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Rellich Kondrachov and Sobolev Compactness
1 · Prerequisites
- Absolute and Conditional Convergence; Rearrangement; Products
- Absolute Continuity and the Sharp Fundamental Theorem of Calculus
- Approximation and Compactness in C(K)
- Arc Length and Rectifiable Curves
- Areas of Elementary Plane Figures
- Banach Alaoglu Goldstine and Krein Milman
- Banach Valued Integration and the Radon Nikodym Property
- Binary Operations, Monoids, Groups and Subgroups
- Bounded Linear Operators and Quotient Spaces
- Bounded Variation and the Riemann–Stieltjes Integral
- Compact Operators and Riesz Schauder Theory
- Compactness
- Compactness in Metric Spaces
- Complete Metrizability, Čech-Completeness, and Baire Category
- Completeness, Completion, and Uniform Continuity
- Complex Differentiability and the Cauchy–Riemann Equations
- Complex Lp Spaces and Test-Function Conventions
- Connectedness
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Continuity, IVT, EVT, and Uniform Continuity
- Contour Integration
- Convergence: Nets and Filters
- Convex and Semicontinuous Functions on Rⁿ
- Cosets, Index and Lagrange's Theorem
- Countability and Uncountability
- Countability Axioms and Cardinal Functions
- Density Separability and Convolution in Lᵖ
- Determinants of Matrices over a Commutative Ring
- Differentiation of Monotone Functions and the Vitali Covering Theorem
- Distributions Test Functions and Differentiation
- Divisibility, Euclidean Domains, Principal Ideal Domains and Unique Factorisation
- Dual Spaces Adjoint Operators and Annihilators
- Dual Spaces, Bilinear and Quadratic Forms, and Sylvester's Law of Inertia
- Euclidean Surface Measure, Divergence, and Green Identities
- Filters and Ultrafilters
- Finite Counting, Factorials and Binomial Coefficients
- Finite Dimensional Normed Spaces and Riesz Lemma
- Foundations of the Real Numbers for Analysis
- Fubini and Change of Variables
- Gaussian Elimination, Elementary Matrices and Reduced Row Echelon Form
- Group Actions, Orbits, Stabilisers and Cayley's Theorem
- Harmonic Functions and Mean Values in Rn
- Hausdorff via the Diagonal
- Ideals, Quotient Rings and the Isomorphism Theorems for Rings
- Inner Product Spaces, Gram-Schmidt, Projections and Adjoints
- Lebesgue Measure on Euclidean Space
- Limits of Real Functions
- limsup, liminf, and Subsequential Limits
- Line Integrals and the Gradient Theorem
- Linear Independence, Bases and Dimension
- Linear Transformations, Rank-Nullity and Quotient Spaces
- Matrices, the Matrix of a Linear Map, and Change of Basis
- Measurable Functions and Simple Approximation
- Measures and Their Basic Properties
- Metric Spaces
- Mixed Partials, Taylor Formulae, and Extrema
- Modes of Convergence Egorov and Lusin
- Monotone Functions, Discontinuities, and Continuity Sets
- Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness
- Normal Subgroups and Quotient Groups
- Normed and Banach Spaces
- Norming and Separation under Hahn–Banach
- Order, Zorn's Lemma, and the Axiom of Choice
- Outer Measure and the Caratheodory Extension Theorem
- Poisson Problems and Interior Harmonic Estimates
- Polynomial Rings, the Division Algorithm and Roots
- Power Series and Real-Analytic Functions
- Product Measures and the Fubini Tonelli Theorems
- Properties of the Integral and the Working FTC
- Radon Measures and the Riesz Markov Kakutani Theorem
- Reflexivity and Eberlein Smulian
- Relations, Functions, and Quotients
- Rings, Subrings, Integral Domains and Fields
- Rⁿ as a Normed Space; Vector-Valued Functions
- Roots, Rational Powers, and Classical Inequalities
- Schwartz Space and the Plancherel Theorem
- Separation Axioms: the Hierarchy
- Sequences and Limits
- Sequences and Series of Functions; Uniform Convergence
- Series: Convergence and the Nonnegative Tests
- Sigma Algebras and Borel Sets
- Signed and Complex Measures Hahn and Jordan
- Simple Field Extensions and the Construction of the Complex Numbers
- Smooth Approximation and Sobolev Extension
- Smooth Partitions of Unity and Exhaustions
- Sobolev Poincare and Morrey Inequalities
- Sobolev Traces and Zero Boundary Values
- Subspaces, Products, and Quotients
- Suprema and Infima
- Symmetric Groups, Cycle Decomposition and the Sign Homomorphism
- The Analytic Hahn Banach Theorem
- The Baire Principles of Functional Analysis
- The Cantor Set, Baire Category, and Measure Zero in ℝ
- The Complex Exponential and Euler's Formula
- The Derivative and the Mean Value Theorems
- The Determinant of a Linear Operator, Cofactors and Cramer's Rule
- The Duality of Lᵖ and L^q
- The Exponential Function
- The Inverse and Implicit Function Theorems
- The Inverse Function Theorem Completed
- The Lebesgue Integral and the Convergence Theorems
- The Logarithm and General Powers
- The Lᵖ Spaces Holder Minkowski and Riesz Fischer
- The Maximal Function and Lebesgue Differentiation
- The Radon Nikodym Theorem and Lebesgue Decomposition
- The Riemann Integral in Rᵐ and Jordan Content
- The Riemann Integral: Definition and Integrability
- The Topology of Euclidean Space
- The Total Derivative in ℝᵐ → ℝⁿ
- The ZFC Axioms and the Basic Set Constructions
- Topological Spaces and Continuity
- Topology of ℝ
- Triangularisation, Generalised Eigenspaces and Jordan Canonical Form
- Urysohn's Lemma and the Tietze Extension Theorem
- Vector Spaces, Linear Subspaces, Span and Direct Sums
- Weak and Weak Star Topologies
- Weak Derivatives and Sobolev Spaces
2 · Summary
This page proves the compact embedding theorems behind existence theory for PDEs. It begins with the definition of a compact embedding and the translation estimate for functions, the two inputs of the Fréchet–Kolmogorov criterion: an -bounded family with vanishing tails and uniform translation control is totally bounded, and relative compactness forces uniform translation continuity. From the criterion follow the Rellich theorems: on every bounded open set, and on bounded extension domains, with no boundary regularity in the first case. A compactness proof of the Poincaré–Wirtinger inequality and a local -compactness theorem for -bounded sequences follow, together with the subcritical and critical forms of Rellich–Kondrachov on extension domains, the Morrey compactness into for , and the higher-order statements.
The second half develops the compactness machinery for fractional Sobolev spaces that the subcritical trace theorem needs: the level-set kernel estimate, a dyadic summability lemma, the Slobodeckij lower bound for dyadic level sets, the critical fractional Sobolev inequality on , and mollification rates that yield fractional Rellich compactness on bounded supports. These feed the compactness of the Sobolev trace into subcritical boundary spaces, the strictly subcritical character of every theorem on the page, and three downstream consequences: the whole-sequence strong convergence corollary for bounded sequences and their weak limits, the compact operator corollary obtained from a bounded map into , and the lemma that norm-closed target constraints survive compact extraction.
Notes
Every compactness statement carries the exact choice hypothesis of its proof: the Fréchet–Kolmogorov theorem assumes Countable and Dependent Choice; the Rellich theorems, the Morrey branch and the fractional Rellich theorem assume the Axiom of Choice through their named suppliers, while the bounded-support tail lemma and the compact-operator definition are choice-free. The sequential form of compact embedding is stated under Countable and Dependent Choice.
3 · Logical flowchart
4 · Definitions, theorems and proofs
Compactly embedded normed spaces
Definition
Let and be normed spaces over the same field , read in the real case from A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and in the complex case from Real and complex scalar conventions for normed spaces, and suppose with continuous inclusion : there is a real with for every .
One says that is compactly embedded in , written , when the inclusion operator is a compact operator in the sense of Compact linear operator: the image under of every bounded subset of (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) has compact closure in (Open cover, subcover, compact metric space, and compact subset of a metric space).
The sequential form. Assume the Axiom of Countable Choice (The Axiom of Countable Choice ()) and the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain). Then if and only if every bounded sequence in has a subsequence converging in : Indeed, compact closure gives the sequence conclusion by For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice. Conversely, fix bounded and let . For any sequence in , Countable Choice selects with (start at ); a convergent subsequence of gives one of with the same limit, which belongs to the closed set . Thus is sequentially compact and For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice makes it compact. The empty case is immediate. Both readings of are used on this page; the second is the form in which the compactness theorems below are stated.
Continuity of the inclusion is a separate hypothesis and is never inferred from compactness: a compact operator is bounded by Compact linear operator and A bounded linear operator between normed spaces, but the definition above fixes the continuity of in advance. On this page the continuity of every Sobolev inclusion is verified separately, through the corresponding Sobolev embedding theorem.
Remarks
- The symbol is used only for pairs of normed spaces embedded in one another as above; it never abbreviates a claim that some particular Sobolev space is compactly embedded in another, which always requires its own theorem with its own domain, exponent and boundary hypotheses.
- If is finite-dimensional and is a normed space containing it with continuous inclusion, then : the coordinate isomorphism of A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space sends a closed bounded coordinate ball to a compact set containing any prescribed bounded subset of , by Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line (identify with ). Its image in is compact by continuity of the inclusion, and the closure of the bounded image is a closed subset of it. The zero-dimensional case is immediate.
- The two formulations agree without any hypothesis on or beyond their being normed spaces, and the choice cost of the passage between them is bounded above by Countable Choice plus Dependent Choice, as recorded in For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice.
The translation estimate for functions on
Statement
Assume the Axiom of Choice. Let , , and . For let , read on almost-everywhere classes. Then with , and where and is the Euclidean norm of . The estimate is a statement about classes and does not depend on the chosen representatives.
Facts & Assumptions
Given: the Axiom of Choice, , , a scalar field , a class and a vector .
Smooth density in . Under Countable Choice, for every there are with ; Countable Choice is supplied by the assumed Axiom of Choice. (Compactly supported smooth functions are dense in W^{k,p}(R^n), The Axiom of Countable Choice ())
Fundamental theorem of calculus. For a smooth function , , with the complex-valued identity read componentwise. (Fundamental theorem of calculus for absolutely continuous functions)
Minkowski and translation isometry. Minkowski's integral inequality applies to the -integral on , and Lebesgue translation invariance gives for every . (Minkowski's integral inequality, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space as the quotient by null functions)
Vector gradient norm. The norm of the Euclidean magnitude is equivalent, with constants depending only on , to the finite sum of the component norms used in Integer-order Sobolev spaces and their norms. Indeed , so Minkowski bounds above by . Also , which proves convergence in when in . (Integer-order Sobolev spaces and their norms)
Weak derivatives. A locally integrable function has weak -derivative when for every ; if , this places in in that coordinate. (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms)
Proof
By [F1] and Countable Choice, choose with . For a smooth the fundamental theorem [F2] gives By Cauchy--Schwarz and [F3],
For each coordinate and , the change of variables and the weak-derivative identity for give By [F3], , so [F5] proves and . Translation acts on almost-everywhere classes because it preserves null sets; hence both the derivative identity and the estimate are representative-independent.
The convergence in gives and, by [F4], . Translation is an isometry by [F3], so Letting in the inequality of step 1.1 proves .
If the estimate is equality. For every the preceding argument proves the stated estimate and translated derivative identity, with all expressions depending only on the classes in .
Relative compactness forces uniform translation continuity in
Statement
Assume the Axiom of Countable Choice. Let and let be relatively compact, that is, its closure in is compact. In the displayed nonnegative supremum, take the value if . Then This is the necessity of the translation hypothesis in the Fr'echet--Kolmogorov criterion.
Facts & Assumptions
Given: the Axiom of Countable Choice, , and a relatively compact family with closure .
Compact metric spaces are totally bounded. A compact metric space is totally bounded and complete. (A compact metric space is complete and totally bounded, and neither implication uses any choice principle)
Total boundedness is finite-net covering. A metric space is totally bounded when for every there are finitely many points with ; a subset of a totally bounded space which is itself totally bounded as a subspace has the same property for every . (Finite -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space, A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded)
Continuity of translation in . For every , as , where acts on almost-everywhere classes. ( in as , for , Translation of a function on )
Translation is an -isometry. for every and every , because Lebesgue measure is translation invariant. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space as the quotient by null functions)
Proof
If the supremum is and the claim holds. Otherwise the closure is a compact subset of , so is a compact metric space and [F1] makes it totally bounded. Given , [F2] provides finitely many centres with .
For each the centre is an element of , so [F3] gives with whenever ; set , a minimum over the nonempty finite set of indices. By [F4], for every and every .
Fix and . By step 1.1 there is with , and the triangle inequality together with steps 2.1 and 1.1 gives . Hence for all , and since was arbitrary the supremum tends to . Countable Choice enters only through the continuity-of-translation interface [F3].
Uniformly supported families have vanishing tails
Statement
Let , let be bounded and let be a family such that every vanishes Lebesgue-almost everywhere outside . In the displayed nonnegative supremum, take the value if . Then for every there is with In particular the tightness hypothesis of the Fr'echet--Kolmogorov criterion is automatic for families supported in one bounded set.
Facts & Assumptions
Given: , a bounded set , and a family whose every member vanishes almost everywhere outside .
Bounded sets lie in balls. is bounded in the metric sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space if and only if there is a centre and a radius with ; a ball is contained in the ball about the origin of radius . (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space)
Restriction to a measurable tail. If is measurable and vanishes almost everywhere on , then for any measurable representative the function is measurable and zero almost everywhere. (The space as the quotient by null functions, Measure-null sets and almost-everywhere statements relative to a measure)
Zero integral and almost-everywhere vanishing. A nonnegative measurable satisfies if and only if almost everywhere. (A nonnegative measurable function has integral exactly when it vanishes almost everywhere)
Proof
Proof technique: Choose a ball containing and use the given almost-everywhere vanishing on the measurable tail outside that ball.
By [F1] fix with and put . The set is measurable and is contained in . If then its tail supremum is . Otherwise, for each the hypothesis says that vanishes almost everywhere on , hence on ; by [F2] the measurable function is zero almost everywhere for any representative of , so [F3] gives .
Every member of has tail integral by step 1.1, so . Since was arbitrary, the tightness condition of the Fr'echet--Kolmogorov criterion holds. The argument uses no choice principle: is obtained from the single bounded set and the supremum is evaluated at the constant value .
The Fr'echet--Kolmogorov compactness criterion in
Statement
Assume the Axiom of Countable Choice and the Axiom of Dependent Choice. Let and let . In each of the three displayed nonnegative suprema, take the value if . Assume satisfies: (i) ; (ii) for every there is with ; and (iii) for every there is with whenever . Then is totally bounded in , the closure of is compact, and every sequence in has a subsequence converging in .
Facts & Assumptions
Given: the Axioms of Countable and Dependent Choice, , and a family satisfying conditions (i)--(iii) of the statement; write . For and put , and for each let be the radial mollifier at scale of A radial mollifier family in Rn.
Mollification. For , the function is smooth and . (Convolution with a mollifier is smooth, and derivatives pass under the integral sign)
H"older's inequality. For conjugate exponents , ; in particular and pointwise. (Holder's inequality for integrals, including the endpoint cases)
Minkowski's integral inequality. For measurable on a product of sigma-finite measure spaces with , . (Minkowski's integral inequality)
Translations are -isometries. for every and every , since Lebesgue measure is translation invariant. (Translation of a function on , Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space as the quotient by null functions)
Finite nets for equicontinuous families. An equicontinuous pointwise bounded family in , a compact metric space, is totally bounded for the supremum metric; the same holds after applying the statement to real and imaginary parts of a complex-valued family, and Arzelà--Ascoli for real under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded records the equivalent compact-closure form. (An equicontinuous pointwise-bounded family in has a finite net in the supremum metric, Arzelà--Ascoli for real under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded)
Total boundedness and its closure. A metric space is totally bounded when it has a finite -net for every ; total boundedness passes to subsets and to closures, and for supported in a compact ball one has . (Finite -net and totally bounded metric space, A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded, Open ball, closed ball and sphere in a metric space)
Completeness and compactness of . is complete; a closed subspace of a complete metric space is complete, and a complete and totally bounded metric space is compact under Countable Choice; compactness and sequential compactness of a metric space are equivalent under Countable and Dependent Choice. (Riesz-Fischer completeness of for , Closed subspaces of complete metric spaces are complete; the converse under countable choice, A complete, totally bounded metric space is compact, proved from countable choice used exactly once, For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Proof
If it is totally bounded and its closure is empty. Otherwise fix . By (ii) choose with for every , and by (iii) choose with for every and every ; let be the radial mollifier at scale , so , and .
For every and every , by step 1.1 and [F4]. Since , [F3] gives , and therefore .
By [F1] and [F2], every satisfies and , and vanishes off the ball of radius because vanishes off the ball of radius ; hence the family is uniformly bounded and uniformly Lipschitz, and all its elements are supported in the compact ball .
The real parts are equicontinuous and pointwise bounded on the compact metric space , and likewise the imaginary parts; by [F5] both families are totally bounded in the supremum metric, and combining the finitely many real and imaginary sup-balls, is totally bounded in the supremum metric over . Since every is supported in , the supremum over equals the supremum over , so for every the family has a finite covering by -balls of radius by [F6].
Let and apply steps 1.1--3.1 with , covering by finitely many -balls of radius . Step 2.1 then covers by the same centres with radius . Discard empty intersections with and choose one point of in each remaining ball. Their radius- balls cover by the triangle inequality, so the centres belong to as required by [F6]. Thus is totally bounded.
By [F6] the closure is totally bounded, and it is closed in the complete space , hence complete by [F7]; a complete and totally bounded metric space is compact by [F7], and compactness is equivalent to sequential compactness by [F7], so every sequence in has a subsequence converging in . The empty case was disposed of in step 1.1, and no other choice principle is used: Countable Choice covers the completeness-to-compactness step and the finite-net selection, Dependent Choice covers the compactness-sequential equivalence.
Compactness of on bounded open sets
Statement
Assume the Axiom of Choice. Let , let be a bounded open set and let . Then is compactly embedded in : the inclusion is bounded, and every sequence bounded in has a subsequence converging in . No regularity of is needed.
Facts & Assumptions
Given: the Axiom of Choice, , a bounded open , , and a sequence with .
Zero extension. For each the zero extension of lies in with the zero extension of and ; the extension vanishes outside . (Zero extension of W_0^{1,p} has no boundary derivative, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms)
Translation estimate. , under the Axiom of Choice. (The translation estimate for functions on )
Automatic tails. A family in whose members all vanish almost everywhere outside the fixed bounded set satisfies the tightness condition of the Fr'echet--Kolmogorov criterion. (Uniformly supported families have vanishing tails, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space)
The Fr'echet--Kolmogorov criterion. A bounded family in with vanishing tails and uniform translation control is totally bounded, its closure is compact, and every sequence in it has an -convergent subsequence. (The Fr'echet--Kolmogorov compactness criterion in , The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Restriction is contractive. , and the norm of the restriction never exceeds the norm of an extension. (Integer-order Sobolev spaces and their norms, The space as the quotient by null functions)
Proof
By [F1] the extensions satisfy , and the finite-dimensional equivalence of norms gives since each component is bounded in by ; moreover each vanishes outside the fixed bounded set .
The family meets the hypotheses of [F4]: it is bounded by step 1.1; its tails vanish by [F3]; and by [F2] , a bound uniform in that tends to with .
By [F4] there is a subsequence converging in , say to ; restricting to gives by [F5], so converges in . Boundedness of the inclusion is the inequality of [F5]; the extraction uses the Countable and Dependent Choice of [F4], and the Axiom of Choice is inherited through the translation estimate [F2].
Compactness of on bounded extension domains
Statement
Assume the Axiom of Choice. Let , let be a bounded -extension domain (Sobolev extension domains and extension operators) and let . Then is compactly embedded in : every sequence bounded in has a subsequence converging in . Every bounded domain is an example.
Facts & Assumptions
Given: the Axiom of Choice, a bounded -extension domain , , and a sequence with .
Extension operator. There is a bounded linear with and . (Sobolev extension domains and extension operators)
Cutoff. Since is compact and contained in the open set , with large enough to contain , there is with on and , a fixed bounded set. This construction also works for the empty domain, choosing any ball . (A Euclidean bump for a compact set inside an open set, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space)
Products with a smooth cutoff. If and , then with for a constant depending only on and . (Weak Leibniz rule with a smooth factor)
Translation estimate and automatic tails. Under the Axiom of Choice, ; a family supported in one fixed bounded set has vanishing tails. (The translation estimate for functions on , Uniformly supported families have vanishing tails)
The Fr'echet--Kolmogorov criterion. A bounded family in with vanishing tails and uniform translation control has an -convergent subsequence. (The Fr'echet--Kolmogorov compactness criterion in , The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Restriction and norms. ; and almost everywhere. (Integer-order Sobolev spaces and their norms, The space as the quotient by null functions)
Bounded domains are extension domains. (Bounded C^k domains admit integer-order Sobolev extension)
Proof
Fix as in [F1] and as in [F2], and put . By [F3] each lies in and is supported in the fixed bounded set ; moreover , and almost everywhere on because there.
The family satisfies the three hypotheses of [F5]: it is bounded in by step 1.1; its tails vanish by [F4] because all members are supported in the fixed bounded set ; and [F4] gives with and independent of , so the translation control is uniform and tends to with .
By [F5] some subsequence converges in , say to ; restricting and using almost everywhere on together with [F6] gives , so converges in . This proves compactness of the inclusion; its boundedness follows from by [F1] and [F6]. Finally, [F7] says every bounded domain carries such an extension operator, giving the stated examples. The Axiom of Choice is inherited through [F1], [F4] and [F7], while the extraction uses the Countable and Dependent Choice of [F5].
Poincare-Wirtinger on bounded connected extension domains by Rellich compactness
Statement
Assume the Axiom of Choice. Let , let be a nonempty bounded connected extension domain, and let . Then there is such that every satisfies
Facts & Assumptions
Given: the Axiom of Choice, a nonempty bounded connected extension domain of finite positive measure, , and the mean of . Nonempty openness supplies a ball inside , and boundedness supplies a containing ball; thus by Euclidean balls have positive finite Lebesgue measure.
Rellich compactness. Every sequence bounded in has a subsequence converging in . (Compactness of on bounded extension domains, Sobolev extension domains and extension operators)
The mean is continuous for the norm. and hence , by H"older's inequality on the finite-measure set . (Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Weak gradients vanish when tested against convergent subsequences. If in and , then for every and every coordinate , . (Integer-order Sobolev spaces and their norms, The space as the quotient by null functions)
Zero gradient implies constancy on components. If and almost everywhere for all , then is almost everywhere constant on each connected component of . (Zero weak gradient gives componentwise constants)
Proof
Suppose the assertion fails: for every there is with , and is not almost everywhere constant. Put ; then the mean of is , , and , so for all .
By [F1] there is a subsequence in . By [F2] and step 1.1 the means pass to the limit, so ; and because .
By [F3] applied to the convergent subsequence of step 2.1 with , for every test function and every , so almost everywhere and ; [F4] then makes an almost everywhere constant on the connected , and since its mean is that constant is , contradicting from step 2.1. Hence the constant exists. Countable Choice selects the violating sequence in step 1.1; the assumed Axiom of Choice also supplies [F1] and [F4].
