L^{2}$ : $\int ∣ h g ∣ \leq ∥ h ∥_{2} ∥ g ∥_{2}$ whenever $h, g \in L^{2}$ . Completed-product Tonelli gives measurable section integrals and $\int_{R^{3}} (\int_{R^{3}} ∣ k (x, y) ∣^{2} d y) d x = ∥ k ∥_{L^{2} (R^{3} \times R^{3})}^{2}$ ; completed-product Fubini gives measurable integrals of integrable real and imaginary components (Cauchy-Schwarz inequality for $L^{2}$ , Tonelli and Fubini for the completed product, with only almost-everywhere section measurability).

[F2]

Density of box step functions: for $1 \leq p < \infty$ the finite linear combinations of indicators of half-open boxes in $R^{n}$ with rational endpoints and rational coefficients are dense in $L^{p} (R^{n})$ (Rational box-step functions form a countable dense subset of $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[F3]

A bounded linear operator between normed spaces whose range has a basis of finite length is compact (Bounded finite rank operators are compact). Under Countable Choice, an operator-norm limit of compact operators from a normed space into a Banach space is compact, provided the limit is a bounded linear operator (Norm limit of compact operators is compact). Here the target $L^{2}$ is a Banach space because it is a Hilbert space (Hilbert space), and step 1.1 supplies boundedness for each kernel operator to which these results are applied.

[F4]

$L^{2}$ elements are almost-everywhere classes with the linear-first pairing $⟨ f, g ⟩ = \int f \overline{g}$ . The kernel definition proves that changes of kernel or input representative preserve the section integrals almost everywhere (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions, L^2 kernel operators and Hilbert-Schmidt operators on R^3).

Proof

1.1F1F4algebra

[Boundedness and full domain] Fix $f \in L^{2}$ . For a.e. $x$ the function $y \mapsto k (x, y) f (y)$ is measurable, and by [F1] $\int ∣ k (x, y) f (y) ∣ d y \leq ∥ f ∥_{2} (\int ∣ k (x, y) ∣^{2} d y)^{1 / 2} < \infty$ for a.e. $x$ , the last integral being finite for a.e. $x$ with $\int (\int ∣ k (x, y) ∣^{2} d y) d x = ∥ k ∥_{2}^{2} < \infty$ ; hence $(T_{k} f) (x)$ exists for a.e. $x$ and $∣ (T_{k} f) (x) ∣ \leq ∥ f ∥_{2} (\int ∣ k (x, y) ∣^{2} d y)^{1 / 2}$ . For each bounded box $B \subset R^{3}$ , integrating the preceding bound gives $\int_{B} \int ∣ k (x, y) f (y) ∣ d y d x \leq ∣ B ∣^{1 / 2} ∥ k ∥_{2} ∥ f ∥_{2} < \infty$ . Applying [F1] to the real and imaginary components of $1_{B} (x) k (x, y) f (y)$ shows that $T_{k} f$ is measurable on $B$ up to a null set; a countable box exhaustion proves measurability on $R^{3}$ . Integrating the square and using completed-product Tonelli gives $∥ T_{k} f ∥_{2}^{2} \leq ∥ f ∥_{2}^{2} \int \int ∣ k (x, y) ∣^{2} d x d y = ∥ k ∥_{2}^{2} ∥ f ∥_{2}^{2}$ , so $D (T_{k}) = L^{2}$ , $T_{k} f \in L^{2}$ with the stated bound, and $T_{k}$ is linear because the integral is linear in $f$ for each $x$ . This is (a).

2.1step 1.1F1F3F4algebra

[Finite product kernels] Let $k (x, y) = \sum_{i = 1}^{n} f_{i} (x) \overline{g_{i} (y)}$ a.e. with $f_{i}, g_{i} \in L^{2}$ . By Tonelli and [F1], $\int \int ∣ k ∣^{2} \leq (\sum_{i} ∥ f_{i} ∥_{2} ∥ g_{i} ∥_{2})^{2} < \infty$ , so $k$ is a class in $L^{2} (R^{3} \times R^{3}; C)$ ; moreover for $f \in L^{2}$ step 1.1 applies and, for a.e. $x$ , $(T_{k} f) (x) = \sum_{i} f_{i} (x) \int \overline{g_{i} (y)} f (y) d y = \sum_{i} ⟨ f, g_{i} ⟩ f_{i} (x)$ by linearity of the integral and [F4]; hence the range of $T_{k}$ lies in the finite-dimensional space $span {f_{1}, \dots, f_{n}}$ . By [F3] $T_{k}$ is compact. This is (b).

3.1step 2.1F2F4algebra

[Box step functions are finite product kernels] A half-open box in $R^{6}$ is a product $B \times B^{'}$ of half-open boxes $B, B^{'} \subseteq R^{3}$ , and its indicator satisfies $1_{B \times B^{'}} (x, y) = 1_{B} (x) 1_{B^{'}} (y)$ for every $(x, y)$ ; the indicators of bounded boxes lie in $L^{2} (R^{3}; C)$ . Hence every finite linear combination of indicators of half-open boxes in $R^{6}$ with coefficients in $Q + i Q$ is a finite product kernel as in step 2.1, and its kernel operator is compact by step 2.1.

4.1step 1.1step 3.1F2F3algebra

[Norm approximation and compactness] Let $k \in L^{2} (R^{3} \times R^{3}; C)$ . For each integer $m \geq 0$ , apply [F2] with $n = 6$ , $p = 2$ and tolerance $1 / (m + 1)$ separately to the real $L^{2}$ functions $Re k$ and $Im k$ . By the assumed Countable Choice, select real rational box steps $a_{m}, b_{m}$ for all $m \geq 0$ with $∥ a_{m} - Re k ∥_{2} < 1 / (m + 1)$ and $∥ b_{m} - Im k ∥_{2} < 1 / (m + 1)$ . Then $s_{m} : = a_{m} + i b_{m}$ is a finite box step with coefficients in $Q + i Q$ , and $∥ s_{m} - k ∥_{2}^{2} = ∥ a_{m} - Re k ∥_{2}^{2} + ∥ b_{m} - Im k ∥_{2}^{2} < 2 / (m + 1)^{2}$ . For each $m$ , step 3.1 makes $T_{s_{m}}$ compact, and by linearity of the kernel operator $T_{k} - T_{s_{m}} = T_{k - s_{m}}$ ; step 1.1 applied to the kernel $k - s_{m}$ gives $∥ T_{k} - T_{s_{m}} ∥ \leq ∥ k - s_{m} ∥_{L^{2} (R^{3} \times R^{3})} \to 0$ . By [F3] the operator-norm limit $T_{k}$ is compact, which proves (c). For any sequence $(k_{m})$ as in (d), linearity gives $T_{k_{m}} - T_{k} = T_{k_{m} - k}$ , and step 1.1 gives $∥ T_{k_{m}} - T_{k} ∥ \leq ∥ k_{m} - k ∥_{2} \to 0$ , proving (d).

5.1step 1.1step 2.1step 4.1∎

[Conclusion] Part (a) is step 1.1, part (b) is step 2.1, and parts (c) and (d) are step 4.1.

L^2 kernel operators are bounded and compact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources