[Paper Review] Compactness of Embeddings

А. Г. Рамм|ArXiv.org|Nov 14, 2005

Rings, Modules, and Algebras3 citations

TL;DR

This paper refines necessary and sufficient conditions for the compactness of embedding operators between three nested Banach spaces, proving that compactness of $ i:X_1 \to X_2 $ holds if and only if the embedding $ j:X_1 \to X_3 $ is compact and a specific interpolation-type inequality $ \|u\|_2 \leq s\|u\|_1 + c(s)\|u\|_3 $ holds for all $ s \in (0,1) $. A counterexample is constructed to disprove a widely cited claim in the literature that this inequality holds without norm compatibility assumptions between $ X_2 $ and $ X_3 $.

ABSTRACT

An improvement of the author's result, proved in 1961, concerning necessary and sufficient conditions for the compactness of embedding operators is given. A counterexample to a published statement concerning compactness of embedding operators is constructed.

Motivation & Objective

• To refine and generalize the author's 1961 result on compact embeddings by removing the restriction that $ X_2 $ be a Hilbert space.
• To establish a sharp criterion for compactness of $ i:X_1 \to X_2 $ using compactness of $ j:X_1 \to X_3 $ and a parameterized norm inequality.
• To demonstrate that the compatibility of norms in $ X_2 $ and $ X_3 $ is essential for the necessity of the norm inequality condition.
• To construct a counterexample invalidating a commonly cited claim (Lions’ lemma) that the norm inequality holds without norm compatibility.

Proposed method

• Prove sufficiency by assuming compactness of $ j:X_1 \to X_3 $ and the interpolation inequality $ \|u\|_2 \leq s\|u\|_1 + c(s)\|u\|_3 $, then showing any bounded sequence in $ X_1 $ has a convergent subsequence in $ X_2 $.
• Prove necessity by assuming $ i:X_1 \to X_2 $ is compact and deriving a contradiction if the norm inequality fails, using the existence of a sequence with $ \|u_n\|_2 \geq s_0 + n\|u_n\|_3 $.
• Use the compatibility condition between $ X_2 $ and $ X_3 $ norms to ensure convergence in $ X_3 $ implies convergence in $ X_2 $, which is essential for the contradiction argument.
• Construct a counterexample using $ X_1 $ as a one-dimensional space of functions vanishing a.e. but pointwise non-zero at a point, with $ X_2 $ norm including a point evaluation.
• Define $ X_3 = L^2(0,1) $, $ X_2 = L^2(0,1) + \text{point evaluation} $, and $ X_1 $ as span of a function with zero $ L^2 $-norm but non-zero point value.
• Show that the claimed inequality $ \|u\|_2 \leq \epsilon\|u\|_1 + K(\epsilon)\|u\|_3 $ fails in the counterexample because $ \|u\|_3 = 0 $ and $ \|u\|_2 > 0 $, leading to $ |\lambda| \leq \epsilon|\lambda| $, which fails for $ \epsilon < 1 $.

Experimental results

Research questions

• RQ1 Under what conditions is the embedding operator $ i:X_1 \to X_2 $ compact when $ X_1 \subset X_2 \subset X_3 $ are nested Banach spaces with comparable norms?
• RQ2 Is the norm inequality $ \|u\|_2 \leq s\|u\|_1 + c(s)\|u\|_3 $ necessary for compactness of $ i:X_1 \to X_2 $, and does it hold without norm compatibility between $ X_2 $ and $ X_3 $?
• RQ3 Can the claim in [1], p.35 (Lions’ lemma), that such an inequality holds for any compact embedding, be valid without norm compatibility?
• RQ4 What is the role of norm compatibility between $ X_2 $ and $ X_3 $ in the necessity proof of the norm inequality condition?
• RQ5 Does the failure of norm compatibility invalidate the standard interpolation-type inequality used in compactness criteria?

Key findings

• The embedding $ i:X_1 \to X_2 $ is compact if and only if the embedding $ j:X_1 \to X_3 $ is compact and the inequality $ \|u\|_2 \leq s\|u\|_1 + c(s)\|u\|_3 $ holds for all $ s \in (0,1) $, with $ c(s) > 0 $.
• The proof of sufficiency does not require norm compatibility between $ X_2 $ and $ X_3 $, but the necessity proof does.
• A counterexample is constructed where $ X_1 $ is one-dimensional, $ X_3 = L^2(0,1) $, $ X_2 $ includes a point evaluation, and $ \|u\|_3 = 0 $, $ \|u\|_2 = 1 $ for a non-zero $ u \in X_1 $.
• In this counterexample, the embedding $ i:X_1 \to X_2 $ is compact because $ X_1 $ is finite-dimensional, satisfying all assumptions of the claimed lemma in [1].
• However, the inequality $ \|u\|_2 \leq \epsilon\|u\|_1 + K(\epsilon)\|u\|_3 $ fails for $ \epsilon < 1 $, since it reduces to $ |\lambda| \leq \epsilon|\lambda| $, which is false unless $ \lambda = 0 $.
• Thus, the claim in [1], p.35, is invalid without the assumption of norm compatibility between $ X_2 $ and $ X_3 $, and the counterexample proves this explicitly.