Local compactness of -bounded sequences
Statement
Assume the Axiom of Choice. Let , let be open and let . Let be a sequence that is bounded in : for every there is with for all . Then has a subsequence converging in .
If additionally , the limit of that subsequence can be chosen in , and then it is the limit in . For membership of the limit in is not asserted: the weak-compactness argument below uses the reflexivity of , which fails at , and a weak limit of gradients need not be an function. The compactness conclusion itself is proved for every .
Facts & Assumptions
Given: the Axiom of Choice, an open set , , and a sequence bounded in .
Countable relatively compact ball cover. The rational balls with , and form a countable cover of : given , openness and density of and give such a ball containing . Countability follows from countability of and finite products, and each closed ball is compact by Heine--Borel. (The rationals embed densely in the reals, is countably infinite, A product of two at most countable sets is at most countable, Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line)
Rellich on smooth balls. Every bounded ball is a bounded smooth extension domain, and every sequence bounded in on such a ball has a subsequence converging in . (Bounded C^k domains admit integer-order Sobolev extension, Compactness of on bounded extension domains)
Reflexivity of . Assume Countable Choice. For and every measure space, is reflexive, so every norm-bounded sequence in has a weakly convergent subsequence under the ultrafilter lemma, Dependent Choice and Hahn--Banach, all supplied by the Axiom of Choice. (Reflexivity of Lp for one less p less infinity, Reflexivity is equivalent to weak subsequential compactness of bounded sequences, Weak convergence of nets and sequences, The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Weak derivatives pass to weak limits. If and in , , then with : test against and pass to the limit in . (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function, Weak convergence of nets and sequences)
A compact subset of has a finite subcover from any open cover of , in particular from the ball cover in [F1]. (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
If , the assertions are immediate. Otherwise enumerate the countable cover in [F1] as . The local boundedness hypothesis makes bounded in , so [F2] gives a subsequence converging in . Recursively, after obtaining a subsequence converging on , apply [F2] to that subsequence on and retain a further subsequence converging there. Countable and Dependent Choice select these nested subsequences. The diagonal sequence , taking the -th term of the -th subsequence, is eventually a subsequence of each stage; hence it converges in for every .
Let be the limit of . On every overlap the limits and agree almost everywhere, by uniqueness of limits of the same sequence in . Since the cover is countable, these compatible classes patch to a class . If , compactness and [F5] give a finite subcover ; therefore Thus in .
Suppose and fix . The sequence is bounded in . By [F3], after finitely many further subsequence extractions, its function and each of its weak derivatives converge weakly in , say and . Step 2.1 gives strong convergence to in , so the weak limit is . Passing to the limit in the weak-derivative identities against each and using [F4] gives with . This argument may use a further subsequence depending on : it identifies the already fixed strong limit , so it identifies the derivatives of the already fixed limit on every ball without changing the diagonal sequence of step 1.1. On overlaps these derivative classes agree by the weak test identity. For any , choose a finite ball subcover of and an ambient smooth partition equal to one near that compact set (Finite ambient partitions near compact sets). Testing after multiplication by the partition pieces proves the weak derivative identity on ; the partition-gradient terms sum to zero, and the finitely many local bounds give global bounds. Thus . For this weak-compactness step is unavailable, and no membership of the limit in is asserted. The assumed Axiom of Choice supplies the countable selections in step 1.1 and the Countable and Dependent Choice interfaces of [F2] and [F3].
Subcritical compactness for on arbitrary bounded open sets
Statement
Assume the Axiom of Choice. Let , let be a bounded open set with no boundary regularity assumed, let and . For every the space is compactly embedded in : every sequence bounded in has a subsequence converging in .
Facts & Assumptions
Given: the Axiom of Choice, a bounded open set , , , a target exponent , and a sequence with .
-compactness. Some subsequence converges in ; the extraction uses Countable and Dependent Choice. (Compactness of on bounded open sets, The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Sobolev inequality for of an arbitrary bounded open set. for all , with ; the zero extensions of the therefore satisfy . (The Sobolev inequality for zero-boundary Sobolev closures on open sets, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The Sobolev conjugate exponent and the scaling identity)
For , supply the endpoint separately: the zero extension of a zero-boundary class belongs to . Choose smooth compactly supported in by Compactly supported smooth functions are dense in W^{k,p}(R^n). The endpoint The p=1 Gagliardo-Nirenberg-Sobolev inequality on differences makes Cauchy in ; completeness and the almost-everywhere subsequence theorem identify this limit with , since in as well. Passing to the limit in the endpoint inequality gives , and restriction supplies the bound asserted in [F2]. (Riesz-Fischer completeness of for , Assuming Countable Choice, -convergent sequences have almost-everywhere convergent subsequences)
Lyapunov interpolation and H"older. For and , ; for , . (Lyapunov interpolation inequality for norms, Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Under Countable Choice, each , , is complete. (Riesz-Fischer completeness of for )
Proof
If , all classes are zero and the claim is immediate. Otherwise, by [F1] extract a subsequence converging in ; write . By [F2] the differences satisfy for all .
Fix . If , then [F3] gives by step 1.1; at the factor is . If , choose with ; [F3] gives by step 1.1. In both cases is Cauchy in , hence converges there by [F4].
Every bounded sequence in therefore has a subsequence convergent in , which is the asserted compact embedding; no property of was used. The Axiom of Choice is inherited through [F1] and the supplier [F2]; the extraction uses the Countable and Dependent Choice of [F1].
The Rellich--Kondrachov theorem for on bounded extension domains
Statement
Assume the Axiom of Choice. Let , let be a bounded extension domain, let and . Then for every the inclusion is bounded and compact: every sequence bounded in has a subsequence converging in .
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , , , a target exponent , and a sequence with .
-compactness. Some subsequence converges in . (Compactness of on bounded extension domains, Sobolev extension domains and extension operators, Compactly embedded normed spaces)
Sobolev embedding on bounded extension domains. for all , , hence and . (Sobolev embedding on bounded extension domains for , The Sobolev conjugate exponent and the scaling identity, Integer-order Sobolev spaces and their norms)
For , supply the endpoint separately: take the given extension and smooth compactly supported in by Compactly supported smooth functions are dense in W^{k,p}(R^n). The endpoint The p=1 Gagliardo-Nirenberg-Sobolev inequality on differences makes Cauchy in ; completeness and almost-everywhere subsequences identify this limit with , since in . Passing to the limit gives , and restriction gives [F2]. (Riesz-Fischer completeness of for , Assuming Countable Choice, -convergent sequences have almost-everywhere convergent subsequences)
Interpolation and H"older. For , with ; for , . (Lyapunov interpolation inequality for norms, Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Under Countable Choice, each , , is complete. (Riesz-Fischer completeness of for )
Proof
If , all classes are zero and the claim is immediate. Otherwise, by [F1] extract a subsequence converging in and write ; by [F2] the differences satisfy .
Fix . If then by [F3] and step 1.1; if then [F3] gives . Completeness [F4] therefore makes converge in .
Boundedness of the inclusion holds for every by [F2] and the interpolation bound of [F3], and for by H"older's inequality of [F3]; compactness is the extraction just proved, so in the sense of Compactly embedded normed spaces. The Axiom of Choice is inherited through [F1] and the supplier [F2].
Rellich--Kondrachov at the critical source exponent
Statement
Assume the Axiom of Choice. Let and let be a bounded extension domain. Then is compactly embedded in for every finite : for each fixed , every sequence bounded in has a subsequence converging in . There is no claim of compactness into .
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , , and a sequence with .
-compactness. Some subsequence converges in . (Compactness of on bounded extension domains, Sobolev extension domains and extension operators, Compactly embedded normed spaces)
Critical embedding into every finite . For every finite there is with for all ; the sequence is therefore uniformly bounded in for each fixed finite . (Higher-order Sobolev embedding (case , ), Integer-order Sobolev spaces and their norms)
Interpolation and H"older. For , with ; for , . (Lyapunov interpolation inequality for norms, Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Under Countable Choice, each , , is complete. (Riesz-Fischer completeness of for )
Proof
If , all classes are zero and the claim is immediate. Otherwise fix and choose . By [F1] extract a subsequence converging in and write ; by [F2] the differences satisfy .
