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Sobolev Poincare and Morrey Inequalities — Examples

1 · Prerequisites

2 · Summary

This companion page carries the computations and witnesses that pin down the constants, exponents and hypotheses of the Sobolev, Poincare and Morrey inequalities on the parent page. The dilation computations show both directions of the scaling argument: writing $u_{λ} (x) = u (λ x)$ forces the balance $1 - n / p + n / q = 0$ and hence the Sobolev conjugate $q = p^{*}$ , while the opposite concentration $φ_{λ} (x) = φ (x / λ)$ makes the $W^{1, p} \to L^{q}$ ratio diverge for every $q > p^{*}$ , so the exponent is sharp. The outward dilation $φ_{k} (x) = φ (x / k)$ shows that on $R^{n}$ the full $W^{1, p}$ norm does not bound $L^{q}$ for $0 < q < p$ , and an explicit power tail shows that set inclusion fails as well. The same mean-zero dilates defeat the homogeneous Poincare estimate, while the critical whole-space inequality is unaffected.

The Poincare examples calibrate the constants: on an interval the mean-zero inequality is linear in the length, and the affine function attains the matching positive multiple of the length, so the dependence cannot be improved. The two disjoint unit balls with the indicator function of one of them show that connectedness is essential, and the constant function on a ball shows that a gradient-only estimate without a mean, trace or positive-measure zero-set normalisation fails for both the critical and the same-exponent bounds. On the Morrey side, the radial powers $∣ x ∣^{α_{0} + δ}$ with $α_{0} = 1 - n / p$ attain the Holder exponent exactly and exhibit the borderline for fixed $p > n$ : the energy diverges as $δ ↓ 0$ and $∣ x ∣^{α_{0}}$ is not a Sobolev function. The failure at $p = n$ is supplied by the localised double logarithm $\log \log (1 + 1 / ∣ x ∣)$ is the unbounded $W^{1, n}$ witness: this compactly supported function belongs to every finite $L^{q}$ but into neither $L^{\infty}$ nor any Holder class.

Conventions: $n \geq 2$ , $1 \leq p < n$ (or $p > n$ in the Morrey examples), and all functions are scalar. Each computation is independent of the parent page's constants and records the exact exponents and normalisations it uses.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Dilations force the Sobolev conjugate

Example

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 2$ , $1 \leq p < n$ , and let $1 \leq q \leq \infty$ and $u \in C_{c}^{\infty} (R^{n}) ∖ {0}$ ; for $λ > 0$ put $u_{λ} (x) : = u (λ x)$ . Then $∥ u_{λ} ∥_{L^{q}} = λ^{- n / q} ∥ u ∥_{L^{q}}, ∥ D u_{λ} ∥_{L^{p}} = λ^{1 - n / p} ∥ D u ∥_{L^{p}} .$ If the inequality $∥ u_{λ} ∥_{L^{q}} \leq C ∥ D u_{λ} ∥_{L^{p}}$ is to hold for all $λ > 0$ with a constant $C$ independent of $λ$ , then the two powers of $λ$ must balance: $- \frac{n}{q} - (1 - \frac{n}{p}) = 0, that is 1 - \frac{n}{p} + \frac{n}{q} = 0,$ which is equivalent to $q = \frac{n p}{n - p} = p^{*}$ . The exponent $p^{*}$ is therefore forced by scaling alone.

Facts & Assumptions

Given: Countable Choice; $1 \leq q \leq \infty$ ; $n \geq 2$ ; $1 \leq p < n$ ; a fixed nonzero $u \in C_{c}^{\infty} (R^{n})$ ; and $λ > 0$ .

[F1]

The Sobolev conjugate is $p^{*} = n p / (n - p)$ and satisfies $\frac{1}{p^{*}} = \frac{1}{p} - \frac{1}{n}$ (The Sobolev conjugate exponent and the scaling identity).

[F2]

An invertible linear map scales Lebesgue measure by $∣ \det ∣$ : substituting $y = λ x$ gives $\int_{R^{n}} g (λ x) d x = λ^{- n} \int_{R^{n}} g (y) d y$ ; an $L^{q}$ class is determined up to null sets (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

The classical chain rule computes $D (u \circ T)$ for the linear map $T (x) = λ x$ : $D (u_{λ}) (x) = λ (D u) (λ x)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F4]

The fundamental theorem on smooth line segments (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz) implies that a smooth function with zero gradient is constant.

Verification

technique · direct

1.1F2F3F4givenalgebra

The norm scalings. For finite $q$ , by [F2] with $y = λ x$ , $∥ u_{λ} ∥_{L^{q}}^{q} = \int_{R^{n}} ∣ u (λ x) ∣^{q} d x = λ^{- n} \int_{R^{n}} ∣ u (y) ∣^{q} d y$ , so $∥ u_{λ} ∥_{L^{q}} = λ^{- n / q} ∥ u ∥_{L^{q}} > 0$ . For $q = \infty$ , the superlevel sets scale in measure by $λ^{- n}$ , so the essential supremum is unchanged; take $1 / q = 0$ . By the chain rule [F3], $D (u_{λ}) (x) = λ (D u) (λ x)$ , so $∣ D u_{λ} (x) ∣ = λ ∣ D u (λ x) ∣$ and $∥ D u_{λ} ∥_{L^{p}}^{p} = λ^{p} λ^{- n} \int_{R^{n}} ∣ D u (y) ∣^{p} d y = λ^{p - n} ∥ D u ∥_{L^{p}}^{p}$ , that is $∥ D u_{λ} ∥_{L^{p}} = λ^{1 - n / p} ∥ D u ∥_{L^{p}} > 0$ (if this norm were zero, continuity would give $D u = 0$ everywhere; [F4] would make $u$ constant, and compact support would force $u = 0$ ).

2.1F1step 1.1givenalgebra∎

The balance condition. Dividing the two identities of step 1.1, the proposed inequality reads $λ^{- n / q} ∥ u ∥_{L^{q}} \leq C λ^{1 - n / p} ∥ D u ∥_{L^{p}}$ for every $λ > 0$ ; after multiplying by $λ^{n / q}$ this is $∥ u ∥_{L^{q}} \leq C λ^{1 - n / p + n / q} ∥ D u ∥_{L^{p}}$ . Since $∥ u ∥_{L^{q}}$ and $∥ D u ∥_{L^{p}}$ are fixed positive numbers, a finite $C$ satisfying this for all $λ > 0$ exists exactly when the exponent vanishes, i.e. $1 - \frac{n}{p} + \frac{n}{q} = 0$ ; solving, $\frac{n}{q} = \frac{n}{p} - 1 = \frac{n - p}{p}$ , so $q = \frac{n p}{n - p} = p^{*}$ by [F1].

Source notes

The computation is the dilation argument that opens Kinnunen's Chapter 3, printed p. 61, and the same scaling discussion precedes Hunter's Theorem 3.28. For $u_{λ} (x) = u (λ x)$ the conjugate exponent is selected by requiring the two sides to carry the same power of $λ$ ; the exponent $p^{*}$ arises from the balance $1 - \frac{n}{p} + \frac{n}{q} = 0$ . No optimality of constants beyond this power balance is claimed, and the example does not construct a counterexample for other exponents, which is done on the companion page by the opposite dilation convention.

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Poincare on an interval: the length dependence is linear

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $I = (0, L)$ with $L > 0$ and let $1 \leq p < \infty$ . For every $u \in W^{1, p} (I)$ with mean $u_{I} = ∣ I ∣^{- 1} \int_{I} u$ one has $∥ u - u_{I} ∥_{L^{p} (I)} \leq L ∥ u^{'} ∥_{L^{p} (I)} .$ The dependence on $L$ cannot be improved below a positive multiple of $L$ : for $u (x) = x - \frac{L}{2}$ the weak derivative is $u^{'} \equiv 1$ , the mean vanishes, and $∥ u - u_{I} ∥_{L^{p} (I)} = L^{1 + 1 / p} 2^{- 1} (p + 1)^{- 1 / p}, ∥ u^{'} ∥_{L^{p} (I)} = L^{1 / p},$ so the ratio $∥ u - u_{I} ∥_{L^{p}} / ∥ u^{'} ∥_{L^{p}}$ equals $c_{p} L$ with $c_{p} = \frac{1}{2} (p + 1)^{- 1 / p} > 0$ . Consequently every admissible constant for this family is at least $c_{p} L$ , while the inequality above shows that $L$ itself is admissible: the optimal constant is of order $L$ .

Facts & Assumptions

Given: The Axiom of Choice (used only through the cited absolutely-continuous-representative interface); reals $0 < L < \infty$ and $1 \leq p < \infty$ ; the interval $I = (0, L)$ ; and a class $u \in W^{1, p} (I; K)$ with $K \in {R, C}$ .

[F1]

Every $u \in W^{1, p} (I)$ has exactly one continuous locally absolutely continuous representative $u^{*}$ with $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ for all $x, y \in I$ , and $u^{*} = u$ almost everywhere (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives).

[F2]

$W^{1, p}$ consists of the $L^{p}$ classes whose weak derivative exists as an $L^{p}$ class, and equality of classes is equality almost everywhere (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

On a finite measure space $L^{r}$ includes into $L^{1}$ for $1 \leq r < \infty$ , so $u \in L^{1} (I)$ and the mean $u_{I} = ∣ I ∣^{- 1} \int_{I} u$ is a well-defined scalar (Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$ ); the interval has $∣ I ∣ = L$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F4]

Holder's inequality gives $\int_{I} ∣ g ∣ \leq ∣ I ∣^{1 - 1 / p} ∥ g ∥_{L^{p} (I)}$ for $g \in L^{p} (I)$ and $1 < p < \infty$ , while for $p = 1$ the same display holds with $∣ I ∣^{0} = 1$ (Holder's inequality for integrals, including the endpoint cases).

[F5]

A function with $C^{k}$ real and imaginary parts has its classical partial derivatives as weak derivatives (Classical derivatives agree with weak derivatives).

[F6]

Newton-Leibniz with finitely many exceptional points: if $G$ is continuous on $[a, b]$ , differentiable off a finite set, and $f$ is Riemann integrable with $f = G^{'}$ off that set, then $\int_{a}^{b} f = G (b) - G (a)$ (Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous).

[F7]

A bounded Riemann integrable function on $[a, b]$ is Lebesgue integrable there and its Lebesgue integral equals its Riemann integral (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[F8]

For real $α$ the function $x \mapsto x^{α}$ is differentiable on $(0, \infty)$ with derivative $α x^{α - 1}$ (Continuity and derivatives of positive-base real powers).

Verification

technique · direct

1.1F1F2F3givenalgebra

By [F1] the class $u$ has a continuous representative $u^{*}$ on $I$ that is locally absolutely continuous and satisfies $u^{*} (x) - u^{*} (y) = \int_{y}^{x} u^{'}$ for all $x, y \in I$ , and $u^{*} = u$ almost everywhere. Since $u \in L^{p} (I)$ and $∣ I ∣ = L < \infty$ , [F3] gives $u \in L^{1} (I)$ , so $u_{I} = ∣ I ∣^{- 1} \int_{I} u$ is well defined and equals $L^{- 1} \int_{I} u^{*}$ by [F2]; the norms $∥ u - u_{I} ∥_{L^{p} (I)} = ∥ u^{*} - u_{I} ∥_{L^{p} (I)}$ agree because the two integrands coincide almost everywhere.

1.2F3F5F6F7F8givenalgebra

The linear function attains the scaling. Let $u (x) = x - \frac{L}{2}$ on $I$ . Its real and imaginary parts are $C^{\infty}$ , so by [F5] $u \in W^{1, p} (I)$ with weak derivative $u^{'} \equiv 1$ , and $∥ u^{'} ∥_{L^{p} (I)} = (L)^{1 / p} = L^{1 / p}$ by [F3]. The antiderivative $G (x) = \frac{x^{2}}{2} - \frac{L x}{2}$ is continuous and $G^{'} = u$ on $I$ , so [F6], [F7] and [F3] give $u_{I} = L^{- 1} (G (L) - G (0)) = L^{- 1} (L^{2} / 2 - L^{2} / 2) = 0$ . Finally $∣ u ∣^{p}$ has the continuous majorant-free antiderivative identity $\int_{0}^{L / 2} s^{p} d s = (L / 2)^{p + 1} / (p + 1)$ by [F6], [F7] and [F8] (the derivative of $s^{p + 1} / (p + 1)$ is $s^{p}$ on $(0, L / 2)$ and the endpoint values are limits), so $\int_{I} ∣ u - u_{I} ∣^{p} d x = 2 \int_{0}^{L / 2} s^{p} d s = 2 (L / 2)^{p + 1} / (p + 1) = L^{p + 1} 2^{- p} (p + 1)^{- 1}$ . Taking $p$ -th roots gives $∥ u - u_{I} ∥_{L^{p} (I)} = L^{1 + 1 / p} 2^{- 1} (p + 1)^{- 1 / p}$ .

2.1F4step 1.1algebra

Oscillation bound. For all $x, y \in I$ the representative satisfies $∣ u^{*} (x) - u^{*} (y) ∣ = ∣ \int_{y}^{x} u^{'} ∣ \leq \int_{I} ∣ u^{'} ∣$ ; by [F4], applied to $g = u^{'}$ , this is at most $L^{1 - 1 / p} ∥ u^{'} ∥_{L^{p} (I)}$ for $1 < p < \infty$ and at most $∥ u^{'} ∥_{L^{1} (I)}$ for $p = 1$ , that is, at most $L^{1 - 1 / p} ∥ u^{'} ∥_{L^{p} (I)}$ in both cases.

3.1F4step 1.1step 2.1algebra

The mean-zero bound. For every $x \in I$ the identity $u^{*} (x) - u_{I} = L^{- 1} \int_{I} (u^{*} (x) - u^{*} (y)) d y$ and [F4] with $g (y) = u^{*} (x) - u^{*} (y)$ give $∣ u^{*} (x) - u_{I} ∣^{p} \leq L^{- 1} \int_{I} ∣ u^{*} (x) - u^{*} (y) ∣^{p} d y$ ; by step 2.1 the integrand is at most $(L^{1 - 1 / p} ∥ u^{'} ∥_{L^{p} (I)})^{p} = L^{p - 1} ∥ u^{'} ∥_{L^{p} (I)}^{p}$ for every $y$ . Integrating over $x \in I$ therefore yields $\int_{I} ∣ u - u_{I} ∣^{p} \leq L \cdot L^{p - 1} ∥ u^{'} ∥_{L^{p} (I)}^{p} = L^{p} ∥ u^{'} ∥_{L^{p} (I)}^{p}$ , and taking $p$ -th roots gives the asserted inequality.

4.1step 3.1step 1.2givenalgebra∎

Sharpness. Dividing the two norms computed in step 1.2 gives $∥ u - u_{I} ∥_{L^{p} (I)} / ∥ u^{'} ∥_{L^{p} (I)} = L \cdot \frac{1}{2} (p + 1)^{- 1 / p} = c_{p} L$ with $c_{p} = \frac{1}{2} (p + 1)^{- 1 / p} > 0$ . Hence for every $L > 0$ there is a class in $W^{1, p} (I)$ for which the ratio of the left side to $∥ u^{'} ∥_{L^{p} (I)}$ equals $c_{p} L$ , so no constant smaller than $c_{p} L$ can be admissible for all $u \in W^{1, p} (I)$ ; combined with step 3.1 the optimal constant for this family lies between $c_{p} L$ and $L$ , and in particular is a positive multiple of $L$ .

Source notes

The upper bound is the one-dimensional instance of the Poincare inequality for $W^{1, p}$ functions on bounded open sets; Kinnunen, Theorem 3.12 treats cube mean oscillations (an interval when $n = 1$ ); the present proof works directly on the given interval, and Hunter's Theorem 4.9 gives the related p=2 zero-boundary slab estimate; here the cited one-dimensional representative supplier gives the mean-zero interval argument for all finite p. The computation of the extremal ratio for the affine function is included to fix the linear dependence on the interval length, which the statement of the companion theorem does not quantify; no claim about the exact optimal constant beyond the two-sided order $c_{p} L \leq C^{*} (L) \leq L$ is made.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Poincare-Wirtinger fails on disconnected bounded domains

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Ω = B (0, 1) \cup B (3 e_{1}, 1) \subset R^{2}$ (two disjoint unit balls) and $u = 1_{B (3 e_{1}, 1)}$ . Then $D u = 0$ almost everywhere and $u$ is not almost everywhere constant on $Ω$ , so $u - u_{Ω}$ is nonzero on a set of positive measure and $∥ u - u_{Ω} ∥_{L^{p} (Ω)} > 0$ : the mean-zero Poincare-Wirtinger inequality is false on disconnected bounded open sets, and connectedness is essential for the single-global-mean normalisation.

Facts & Assumptions

Given: Countable Choice; the open bounded set $Ω = B (0, 1) \cup B (3 e_{1}, 1) \subseteq R^{2}$ ; the indicator $u = 1_{B (3 e_{1}, 1)}$ ; and $1 \leq p < \infty$ .

[F1]

Every Euclidean ball has positive finite Lebesgue measure, measures are monotone and countably additive on disjoint measurable sets (Euclidean balls have positive finite Lebesgue measure, Measures are monotone, Measures on sigma-algebras).

[F2]

$D_{i} u$ is the weak derivative if $\int_{Ω} u \partial_{i} φ = - \int_{Ω} D_{i} u φ$ for every test function $φ$ , and constant classes have zero weak derivative (Weak derivative of a locally integrable function).

[F3]

The two open balls are disjoint because $∣ 3 e_{1} ∣ = 3 > 2$ ; their closures are compact, and each open ball has positive finite measure; the ball average is the normalized integral (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, The average of a locally integrable function over a Euclidean ball, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

Countable Choice is assumed; classical smooth derivatives are weak derivatives (Classical derivatives agree with weak derivatives). Translated balls have equal measure (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F5]

With the additional Axiom of Choice (The Axiom of Choice), the mean-zero Poincare-Wirtinger inequality holds on bounded John domains in dimensions $n \geq 2$ for every $1 \leq p < \infty$ (The mean-zero Poincare inequality on bounded John domains). This positive comparison uses the stronger hypothesis; the two-ball counterexample needs only Countable Choice.

Counterexample

technique · direct

1.1F1F2F3F4givenalgebra

The function $u$ is locally constant, hence smooth on $Ω$ , with all classical partial derivatives zero. By [F4] its weak gradient is zero. Since $∣ u ∣ \leq 1$ and $Ω$ has finite measure, $u \in W^{1, p} (Ω)$ and $∥ D u ∥_{p} = 0$ . The two balls have equal measure by [F4], so $∣ Ω ∣ = 2 ∣ B (0, 1) ∣$ by [F1].

2.1F1F3step 1.1givenalgebra

The mean and the oscillation. Since $u = 0$ on $B (0, 1)$ and $u = 1$ on $B (3 e_{1}, 1)$ and the two balls have equal measure, $u_{Ω} = ∣ Ω ∣^{- 1} \int_{Ω} u = ∣ B (3 e_{1}, 1) ∣ / (2 ∣ B (0, 1) ∣) = 1 / 2$ . Therefore $∣ u - u_{Ω} ∣ = 1 / 2$ on both balls, and $∥ u - u_{Ω} ∥_{L^{p} (Ω)}^{p} = \int_{Ω} (1 / 2)^{p} = 2 ∣ B (0, 1) ∣ 2^{- p} > 0$ , while $u$ is not almost everywhere constant on $Ω$ (it takes the values $0$ and $1$ on sets of positive measure).

3.1F1F5step 1.1step 2.1givenalgebra∎

Failure and the role of connectedness. The mean-zero Poincare-Wirtinger inequality would require $∥ u - u_{Ω} ∥_{L^{p} (Ω)} \leq C ∥ D u ∥_{L^{p} (Ω)}$ for a constant $C$ depending only on the domain and $p$ ; but the left side is $2^{1 / p} ∣ B (0, 1) ∣^{1 / p} / 2 > 0$ by step 2.1 while the right side is $0$ by step 1.1, so no finite $C$ exists. A zero-set normalisation on only one component does not repair the inequality: this very $u$ vanishes on $B (0, 1)$ , a set of half the domain measure. Each component must be normalised separately, or connectedness imposed; with the additional Axiom of Choice, bounded John domains, which are connected, satisfy the inequality by [F5].

Source notes

The counterexample is the standard two-ball two-valued function, matching the "what eliminates constants" discussion in Kinnunen's Remark 3.11 and Hunter's Chapter 4: the mean of a nonzero mean-zero function is the only quantity that can fail, and on a disconnected domain a locally constant function need not be constant. The computation uses only that the two balls have equal positive measure and that the gradient of a locally constant class vanishes.

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A gradient-only Poincare estimate needs normalisation

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 2$ and $1 < p < n$ with $p^{*} = \frac{n p}{n - p}$ , and let $Ω$ be a nonempty bounded connected domain with smooth boundary. The following two assertions are false:

there is a finite constant $C$ with $∥ u ∥_{L^{p^{*}} (Ω)} \leq C ∥ D u ∥_{L^{p} (Ω)}$ for every $u \in W^{1, p} (Ω; K)$ ;
there is a finite constant $C$ with $∥ u ∥_{L^{p} (Ω)} \leq C ∥ D u ∥_{L^{p} (Ω)}$ for every $u \in W^{1, p} (Ω; K)$ .

A condition excluding nonzero constants is necessary; mean subtraction, zero trace, and vanishing on a set of positive measure are standard normalisations.

Facts & Assumptions

Given: Countable Choice; $n \geq 2$ ; $1 < p < n$ ; the nonempty bounded connected smooth domain $Ω_{0} = Ω$ of the statement; and a test function $φ \in C_{c}^{\infty} (Ω_{0})$ .

[F1]

$W^{1, p} (Ω_{0})$ consists of the $L^{p}$ classes whose first weak derivatives exist as $L^{p}$ classes; the weak derivative $D_{i} u$ is characterized by $\int_{Ω_{0}} u \partial_{i} φ = - \int_{Ω_{0}} D_{i} u φ$ for every test function (Integer-order Sobolev spaces and their norms, Weak derivative of a locally integrable function).

[F2]

$L^{p}$ consists of almost-everywhere classes, and the constant class $u \equiv 1$ lies in $L^{p} (Ω_{0})$ because $Ω_{0}$ contains a ball and is bounded, so it has finite positive measure (The space $L^{p} (μ)$ as the quotient by null functions, Euclidean balls have positive finite Lebesgue measure, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F3]

On sigma-finite products nonnegative measurable functions may be integrated in either order (Tonelli-Fubini) (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F4]

If $G$ is differentiable everywhere on a compact interval and $G^{'}$ is integrable, then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ). Applied to a compactly supported smooth real section, or separately to both components of a complex section, its derivative integral is zero.

[F5]

The Sobolev conjugate satisfies $p^{*} > p$ and $\frac{1}{p^{*}} = \frac{1}{p} - \frac{1}{n}$ , so the two exponents in the refuted assertions are distinct and both finite (The Sobolev conjugate exponent and the scaling identity).

Counterexample

technique · direct

1.1F1F2F3F4givenalgebra

The constant function is a Sobolev function with zero gradient. Take $u \equiv 1$ on $Ω_{0}$ . By [F2], $u \in L^{p} (Ω_{0}; K)$ (the class is represented by the constant function; for $K = C$ it is the complex constant). Fix $i$ and let $φ \in C_{c}^{\infty} (Ω_{0})$ ; extend $φ$ by zero to $R^{n}$ . Writing ${\hat{x}}_{i}$ for the coordinates other than $x_{i}$ and using Fubini [F3], $\int_{Ω_{0}} \partial_{i} φ = \int_{R^{n}} \partial_{i} φ = \int_{R^{n - 1}} (\int_{R} \partial_{i} φ (\dots, t, \dots) d t) d {\hat{x}}_{i} = 0$ , because for each fixed ${\hat{x}}_{i}$ the inner function $t \mapsto φ (\dots, t, \dots)$ is compactly supported and smooth, so [F4] gives vanishing integral. Hence $\int_{Ω_{0}} u \partial_{i} φ = 0 = \int_{Ω_{0}} 0 \cdot φ$ for every test function, i.e. the zero class is the weak derivative $D_{i} u$ by [F1]; in particular $D u = 0$ as an element of $L^{p} (Ω_{0}; K)$ .

2.1F2F5step 1.1givenalgebra∎

Both proposed inequalities fail. Since $u \equiv 1$ and $D u = 0$ , the two sides of the first proposed estimate are $∥ u ∥_{L^{p^{*}} (Ω_{0})} = ∣ Ω_{0} ∣^{1 / p^{*}} > 0$ by [F2] and $∥ D u ∥_{L^{p} (Ω_{0})} = 0$ ; the second has $∥ u ∥_{L^{p} (Ω_{0})} = ∣ Ω_{0} ∣^{1 / p} > 0$ and again $∥ D u ∥_{L^{p} (Ω_{0})} = 0$ . No finite $C$ can satisfy $∣ Ω_{0} ∣^{1 / p^{*}} \leq C \cdot 0$ or $∣ Ω_{0} ∣^{1 / p} \leq C \cdot 0$ . Finally $u - u_{Ω_{0}} = 0$ and $u$ does not vanish on any set of positive measure and satisfies no vanishing trace condition, so mean subtraction annihilates precisely this witness while trace or positive-measure zero-set normalisations exclude it; a condition excluding nonzero constants is therefore necessary.

Source notes

The refutation is the explicit constant-function witness against the un-normalised inequalities. Kinnunen's Theorem 3.47, printed pp. 90-91, states the mean-zero form; the constant witness directly shows why that normalisation matters. The computation of the weak gradient of a constant uses only Fubini and the one-dimensional fundamental theorem, so it applies to every nonempty open $Ω_{0}$ of finite measure, not only to a ball; the same computation proves the claim on the arbitrary finite-measure domain in the statement.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6.1-sol)Open item page →

The $W^{1, p} \to L^{q}$ bound fails for $q > p^{*}$ by dilation

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 2$ , $1 \leq p < n$ and $q > p^{*} = \frac{n p}{n - p}$ . The assertion that there is a finite constant $C$ with $∥ u ∥_{L^{q} (R^{n})} \leq C ∥ D u ∥_{L^{p} (R^{n})} for every u \in C_{c}^{\infty} (R^{n})$ is false. The stronger full-norm assertion $∥ u ∥_{L^{q}} \leq C ∥ u ∥_{W^{1, p}}$ also fails for this family, so the exponent $p^{*}$ is sharp for both estimates.

Facts & Assumptions

Given: Countable Choice; $n \geq 2$ ; $1 \leq p < n$ ; a fixed nonzero $φ \in C_{c}^{\infty} (B (0, 1))$ ; and $0 < λ \leq 1$ .

[F1]

The Sobolev conjugate satisfies $\frac{1}{p^{*}} = \frac{1}{p} - \frac{1}{n}$ , so for finite $q > p^{*}$ one has $\frac{1}{q} < \frac{1}{p} - \frac{1}{n}$ ; the case $q = \infty$ is treated separately with $1 / q = 0$ (The Sobolev conjugate exponent and the scaling identity).

[F2]

An invertible linear map scales Lebesgue measure by $∣ \det ∣$ , so for $a > 0$ the substitution $y = x / a$ gives $\int_{R^{n}} g (x / a) d x = a^{n} \int_{R^{n}} g (y) d y$ ; an $L^{q}$ class is determined by its values almost everywhere (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

The closed unit ball is compact, so a continuous function on it is bounded and the support of $φ$ is compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

[F4]

A nonzero smooth bump in $B (0, 1)$ exists (A Euclidean bump for a compact set inside an open set); its gradient scales by the chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). Its gradient norm is positive: otherwise continuity gives zero gradient everywhere, the fundamental theorem along segments makes it constant, and compact support makes it zero (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz).

Counterexample

technique · direct

1.1F2F3F4givenalgebra

The norm scalings. For $0 < λ \leq 1$ put $φ_{λ} (x) : = φ (x / λ)$ , so $φ_{λ} \in C_{c}^{\infty} (B (0, λ)) ∖ {0}$ . If $q < \infty$ , substituting $y = x / λ$ and using [F2] gives $∥ φ_{λ} ∥_{L^{q}}^{q} = λ^{n} ∥ φ ∥_{L^{q}}^{q}$ , hence $∥ φ_{λ} ∥_{L^{q}} = λ^{n / q} ∥ φ ∥_{L^{q}} > 0$ . If $q = \infty$ , then for each $t > 0$ , [F2] gives $λ_{n} ({∣ φ_{λ} ∣ > t}) = λ^{n} λ_{n} ({∣ φ ∣ > t})$ , so $∥ φ_{λ} ∥_{L^{\infty}} = ∥ φ ∥_{L^{\infty}} > 0$ . In either case $D φ_{λ} (x) = λ^{- 1} (D φ) (x / λ)$ and $∥ D φ_{λ} ∥_{L^{p}} = λ^{n / p - 1} ∥ D φ ∥_{L^{p}} > 0$ ; the support statement uses [F3].

2.1F1step 1.1givenalgebra∎

The ratio diverges above $p^{*}$ . For finite $q > p^{*}$ , dividing the two scalings of step 1.1 gives $∥ φ_{λ} ∥_{L^{q}} / ∥ D φ_{λ} ∥_{L^{p}} = λ^{1 + n / q - n / p} ∥ φ ∥_{L^{q}} / ∥ D φ ∥_{L^{p}}$ ; the exponent is negative exactly when $\frac{1}{q} < \frac{1}{p} - \frac{1}{n} = \frac{1}{p^{*}}$ by [F1], so this ratio tends to $+ \infty$ . For $q = \infty$ , step 1.1 gives $∥ φ_{λ} ∥_{L^{\infty}} / ∥ D φ_{λ} ∥_{L^{p}} = λ^{1 - n / p} ∥ φ ∥_{L^{\infty}} / ∥ D φ ∥_{L^{p}} \to + \infty$ because $p < n$ . In either case any finite $C$ satisfying the proposed inequality for every compactly supported smooth $u$ would have to dominate this unbounded ratio, which is impossible. Also $∥ φ_{λ} ∥_{W^{1, p}}^{p} = λ^{n} ∥ φ ∥_{p}^{p} + λ^{n - p} \sum_{i} ∥ D_{i} φ ∥_{p}^{p} \leq λ^{n - p} (∥ φ ∥_{p}^{p} + \sum_{i} ∥ D_{i} φ ∥_{p}^{p})$ for $λ \leq 1$ . Thus the full-norm ratio has the same divergent lower bound $c λ^{1 + n / q - n / p}$ , with $1 / q = 0$ for $q = \infty$ . Both bounds fail for every $q > p^{*}$ .

Source notes

The dilation computation is Kinnunen's, printed pp. 61-62, and Laugesen's sharpness discussion, printed pp. 65-66; the counterexample is the standard concentrated-bump family.

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$W^{1, n}$ is not contained in $L^{\infty}$

Statement refuted

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 2$ and let $φ \in C_{c}^{\infty} (R^{n})$ satisfy $0 \leq φ \leq 1$ , $φ = 1$ on $B (0, \frac{1}{2})$ and $supp φ \subseteq B (0, 2)$ , and set $f (x) = φ (x) \log \log (1 + \frac{1}{∣ x ∣})$ for $x \neq 0$ , $f (0) = 0$ . Then $f \in W^{1, n} (R^{n})$ but $f$ is unbounded near the origin; therefore this $W^{1, n}$ class admits no bounded (and no continuous, let alone Holder) representative, and the endpoint $p = n$ has no $L^{\infty}$ embedding, while this compactly supported witness belongs to $L^{q}$ for every finite $q$ (the general bounded-domain finite- $q$ embedding is stated separately).

Facts & Assumptions

Given: The Axiom of Choice; $n \geq 2$ ; a cutoff $φ$ as in the statement (A Euclidean bump for a compact set inside an open set); and $f = φ g$ with $g (x) = \log \log (1 + 1 / ∣ x ∣)$ for $x \neq 0$ , $g (0) = 0$ .

[F1]

The witness $g$ lies in $W^{1, n} (B (0, 1))$ and has no bounded representative; consequently no representative of $g$ is bounded on any neighbourhood of the origin (Morrey's inequality has no $p = n$ endpoint).

[F2]

$W^{1, n}$ consists of the $L^{n}$ classes with weak gradient in $L^{n}$ ; multiplication by the smooth compactly supported cutoff $φ$ preserves $W^{1, n}$ classes and $L^{p}$ classes are determined up to null sets (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

The critical embedding gives $W^{1, n} (Ω) ↪ L^{q} (Ω)$ for every finite $q$ on nonempty bounded domains that are $W^{1, s}$ -extension domains for every $s \in (n / 2, n)$ (The critical Sobolev embedding into every finite $L^{q}$ ).

[F4]

A compact ball inside an open ball admits a smooth cutoff equal to $1$ on the smaller ball and supported in the larger one (A Euclidean bump for a compact set inside an open set); weak derivatives are characterized by the test-function identity (Weak derivative of a locally integrable function).

[F5]

Balls are bounded smooth domains, so under AC they have a bounded extension operator at every Sobolev index (Bounded C^k domains admit integer-order Sobolev extension). Smooth classical derivatives are weak derivatives under CC (Classical derivatives agree with weak derivatives).

Counterexample

technique · direct

1.1F1F2F4F5givenalgebra

Membership in $W^{1, n}$ . On $B (0, 1)$ , $∣ f ∣ \leq ∣ g ∣$ because $0 \leq φ \leq 1$ , so $f \in L^{n} (B (0, 1))$ by [F1]. On $R^{n} ∖ B (0, 1)$ , the function $g$ and its derivatives are smooth and bounded on the compact support of $φ$ , so $f \in L^{n} (R^{n})$ . Choose $χ \in C_{c}^{\infty} (B (0, 1))$ equal to $1$ on $B (0, 1 / 4)$ , and write $a : = χ φ$ and $b : = (1 - χ) φ g$ . Then $a$ is supported in $B (0, 1)$ , while $b$ is smooth and compactly supported away from the origin. For a test function $ψ \in C_{c}^{\infty} (R^{n})$ , $a ψ \in C_{c}^{\infty} (B (0, 1))$ , so the weak derivative identity for $g$ from [F1] gives $\int_{R^{n}} a g \partial_{i} ψ = \int_{B (0, 1)} g \partial_{i} (a ψ) - \int_{B (0, 1)} g (\partial_{i} a) ψ = - \int_{B (0, 1)} (a D_{i} g + g \partial_{i} a) ψ .$ The summand $a g$ therefore has weak derivative $a D_{i} g + g \partial_{i} a$ , which lies in $L^{n}$ by [F1] and boundedness of $a, D a$ ; the summand $b$ has its classical, compactly supported $L^{n}$ weak derivative by [F5]. Thus $f = a g + b \in W^{1, n} (R^{n})$ by [F2].

2.1F1F2step 1.1givenalgebra

No bounded representative. For every $M > 0$ the set ${x \in B (0, \frac{1}{2}) : f (x) > M}$ is a punctured neighbourhood of $0$ and has positive measure, because $f = g \to + \infty$ as $x \to 0$ ; hence every representative of the $W^{1, n}$ class of $f$ is unbounded on every neighbourhood of $0$ . In particular $f$ has neither a bounded, nor a continuous, nor a Holder representative.

3.1F3F5step 1.1step 2.1givenalgebra∎

The endpoint contrast. By [F3], applied on $Ω = B (0, 2)$ , whose all-exponents extension hypothesis is supplied by [F5], the class $f$ lies in $L^{q}$ for every finite $q$ and satisfies $∥ f ∥_{L^{q}} \leq C (n, q) ∥ f ∥_{W^{1, n}}$ . But no finite constant bounds $∥ f ∥_{L^{\infty}}$ , since any representative is unbounded by step 2.1; hence $W^{1, n}$ admits no embedding into $L^{\infty}$ or a Holder class; the finite- $q$ statement here concerns this compactly supported witness and the bounded domains covered by [F3].

Source notes

The witness is Hunter's Example 3.30 and Kinnunen's Remark 3.16(1), printed at p. 66 and p. 73. The membership in $W^{1, n}$ and the absence of a bounded representative are established for the localised logarithm in Morrey's inequality has no $p = n$ endpoint; multiplying by the cutoff $φ$ does not change the behaviour near the origin and makes the function compactly supported, and the finite- $q$ embeddings are the complement recorded in The critical Sobolev embedding into every finite $L^{q}$ .

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Morrey's inequality has no $p = n$ endpoint

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 2$ and let $B = B (0, 1)$ . There are no constants $α > 0$ and $C$ with $[u]_{C^{0, α} (B)} \leq C ∥ u ∥_{W^{1, n} (B)} for every u \in W^{1, n} (B);$ that is, Morrey's inequality has no endpoint at $p = n$ . The witness is $f (x) = \log \log (1 + \frac{1}{∣ x ∣}) (x \in B ∖ {0}), f (0) = 0,$ which lies in $W^{1, n} (B)$ but has no bounded representative. The hypothesis $p > n$ in Morrey's inequality is therefore essential.

Facts & Assumptions

Given: Countable Choice; $n \geq 2$ ; the unit ball $B = B (0, 1)$ ; and the function $f : B \to R$ defined by $f (x) = \log \log (1 + 1 / ∣ x ∣)$ for $x \neq 0$ and $f (0) = 0$ .

[F1]

Polar coordinates: for nonnegative Borel $h$ , $\int_{B} h d λ_{n} = \int_{0}^{1} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ with $0 < σ (S^{n - 1}) < \infty$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere).

[F2]

$W^{1, n} (B)$ consists of the $L^{n}$ classes whose first weak derivatives exist as $L^{n}$ classes; $L^{n}$ consists of almost-everywhere classes (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

Holder's inequality with conjugate exponents on $R$ and Tonelli's theorem (Holder's inequality for integrals, including the endpoint cases); a function with finite global $α$ -Holder seminorm on the bounded ball is bounded, since fixing $y_{0} \in B$ gives $∣ v (x) ∣ \leq ∣ v (y_{0}) ∣ + [v]_{C^{0, α} (B)} diam (B)^{α}$ .

[F5]

Countable Choice is assumed; the fundamental theorem (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz) gives integration by parts on smooth one-dimensional sections, and Fubini (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability) integrates those identities. Classical smooth derivatives are weak derivatives (Classical derivatives agree with weak derivatives).

Counterexample

technique · direct

1.1F1F2F4F5givenalgebra

Membership in $W^{1, n}$ . Off the origin $f$ is smooth and radial with $∣ D f (x) ∣ = \frac{1}{∣ x ∣ (∣ x ∣ + 1) \log (1 + 1 / ∣ x ∣)} .$ For $0 < r = ∣ x ∣ \leq \frac{1}{2}$ , this is at most $1 / (r \log (1 / r))$ , so polar coordinates [F1] give $\int_{B (0, 1 / 2)} ∣ D f ∣^{n} d x \leq σ (S^{n - 1}) \int_{0}^{1 / 2} \frac{d r}{r (\log (1 / r))^{n}} = σ (S^{n - 1}) \int_{\log 2}^{\infty} s^{- n} d s < \infty,$ where $s = \log (1 / r)$ and $n \geq 2$ . On $1 / 2 \leq r < 1$ , the exact derivative is bounded by $4 / (3 \log 2)$ , since $r \geq 1 / 2$ , $r + 1 \geq 3 / 2$ , and $\log (1 + 1 / r) \geq \log 2$ ; hence its $n$ th-power polar integral on this annulus is finite as well. Also $\int_{B} ∣ f ∣^{n} d x < \infty$ : near $0$ , $\log \log (1 + 1 / r) = O (r^{- 1 / 2})$ , so the radial $L^{n}$ majorant is $O (r^{n / 2 - 1})$ , which is integrable for $n \geq 2$ , and $f$ is bounded on $1 / 2 \leq r < 1$ . To identify the weak gradient, fix a test $φ \in C_{c}^{\infty} (B)$ and a coordinate $i$ . For every transverse coordinate other than the zero vector, the coordinate line avoids the origin; its intersection with $B$ is an interval on which $f$ is smooth, and $φ$ vanishes near the endpoints. The fundamental theorem applied to $f φ$ gives $\int f \partial_{i} φ = - \int (\partial_{i} f) φ$ on that line. The omitted transverse singleton is null because $n - 1 \geq 1$ ; both integrands are globally integrable since $f, D f \in L^{n} (B) \subseteq L^{1} (B)$ . Fubini [F5] therefore integrates the section identities into the weak derivative identity. Thus $D f$ is the weak gradient and $f \in W^{1, n} (B)$ by [F2].

