LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Continuous convolution operators are Hilbert–Schmidt

Statement

Assume the Axiom of Choice. Let $G$ be a compact Lie group with normalized Haar measure $d g$ and let $k \in C (G, C)$ . Then $T_{k}$ has the square-integrable kernel $K (x, y) = k (x^{- 1} y)$ on $G \times G$ , is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ , and is compact.

Facts & Assumptions

Given: Assume the Axiom of Choice, a compact Lie group $G$ with normalized Haar measure $d g$ , and $k \in C (G, C)$ .

[A1]

The Axiom of Choice is The Axiom of Choice; it enters through the Haar measure and the Hilbert–Schmidt theory cited.

[L1]

$(T_{k} f) (x) = \int_{G} k (x^{- 1} y) f (y) d y$ defines a bounded operator on $L^{2} (G)$ , and $K (x, y) = k (x^{- 1} y)$ is continuous (Convolution operators).

[L2]

If $k \in L^{2} (G \times G)$ is a kernel class, the associated operator $T$ on $L^{2} (G)$ is Hilbert–Schmidt with $∥ T ∥_{H S} = ∥ k ∥_{2}$ ; Hilbert–Schmidt operators are compact (L two kernels give Hilbert–Schmidt operators, Hilbert–Schmidt operators are compact).

[L3]

Fubini's theorem identifies the product integral of an integrable function on two sigma-finite measure spaces with either iterated integral (Fubini's theorem for L^1 functions on a sigma-finite product), and Haar measure is translation invariant (Haar integration is translation and conjugation invariant).

Proof

technique · direct

1.1

The kernel $K (x, y) = k (x^{- 1} y)$ is continuous on the compact product $G \times G$ by [L1], hence bounded and measurable; its squared modulus is integrable, and by Fubini and translation invariance $\int ⁣ ⁣ \int_{G \times G} ∣ K (x, y) ∣^{2} d x d y = \int_{G} (\int_{G} ∣ k (x^{- 1} y) ∣^{2} d x) d y = \int_{G} ∥ k ∥_{2}^{2} d y = ∥ k ∥_{2}^{2} < + \infty$ .

L1L3

2.1

The operator with kernel $K$ is $T_{k}$ by [L1], so by [L2] the operator $T_{k}$ is Hilbert–Schmidt with $∥ T_{k} ∥_{H S} = ∥ k ∥_{2}$ and is compact.

A1L1L2step 1.1∎

Depends on

Used by

Spectral convolution eigenspaces are finite-dimensional and invariant Lemma

Dependency tree · two levels

61 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)