If , then by [F3] and step 1.1. If , then and [F3] gives for the corresponding . In both cases is Cauchy, hence convergent by [F4], in .
Every bounded sequence in therefore has a subsequence converging in for each fixed finite , so in the sense of Compactly embedded normed spaces. No compactness into is asserted. The proof uses only finite target exponents. The Axiom of Choice is inherited through [F1] and the supplier [F2].
Morrey--Rellich compactness for
Statement
Assume the Axiom of Choice. Let , let be a bounded extension domain, let and . Replace every by its continuous Morrey representative (Morrey's inequality for ). Then every sequence bounded in has a subsequence whose representatives converge in for every ; in particular is compactly embedded in every , , and in for every .
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , , , and a sequence with .
Extension and cutoff. There are a bounded extension operator and a fixed with on ; the products lie in , are supported in the fixed compact set , satisfy , and equal almost everywhere on . (Sobolev extension domains and extension operators, A Euclidean bump for a compact set inside an open set, Weak Leibniz rule with a smooth factor, Integer-order Sobolev spaces and their norms)
Morrey's inequality on the compact set used here. Each in [F1] is supported in a fixed compact set . Choose one ball containing . The supplier's local estimate on , with , gives for the continuous representative. Its average on has modulus at most , so the same oscillation estimate also bounds by . Continuous representatives are unique because continuous functions equal almost everywhere on an open ball are equal everywhere there. Thus both norms on are bounded by , with constants depending only on . (Morrey's inequality for , Local Hölder and scaled C-two-alpha norms on balls, The space as the quotient by null functions)
Arzel`a--Ascoli. A uniformly bounded, equicontinuous family of real functions on a compact metric space has a uniformly convergent subsequence; for complex-valued families apply this to real and imaginary parts. (Arzelà--Ascoli for real under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded)
H"older interpolation. For a bounded function on any set , write and . For and , the bound gives and hence These follow from for ; the seminorm and supremum conventions agree with Local Hölder and scaled C-two-alpha norms on balls.
Proof
If the claim is immediate. Otherwise, by [F1] and [F2] each satisfies and ; hence the family is uniformly bounded and -H"older, in particular equicontinuous, on the compact set .
By [F3] applied to the real and imaginary parts on the compact metric space , a subsequence of converges uniformly, that is, in ; along it the seminorms stay bounded by step 1.1.
Fix and put . For the differences of the uniformly convergent subsequence, step 1.1 gives , while . By [F4], If is the uniform limit, passing to the limit in each difference quotient shows . Apply the same estimate to to obtain convergence in ; the case is step 2.1. Since almost everywhere on by [F1] and [F2], this is the convergence of the Morrey representatives of the , and uniform convergence on the bounded implies convergence in for every , so is compactly embedded in each , , and in each , . The Axiom of Choice is inherited through the extension operator and Morrey's inequality.
Higher-order Rellich--Kondrachov compactness
Statement
Assume the Axiom of Choice. Let , let be a bounded extension domain, let be integers and ; write when .
- (a) If , then is compactly embedded in for every .
- (b) If , then is compactly embedded in for every finite .
- (c) If and with , then every sequence bounded in has a subsequence whose representatives converge in .
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , integers , , and a sequence with .
Lower-order derivatives. For every , the sequence is bounded in , with norm at most , because . (Weak partial derivatives lower the Sobolev order, Integer-order Sobolev spaces and their norms, The notation and the reserved zero-boundary symbol)
First-order compactness. On a bounded extension domain, every sequence bounded in has a subsequence converging in . (Compactness of on bounded extension domains)
Higher-order continuous embeddings. Applying the higher-order Sobolev embedding on the support ball to each , where is the compactly supported extension in step 1.1, gives, uniformly in and , an bound when , a bound in every finite when , and a bound for by restriction when and . For derivatives with , the remaining Sobolev order is larger; finite-measure inclusion handles any stronger resulting integrability. (Weak partial derivatives lower the Sobolev order, Higher-order Sobolev embedding, Sobolev extension domains and extension operators, Compactly embedded normed spaces)
Finite-measure inclusion, interpolation, and completeness. If , then ; if and , then . Each is complete for . (Holder's inequality for integrals, including the endpoint cases, Lyapunov interpolation inequality for norms, Riesz-Fischer completeness of for , The space as the quotient by null functions)
Weak derivatives pass to strong limits. If and in for , passing to the limit in the test identity shows weakly. Iterating gives the same conclusion for all derivatives of order at most . (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms)
Arzel`a--Ascoli. A uniformly bounded equicontinuous family on a nonempty compact metric space has compact closure in the supremum norm; under Countable and Dependent Choice this gives a uniformly convergent subsequence. For complex functions apply the real result to real and imaginary parts. (Arzelà--Ascoli for real under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded, For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, The Axiom of Countable Choice (), The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain)
Uniform limits preserve classical derivatives. If functions and their first derivatives converge uniformly on compact balls, the limit is there and its derivatives are the corresponding limits; apply the fundamental theorem of calculus on line segments, coordinate by coordinate. (Fundamental theorem of calculus for absolutely continuous functions, maps and multi-index derivative notation in Euclidean space)
H"older interpolation. For , a uniformly convergent sequence with uniformly bounded seminorms converges in . Indeed, the difference quotient is bounded by the minimum of and , yielding the usual interpolation estimate with a constant depending on . (Local Hölder and scaled C-two-alpha norms on balls)
Proof
If , all target spaces are trivial and the assertions hold. Otherwise choose the bounded extension at and a smooth cutoff equal to one near , with compact support in a ball . By the weak Leibniz rule, is bounded in and equals on (A Euclidean bump for a compact set inside an open set, Weak Leibniz rule with a smooth factor). For each , is bounded in by [F1]. The smooth ball is a -extension domain (Bounded C^k domains admit integer-order Sobolev extension), so [F2] applied on , followed by restriction to , gives a common subsequence on which all converge in . Only finitely many derivative sequences are extracted.
Consider cases (a) and (b). By [F3], for every the sequence is bounded in in case (a), and in every finite in case (b). If , finite- measure inclusion [F4] and step 1.1 make each derivative sequence Cauchy in . If in case (a), choose ; if in case (b), choose any finite . Lyapunov interpolation [F4] applied to differences, whose norms tend to zero by step 1.1 and whose norms are uniformly bounded by [F3], makes every derivative sequence Cauchy in . Completeness of gives limits . Passing to the limit in the weak derivative test identities by [F5] shows that for all , so in . The higher-order embedding [F3] also gives boundedness of the inclusion into each stated target, so this is compact embedding.
Consider case (c), and fix . Choose with . By [F3] the sequence is bounded in , so each of its finitely many derivative families of orders at most is uniformly bounded and equicontinuous on the compact set . By [F6], applying Arzel`a--Ascoli successively to these derivative families gives a common subsequence on which every converges uniformly to a continuous function on . On each ball compactly contained in , [F7] applied to line segments shows that is the classical -th derivative of whenever ; hence is a representative in whose derivatives through order extend continuously to . For the uniform convergence is the desired convergence. For , [F8] applied to each difference , using uniform convergence and the uniform bounds, gives convergence in to : each uniform limit retains the bounded seminorm by passage to the limit in the pointwise difference quotients, so [F8] applies directly to . Thus the asserted compact embedding holds, and the Axiom of Choice supplies the subsequence and the choice interfaces of [F2] and [F6].
Bounded Sobolev sequences have strongly convergent subsequences with the weak limit as limit
Statement
Assume the Axiom of Choice. Let , let be a bounded extension domain, let , , and let be bounded in with weakly in . Then in for every ; the convergence is of the whole sequence, not merely of a subsequence.