1.2F2givenalgebra

No bounded representative. For every $M > 0$ the set ${x \in B ∖ {0} : f (x) > M}$ is a punctured neighbourhood of $0$ (since $f (x) \to + \infty$ as $x \to 0$ ) and has positive measure; hence if a measurable $g$ equals $f$ almost everywhere on $B$ , then for every $ε > 0$ the set ${x \in B (0, ε) : g (x) > M}$ has positive measure, so $g$ is unbounded on every neighbourhood of $0$ . In particular no representative of $f$ is bounded, and a fortiori none is uniformly continuous or Holder on $B$ .

2.1F4step 1.1step 1.2givenalgebra∎

Failure of every Holder bound. Suppose there were $α > 0$ and $C$ with $[u]_{C^{0, α} (B)} \leq C ∥ u ∥_{W^{1, n} (B)}$ for every $u \in W^{1, n} (B)$ . Applying this to the class of $f$ would produce a representative $f^{*}$ with $∣ f^{*} (x) - f^{*} (y) ∣ \leq C ∥ f ∥_{W^{1, n}} ∣ x - y ∣^{α}$ for all $x, y \in B$ ; such an $f^{*}$ is continuous and bounded on the bounded set $B$ . But $f^{*}$ is a representative of $f$ , contradicting step 1.2. Hence no such constants exist, Morrey's inequality has no $p = n$ endpoint, and the hypothesis $p > n$ in Morrey's inequality for $p > n$ is essential.

Source notes

The witness is Hunter's Example 3.30 and Kinnunen's Remark 3.16(1), printed at p. 66 and p. 73: the double logarithm is the borderline function whose gradient just fails to be $L^{n}$ -integrable when $\log \log$ is replaced by $\log$ . The computation above records membership in $W^{1, n}$ by splitting at $r = 1 / 2$ : the logarithmic majorant is integrated only near the origin, and the exact derivative is bounded on the remaining annulus. It also records the absence of any bounded representative; the contradiction with a hypothetical Holder bound is then immediate because Holder functions on a bounded domain are bounded.

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Radial powers approach the Morrey borderline exponent

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 2$ , $n < p < \infty$ , $α_{0} = 1 - \frac{n}{p} \in (0, 1)$ , and for $0 < δ < \frac{n}{p}$ define $u_{δ} (x) = ∣ x ∣^{α_{0} + δ}$ on $B = B (0, 1)$ , with $u_{δ} (0) = 0$ . Then $u_{δ} \in W^{1, p} (B)$ and $∥ D u_{δ} ∥_{L^{p} (B)} = (α_{0} + δ) (\frac{σ (S^{n - 1})}{p δ})^{1 / p} .$ The function $u_{δ}$ is Holder of exponent $α_{0} + δ$ on $B$ with $[u_{δ}]_{C^{0, α_{0} + δ}} = 1$ , and the local Morrey estimate holds on each $B (0, r)$ with $2 r < 1$ ; as $δ ↓ 0$ the energy $∥ D u_{δ} ∥_{L^{p} (B)}^{p}$ diverges like $α_{0}^{p} σ (S^{n - 1}) / (p δ)$ , and the borderline function $∣ x ∣^{α_{0}}$ is not in $W^{1, p} (B)$ . The exponent $α_{0} = 1 - \frac{n}{p}$ is exactly the borderline produced by $p > n$ .

Facts & Assumptions

Given: The Axiom of Choice; $n \geq 2$ ; $n < p < \infty$ ; $α_{0} = 1 - n / p \in (0, 1)$ ; $0 < δ < n / p$ ; and $u_{δ} (x) = ∣ x ∣^{α_{0} + δ}$ on $B = B (0, 1)$ , $u_{δ} (0) = 0$ .

[F1]

Polar coordinates: for nonnegative Borel $h$ , $\int_{B} h = \int_{0}^{1} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ with $0 < σ (S^{n - 1}) < \infty$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere).

[F2]

For real $β$ the function $r \mapsto r^{β}$ is differentiable on $(0, \infty)$ with derivative $β r^{β - 1}$ , so for $β > - 1$ its antiderivative is $r^{β + 1} / (β + 1)$ ; Newton-Leibniz on $[ε, 1]$ combined with monotone convergence as $ε ↓ 0$ gives $\int_{0}^{1} r^{β} d r = 1 / (β + 1)$ (Continuity and derivatives of positive-base real powers, Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Monotone convergence for the integral).

[F3]

The radial derivative off the origin is $D u_{δ} (x) = (α_{0} + δ) ∣ x ∣^{α_{0} + δ - 2} x$ , of modulus $(α_{0} + δ) ∣ x ∣^{α_{0} + δ - 1}$ ; $W^{1, p}$ consists of $L^{p}$ classes with weak gradient in $L^{p}$ (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

For $0 < α < 1$ and $h \geq 0$ , the function $t \mapsto (t + h)^{α} - t^{α}$ decreases for $t > 0$ , since its derivative is $α ((t + h)^{α - 1} - t^{α - 1}) \leq 0$ . Continuity at $t = 0$ gives $(t + h)^{α} - t^{α} \leq h^{α}$ , hence $∣ s^{α} - t^{α} ∣ \leq ∣ s - t ∣^{α}$ for $s, t \geq 0$ ; the case $α = 1$ is equality (Continuity and derivatives of positive-base real powers). The Holder seminorm is the supremum of the difference quotients (Local Hölder and scaled C-two-alpha norms on balls).

[F5]

Morrey gives $[v]_{C^{0, α_{0}} (B (0, r))} \leq C (n, p) ∥ D v ∥_{L^{p} (B (0, 2 r))}$ for $2 r < 1$ (Morrey's inequality for $p > n$ ).

[F6]

Countable Choice is assumed; the vector-valued fundamental theorem (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz) gives integration by parts for smooth products on intervals, and Fubini (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability) integrates the section identities. Classical smooth derivatives are weak derivatives (Classical derivatives agree with weak derivatives).