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , , a bounded sequence with weakly in , and .
Rellich--Kondrachov. Every bounded sequence in has a subsequence converging in . (The Rellich--Kondrachov theorem for on bounded extension domains, Sobolev extension domains and extension operators)
Weak convergence tested against . For with , the functional is bounded on , so . In particular for every measurable of finite measure. For complex-valued classes, these integral identities are read componentwise. (Weak convergence of nets and sequences, Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Strong convergence tested against . If in and , then by H"older's inequality. (Holder's inequality for integrals, including the endpoint cases)
Proof
Proof technique: every subsequence has a further -convergent subsequence; identify its limit with the weak limit by testing against finite-measure set indicators; conclude that the whole sequence converges.
Let be any subsequence. It is bounded in , so by [F1] it has a further subsequence converging in to some .
For every measurable of finite measure, [F2] applied to gives , while [F3] applied to gives ; hence . For each , apply this to the sets where the real or imaginary part of is greater than or less than , intersected with . Each such set has measure zero, since the corresponding signed part of the integral has magnitude at least its measure divided by . As is bounded, these sets cover the nonzero real and imaginary parts, so almost everywhere on .
Thus every subsequence of has a further subsequence converging in to the same limit ; in a metric space this forces the whole sequence to converge to , because otherwise some would admit a subsequence staying -away from , and that subsequence would in turn have a further subsequence converging to . The Axiom of Choice is inherited through [F1], and the weak topology is Hausdorff as recorded in Weak topology is hausdorff.
Remarks
The identification of the strong limit with the weak limit does not use the density of test functions in for : the finite-measure indicator test functions lie in and separate almost-everywhere classes.
Weak convergence plus compactness gives strong convergence
Statement
Assume the Axiom of Choice. Let , let be a bounded extension domain and let be bounded in with weakly in . Then in .
Facts & Assumptions
Given: the Axiom of Choice, a bounded extension domain , , and a bounded sequence with weakly in .
First-order Rellich compactness. Since , every bounded sequence in has a subsequence converging in on this bounded extension domain. (Compactness of on bounded extension domains, The notation and the reserved zero-boundary symbol)
Finite-measure indicators test both limits. For each measurable , the functional is bounded on and on by H"older's inequality. Weak convergence and strong convergence therefore give the same limit for these integrals. Since is bounded, ; if its integral over every measurable set is zero, then its real and imaginary parts vanish almost everywhere. (Weak convergence of nets and sequences, Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Proof
Let be any subsequence. It is bounded in , so by [F1] it has a further subsequence converging in to some .
For each measurable , [F2] gives by weak convergence and by strong convergence. Thus for every such . Applying this to the sets where the real or imaginary part of is greater than or less than , for , shows that each part vanishes almost everywhere; hence in .
Every subsequence of therefore has a further subsequence converging in to . If the whole sequence did not converge to , some would admit a subsequence staying at distance at least from , contradicting the further-subsequence conclusion. Thus in . The Axiom of Choice is inherited through [F1].
Rellich compactness is strictly subcritical
Remarks
The compactness statements of this page are strictly subcritical ; the companion page gives witnesses for the critical target exponents and for escape to infinity.
- (i) For on a bounded domain the continuous embedding is not compact, so the strict inequality in The Rellich--Kondrachov theorem for on bounded extension domains and Subcritical compactness for on arbitrary bounded open sets cannot be replaced by (witness: a concentrating bump sequence ↗).
- (ii) On boundedness in and translation control alone do not give compactness: the family of translates of one compactly supported bump satisfies the boundedness and translation hypotheses of The Fr'echet--Kolmogorov compactness criterion in but not its tightness hypothesis, and has no strongly convergent subsequence (The tightness hypothesis of the Fr'echet--Kolmogorov criterion cannot be dropped ↗); the tightness hypothesis is therefore indispensable.
- (iii) In the Morrey range the endpoint H"older exponent is excluded from Morrey--Rellich compactness for (witness: rescaled H"older spikes ↗).
- (iv) At the critical source exponent , Rellich--Kondrachov at the critical source exponent gives compactness into every finite and makes no compactness claim. When is nonempty, the local log-log test function constructed in the proof of The critical Sobolev embedding into every finite belongs to and is essentially unbounded; this construction needs only an interior ball, independently of the extension hypotheses. For each its superlevel set has positive measure, so forces the best continuous embedding constants to tend to infinity as . On the empty domain all spaces are zero.
For the compactness theorems on extension domains, boundedness and the extension hypothesis are part of the stated setting. The compactness results need only bounded openness and zero extension, with no boundary regularity. On , translations and dilations destroy compactness when the corresponding tail control is absent.
The level-set kernel measure estimate for the Slobodeckij kernel
Statement
Assume the Axiom of Countable Choice. Let , , with , let and let be Lebesgue measurable with . Then One admissible constant is , where denotes the Lebesgue measure of the unit ball.
Facts & Assumptions
Given: the Axiom of Countable Choice, , , with , a point , and a Lebesgue measurable set with . Write for the unit-ball measure and .
The unit ball has positive finite measure. . (Euclidean balls have positive finite Lebesgue measure, Lebesgue measurable sets, the family , and the restricted set function )
change of variables for nonnegative Borel functions. If are open and is a diffeomorphism, then every nonnegative Borel satisfies , with allowed. (Borel change of variables from the compact-support formula and Radon uniqueness)
Polar coordinates. For every nonnegative Borel , , where is the finite Borel surface measure on the unit sphere. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere)
Measures of set differences. If are measurable with , then . (Measure of a set difference when the smaller set has finite measure)
Additivity and monotonicity of the nonnegative integral. For measurable : , and implies ; moreover for real ; for the zero function has integral . (Additivity of the nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral)
Proof
The map is a diffeomorphism of with , so [F2] applied to the indicator of gives . Since and are measurable with , [F4] gives .
Write and ; these are disjoint measurable sets with union . On one has , on and on one has ; hence [F5] gives , where the last equality uses that and partition .
Substituting by [F2] and evaluating the radial integrand by [F3], . Applying [F3] to the indicator of gives , so and the lower bound is with by [F1]; combined with step 2.1 this is the assertion.
A dyadic summability estimate for decreasing level-set sequences
Statement
Let , , with and . Let be a bounded nonnegative nonincreasing sequence of real numbers with for all sufficiently large . Then The constant is explicit: . The argument uses no choice principle; all sums are series of nonnegative terms.
Facts & Assumptions
Given: integers and indices, numbers , with , a real , and a bounded nonnegative nonincreasing sequence with for all sufficiently large . Put , and , so that .
Hölder's inequality. On a measure space , for conjugate exponents and nonnegative measurable with finite respective norms, , which is the finite-norm form used below. Step 1.1 establishes the required finite sums before the application. (Holder's inequality for integrals, including the endpoint cases)
Proof
Since for all and is bounded by some , the sum satisfies , and the sum satisfies because with implies . Also implies , so .
Shifting the index in step 1.1 and dropping exactly the vanishing terms, . For each with the factorization holds, because , and . Applying [F1] with the counting measure on the set to these two factors gives , since raising the first factor-sum to the power returns and raising the second to returns .
If then every and both sides of the asserted inequality are . Otherwise by step 1.1, so dividing step 2.1 by gives , hence and therefore . Substituting gives and , which is the assertion with . No choice principle is used: both series are sums of nonnegative real terms over a countable index set, evaluated as suprema of finite partial sums.
The Slobodeckij seminorm bounds the dyadic level-set sum
Statement
Assume the Axiom of Countable Choice. Let , , with , let have compact support, and put , . Then where is the Slobodeckij seminorm of The Gagliardo--Slobodeckij space on Euclidean space.
Facts & Assumptions
Given: the Axiom of Countable Choice, , , with , a compactly supported , and the sets with . Write , , and , .
Level sets and annuli. , so and ; the are pairwise disjoint, up to a null set, , and for all large because is bounded with compact support. With , for every we have up to a null set. All these sets are measurable. (Lebesgue measurable sets, the family , and the restricted set function , Finite and countable subadditivity of measures, Measure of a set difference when the smaller set has finite measure)
Kernel estimate. If is measurable with and , then with independent of and . (The level-set kernel measure estimate for the Slobodeckij kernel)
Slobodeckij seminorm on disjoint blocks. For measurable , ; sums over pairwise disjoint such blocks of a nonnegative integrand are bounded by the total integral. (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)
Proof
By [F1] the annuli are pairwise disjoint measurable sets with and , and for all large. Let . If and with , then and , so . The same bound holds for , since and . Thus the low positive bands together with cover up to a null set.
If , the conclusion is immediate. Otherwise the disjoint-block sum below is finite by [F3]. The sums are finite: , the are bounded and eventually zero, and , so the negative tail is geometric. Fix with . By [F2] applied to (which has ), for every the integral of over is at least . Since up to a null set, the lower bound from step 1.1 gives where . Writing and summing over with , where and . Swapping the order of summation in and using whenever gives , where . On the other hand, the block bound itself gives , so and therefore , that is .
The blocks with and are pairwise disjoint: the are disjoint in the first coordinate, and for each the second-coordinate bands and are disjoint. Hence [F3] gives . Relabelling in yields , so , which is the assertion with .
The critical fractional Sobolev inequality on
Statement
Assume the Axiom of Countable Choice. Let , , with and . There is such that every measurable, compactly supported satisfies Consequently, for every bounded open and every there is with for every compactly supported measurable .
Facts & Assumptions
Given: Countable Choice, , , with , and ; write , so that and .
Dyadic summability. For a bounded nonnegative nonincreasing sequence vanishing for all large and , . (A dyadic summability estimate for decreasing level-set sequences)
Level-set bound. For compactly supported with , . (The Slobodeckij seminorm bounds the dyadic level-set sum)
Fatou and dominated convergence. Fatou's lemma bounds the integral of a pointwise limit below by the lower limit of the integrals; dominated convergence applies under an integrable dominating function. (Fatou's lemma, Dominated convergence)
Interpolation and H"older. For , let be defined by . Lyapunov interpolation, with its parameter , gives . For on a set of finite measure, . (Lyapunov interpolation inequality for norms, Holder's inequality for integrals, including the endpoint cases)
Seminorm and classes. is the Slobodeckij seminorm, finite on . The scalar truncation is 1-Lipschitz, hence and . (The Gagliardo--Slobodeckij space on Euclidean space, The space as the quotient by null functions)
Proof
Let first have compact support, put and . On one has , and the partition , and vanishes on the remaining set, so ; raising to the power and using the concavity bound for , whose -th powers are , gives . Since the sequence is bounded, nonincreasing and eventually , [F1] followed by [F2] bounds the last sum by a constant times , proving the inequality for this .
For general compactly supported measurable with put . Then pointwise with , so by the pointwise contraction in [F5]; the bounded case of step 1.1 gives , and Fatou's lemma [F3] passes to the limit: , which is the asserted inequality.
Let be bounded open and . If or , the conclusion is immediate. For , Holder [F4] on the finite-measure set gives . If , choose so that . Lyapunov [F4], the embedding of the restricted and norms below their global norms, and step 2.1 give Put and . Weighted AM--GM yields . At , step 2.1 directly gives . These estimates, and the Holder bound, give the claimed consequence with a constant depending on . Countable Choice is inherited through [F1], [F2] and [F3].
Mollification rates for compactly supported Slobodeckij functions
Statement
Assume the Axiom of Countable Choice. Let , , , let have compact support, and let be the mollification of by a radial mollifier with . Then is supported in the -neighbourhood of ; There are constants and such that
Facts & Assumptions
Given: the Axiom of Countable Choice, , , , a compactly supported , a radial mollifier with , and . Write for .
Slobodeckij seminorm as a translation integral. Because the diagonal is null and Tonelli's theorem together with the substitution applies, . (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a function on )
Volume scaling. If , then with for every . (Euclidean balls have positive finite Lebesgue measure, A linear map of sends Lebesgue measurable sets to Lebesgue measurable sets, with when is invertible and Lebesgue null when it is not)
The translation modulus is subadditive. is nondecreasing, and for all , because when and translations are isometries of . (Translation of a function on , Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, The space as the quotient by null functions)
Minkowski's integral inequality. For measurable on a product of sigma-finite measure spaces with , . (Minkowski's integral inequality)
Young's convolution inequality. . (Young's convolution inequality under Countable Choice)
Mollifying a locally integrable function. For the convolution is smooth with ; an function is locally integrable, and a compactly supported function lies in . (Convolution with a mollifier is smooth, and derivatives pass under the integral sign, Holder's inequality for integrals, including the endpoint cases)
Support of a convolution. For Borel representatives of , . (The support of a convolution lies in the closure of the support sumset)
The radial mollifier. satisfies , , and for every , since is radial and its gradient is odd in each coordinate. (A radial mollifier family in Rn)
Proof
Fix and . By [F3], for every we have . Raising to the -th power and averaging over gives because both and lie in and [F2] gives . Taking the supremum over , then using [F1] and for , yields Thus for a constant , which is the translation-modulus estimate needed below.
Since and is supported in , ; taking -norms and applying [F4] with gives , which is (i).
By [F6], is smooth and ; [F6] also gives because has compact support and lies in . By [F8], , so , and [F4] gives because . Moreover by [F5] with , so , which is (ii) after enlarging the constant. Finally, by [F6], and [F7] applied to the Borel representative of and to gives , the -neighbourhood of ; this is compact because is compact, so .
Subcritical compactness for compactly supported Slobodeckij functions
Statement
Assume the Axiom of Choice. Let , , with and . Let be a family of functions all supported in one fixed bounded set. In the displayed nonnegative supremum, take the value if . Assume it satisfies . Then is relatively compact in for every : every sequence in has a subsequence converging in .
Facts & Assumptions
Given: the Axiom of Choice, , , with , , a family supported in one fixed bounded set and bounded in the norm by , and .
Fractional Sobolev inequality. For real compactly supported , ; hence for compactly supported . For complex , apply the real inequality to its real and imaginary parts, whose seminorms are at most , and use the triangle inequality, enlarging the constant by at most . (The critical fractional Sobolev inequality on )
Mollification rates. With the radial mollifier at scale , and ; the mollified functions are supported in the -neighbourhood of the fixed support set, and because and Minkowski's inequality applies in the weighted -space of the seminorm. (Mollification rates for compactly supported Slobodeckij functions, The Gagliardo--Slobodeckij space on Euclidean space)
First-order compactness on a ball. For fixed , the mollified family is bounded in on a smooth ball containing all its supports. Its closure in is compact by the first-order Rellich theorem, including dimension one. (Compactness of on bounded extension domains)
Interpolation, completeness and compactness. Strict interpolation between and is supplied by Lyapunov interpolation inequality for norms; finite-measure inclusion follows from Holder's inequality for integrals, including the endpoint cases. Under Countable Choice, is complete and total boundedness passes to closures, so a totally bounded family has compact closure. Under Countable and Dependent Choice the closure is sequentially compact. (Riesz-Fischer completeness of for , A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded, A complete, totally bounded metric space is compact, proved from countable choice used exactly once, For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, Finite -net and totally bounded metric space)
Proof
The empty family is immediate. Otherwise fix a ball containing the common bounded support and its distance-one neighbourhood, and use only . By [F2], is supported in and bounded in ; [F3] makes it totally bounded in , since restriction to and zero extension preserve distances on this family. Also . Given , choose making this error less than and a finite -net for . Its centres cover with radius ; choosing one point of in every nonempty such ball moves the centres into and gives an -net. Thus is totally bounded in .
Fix . If , step 1.1 applies. If , all members and their differences vanish off , so by [F4]. If , [F1] gives for , and [F4] gives with and . Hence a sufficiently fine finite net with centres in is an net in each case. If , the family contains only the zero class.
By [F4] the totally bounded closure in the complete space is compact and sequentially compact, giving the asserted subsequence for every sequence in . The assumed Axiom of Choice supplies the first-order Rellich interface and Countable and Dependent Choice in [F4].
Subcritical compactness of the Sobolev trace
Statement
Assume the Axiom of Choice. Let , let be a bounded domain, let be the trace of The trace operator on a bounded domain, and let with . Then is compact as a map into for every : every sequence bounded in has a subsequence whose traces converge in . If , then the traces of a suitable subsequence converge in for every , hence also in every , ; the endpoint case is not claimed.
Facts & Assumptions
Given: the Axiom of Choice, a bounded domain , , and , , with a sequence bounded in .
Sharp trace boundedness. For and , the trace satisfies , where the boundary norm is the finite sum over a finite atlas of Euclidean -norms of compactly supported chart representations, and it is independent of the atlas up to equivalence. (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact boundary, Chart independence of the fractional boundary norm)
Fractional compactness in dimension . For and one has and the critical exponent of is ; a family of functions supported in one fixed bounded set and bounded in is therefore relatively compact in for every . (Subcritical compactness for compactly supported Slobodeckij functions, The fractional Sobolev space on a compact boundary)
Trace and chart cutoffs. The trace commutes with multiplication by smooth ambient cutoffs and with the chart parametrisations; on a compact boundary patch the surface-measure density of the parametrisation is continuous and positive, so convergence of the finitely many chart representations gives convergence of their sum. (The trace commutes with smooth cutoffs and is chart local, Surface integration on compact C1 hypersurfaces, Finite ambient partitions near compact sets, Bounded C1 domains and their outward normals)
The Morrey branch. For , the extension theorem at makes the bounded domain a -extension domain; hence a bounded sequence in has a subsequence whose representatives converge in for every , and the trace of such a class is its classical boundary restriction. (Bounded C^k domains admit integer-order Sobolev extension, Morrey--Rellich compactness for , The trace agrees with classical restriction for continuous Sobolev functions)
Proof
Assume . By [F1] the traces satisfy with ; by the definition of the boundary norm this means that each of the finitely many compactly supported chart representations of the traces is bounded in .
Choose exponents with . For each , [F2] applied successively on the finitely many charts supplies a common subsequence converging in every chart in . Dependent Choice selects nested subsequences for ; their diagonal converges in each chart for each . For any , choose with and use finite-measure inclusion on the common bounded chart supports. The chart Jacobian is bounded on each compact support, so [F3] transfers convergence of the finitely many chart pieces to convergence of their sum in . Thus the same subsequence works throughout the stated range.
If , [F4] first verifies the extension-domain hypothesis and then provides a subsequence of the whose representatives converge in for every , and their traces, being the classical boundary restrictions, converge in and hence in every , . The endpoint would require the limiting fractional embedding at in dimension and is deliberately not claimed. The Axiom of Choice is inherited through the published trace theorem [F1] and the Morrey branch [F4].
Strong convergence of subcritical powers
Statement
Assume Countable Choice. Let have finite Lebesgue measure, let and , and let be a sequence in with in and . Then , and for every the nonlinear maps converge: The range is nonempty only when ; the endpoint is not asserted.
Facts & Assumptions
Given: Countable Choice, a finite-measure set , exponents , a real number , and real-valued measurable classes on with in and .
Hölder inclusion on a finite-measure space. If and is measurable on the finite-measure space , then whenever , with ; this is Hölder applied to and the constant function . (Holder's inequality for integrals, including the endpoint cases, The space as the quotient by null functions)
Lyapunov interpolation. If , and , then every lies in with . (Lyapunov interpolation inequality for norms)
Almost-everywhere subsequences. Every sequence converging in , , has a subsequence whose representatives converge almost everywhere to a representative of the limit. (Assuming Countable Choice, -convergent sequences have almost-everywhere convergent subsequences)
Fatou's lemma. For nonnegative measurable functions , . (Fatou's lemma)
Mean value theorem. If is continuous on and differentiable on , then for some . (The mean value theorem, as the case of Cauchy's: for continuous on with and differentiable on there is with )
Hölder's inequality for products. For conjugate exponents and measurable , . (Holder's inequality for integrals, including the endpoint cases)
Proof
Proof technique: Extract an almost-everywhere subsequence to obtain the -bound of the limit, transfer convergence to the exponent by interpolation, and apply the pointwise mean value bound followed by Hölder.
By [F3] fix a subsequence convergent almost everywhere to . Then pointwise almost everywhere, so [F4] gives , hence with . Consequently for every .
Fix and put . If then ; if , then [F1] gives . If , choose with ; [F2] applied to the classes gives by step 1.1. In both cases , and by [F1], while by step 1.1.
If then and , so the claim is step 2.1 itself with . If , consider on ; is differentiable with , and for real every point of the closed interval between them satisfies . By [F5] applied to on that interval, Writing , and applying [F6] with exponents and to the product gives and the right-hand side tends to by the bounds and convergence of step 2.1. Hence in .
A bounded map into yields a compact operator
Statement
Assume the Axiom of Choice. Let be a bounded open set and a bounded linear operator. Then is compact, where is the inclusion. No boundary regularity of is needed.
Facts & Assumptions
Given: the Axiom of Choice, a bounded open set , and a bounded linear operator , with the inclusion.
Zero-boundary Rellich theorem. is compactly embedded in : every sequence bounded in has a subsequence converging in . (Compactness of on bounded open sets, Compactly embedded normed spaces, The notation and the reserved zero-boundary symbol)
Bounded operators map bounded sequences to bounded sequences. If satisfies , then for the operator norm of A bounded linear operator between normed spaces. (A bounded linear operator between normed spaces)
Compactness is the sequential extraction criterion. A bounded operator is compact exactly when the image of every bounded sequence has a convergent subsequence. (Compact linear operator, Compactly embedded normed spaces)
Proof
Let be bounded in with . By [F2], is bounded in , so [F1] supplies a subsequence with in , that is, .
Since every bounded sequence in has an image under with a convergent subsequence, [F3] makes a compact operator. The inclusion is bounded because , and the Axiom of Choice is inherited through the Rellich theorem [F1].
Closed target constraints survive compact extraction
Statement
Assume Countable and Dependent Choice. Let be a normed space with compact continuous inclusion , , and let be closed in norm. Every bounded sequence with admits a subsequence in with . In particular this applies to any of this page's Rellich inclusions, with their stated domain, exponent and choice hypotheses. This statement does not assert that belongs to or that is weakly closed.
Facts & Assumptions
Given: Countable and Dependent Choice, a normed space with compact continuous inclusion , a norm-closed set , and a bounded sequence in with for all .
The sequential form of a compact embedding. Under Countable and Dependent Choice, a compact continuous inclusion sends every bounded sequence in to a sequence with a subsequence converging in . (Compactly embedded normed spaces)
Closed sets contain sequential limits. A closed subset of a metric space contains the limit of every convergent sequence of its points: otherwise the open complement contains a ball about the limit, contradicting eventual membership of the sequence in that ball. (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, The closure of a nonempty is , equals together with its limit points, and is the smallest closed superset)
Proof
By [F1] the bounded sequence has a subsequence converging in to some .
Since for every and is closed, [F2] gives ; the statement makes no claim that lies in the image of . Countable and Dependent Choice are used exactly through the compact-embedding interface [F1].
5 · Examples, counterexamples and false statements
None yet.
Sources
- Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete graduate notes)
- John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes)
- Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes)
- Gerald Teschl, Partial Differential Equations: From Classical to Modern (archived 2025 author manuscript)
- Juha Kinnunen, Sobolev Spaces (Aalto University, complete graduate lecture notes)
- Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete text)
- Eleonora Di Nezza, Giampiero Palatucci and Enrico Valdinoci, Hitchhiker's guide to the fractional Sobolev spaces (arXiv:1104.4345, survey)
- John K. Hunter, Notes on Partial Differential Equations, complete 242-page 2014 notes