Verification

technique · direct

1.1F1F2F3F6givenalgebra

The weak gradient and the energy. Off the origin $u_{δ}$ is smooth with radial derivative $D u_{δ} (x) = (α_{0} + δ) ∣ x ∣^{α_{0} + δ - 2} x$ of modulus $(α_{0} + δ) ∣ x ∣^{α_{0} + δ - 1}$ . For $α : = α_{0} + δ > 0$ , this gradient is in $L^{1} (B)$ because its radial exponent is $n + α - 2 > - 1$ , and $u_{δ}$ is bounded. Every coordinate line with nonzero transverse coordinate avoids the origin; apply the fundamental theorem to $u_{δ} φ$ on its interval in $B$ , with $φ$ a compactly supported test. Since $n \geq 2$ , the omitted transverse singleton is null, and Fubini [F6] gives the global weak derivative identity. With $r = ∣ x ∣$ and $β = p (α_{0} + δ - 1) + n - 1$ , polar coordinates [F1] and the power integral [F2] give $∥ D u_{δ} ∥_{L^{p} (B)}^{p} = σ (S^{n - 1}) (α_{0} + δ)^{p} \int_{0}^{1} r^{β} d r = σ (S^{n - 1}) (α_{0} + δ)^{p} / (β + 1)$ , and $β + 1 = p (α_{0} + δ) - p + n = p δ$ because $p α_{0} = p - n$ . Hence $∥ D u_{δ} ∥_{L^{p} (B)} = (α_{0} + δ) (σ (S^{n - 1}) / (p δ))^{1 / p}$ , and $u_{δ} \in W^{1, p} (B)$ by [F3] since both $u_{δ}$ and this gradient are in $L^{p} (B)$ .

1.2F4givenalgebra

The Holder exponent. For $x \neq y$ in $B$ , writing $s = ∣ x ∣$ , $t = ∣ y ∣$ and using the elementary inequality of [F4] with $α = α_{0} + δ \in (0, 1)$ , $∣ u_{δ} (x) - u_{δ} (y) ∣ = ∣ s^{α} - t^{α} ∣ \leq ∣ s - t ∣^{α} \leq ∣ x - y ∣^{α}$ , so $[u_{δ}]_{C^{0, α_{0} + δ}} \leq 1$ ; the value $1$ is attained by the pair $x = 0$ , $y$ with $∣ y ∣ = t \in (0, 1)$ , where the ratio is $t^{α} / t^{α} = 1$ . Hence $[u_{δ}]_{C^{0, α_{0} + δ}} = 1$ .

2.1F2F5step 1.1step 1.2givenalgebra∎

The borderline. Morrey [F5] applies on every $B (0, r)$ with $2 r < 1$ . The explicit function is also globally $α_{0}$ -Holder on $B$ : step 1.2 gives $[u_{δ}]_{C^{0, α_{0}}} \leq 2^{δ}$ . Step 1.1 gives $∥ D u_{δ} ∥_{p}^{p} = σ (S^{n - 1}) (α_{0} + δ)^{p} / (p δ) \sim σ (S^{n - 1}) α_{0}^{p} / (p δ)$ as $δ ↓ 0$ . For $δ = 0$ the same computation gives $β + 1 = 0$ , so $\int_{B} ∣ D (∣ x ∣^{α_{0}}) ∣^{p} = σ (S^{n - 1}) α_{0}^{p} \int_{0}^{1} r^{- 1} d r = + \infty$ and $∣ x ∣^{α_{0}} \notin W^{1, p} (B)$ ; the failure is exactly the divergence of the energy at the origin.

Source notes

For the positive powers $β > 0$ used here, the borderline computation is $∣ x ∣^{β} \in W^{1, p} (B)$ exactly when $p (β - 1) + n > 0$ , that is $β > 1 - n / p = α_{0}$ , with the derivative energy equal to $σ (S^{n - 1}) (α_{0} + δ)^{p} / (p δ)$ at $β = α_{0} + δ$ . Kinnunen's discussion around Theorem 3.23 and Remarks 3.25 and Laugesen's Theorem 3.21 with its moral give the Morrey estimate used for context; the displayed radial energy is computed directly above. The denominator $p δ$ is the exact value of $p (β - 1) + n$ at $β = α_{0} + δ$ ; the energy diverges like $1 / δ$ as $δ ↓ 0$ , so the $L^{p}$ norm diverges like $δ^{- 1 / p}$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6.1-sol)Open item page →

Outward dilation defeats subcritical inclusion and homogeneous Poincare on Euclidean space

Statement refuted

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 2$ , $1 \leq p < n$ and $0 < q < p$ . The assertions

$W^{1, p} (R^{n}) ↪ L^{q} (R^{n})$ continuously, and
the homogeneous Poincare inequality (even restricted to mean-zero functions) $∥ u ∥_{L^{p} (R^{n})} \leq C ∥ D u ∥_{L^{p} (R^{n})}$ for compactly supported smooth $u$ ,

are both false. In fact $W^{1, p} (R^{n})$ is not even a subset of $L^{q} (R^{n})$ . Although on bounded sets the inclusion $W^{1, p} \subseteq L^{q}$ for $q < p$ follows by applying Holder to $∣ u ∣^{q}$ and $1$ with exponents $p / q$ and $p / (p - q)$ , on the whole of $R^{n}$ even the full $W^{1, p}$ norm does not bound $L^{q}$ for $q < p$ ; additional quantitative decay or integrability assumptions would be needed. The whole-space inequality at $q = p^{*}$ is unaffected.

Facts & Assumptions

Given: The Axiom of Choice (and hence Countable Choice); $n \geq 2$ ; $1 \leq p < n$ ; $0 < q < p$ ; a fixed nonzero mean-zero $φ \in C_{c}^{\infty} (B (0, 1))$ ; and $k \geq 1$ .

[F1]

An invertible linear map scales Lebesgue measure by $∣ \det ∣$ : substituting $y = x / k$ gives $\int_{R^{n}} g (x / k) d x = k^{n} \int_{R^{n}} g (y) d y$ , and $L^{q}$ classes are determined up to null sets (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, The space $L^{p} (μ)$ as the quotient by null functions); for the smooth dilates the chain rule gives $D φ_{k} (x) = k^{- 1} D φ (x / k)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ), and the Sobolev norm is $∥ v ∥_{W^{1, p} (R^{n})} = (∥ v ∥_{L^{p}}^{p} + \sum_{i} ∥ D_{i} v ∥_{L^{p}}^{p})^{1 / p}$ for $1 \leq p < \infty$ (Integer-order Sobolev spaces and their norms).

[F2]

The closed unit ball is compact, so the support of $φ$ is compact and the scaled supports $supp φ_{k} \subseteq B (0, k)$ are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

[F3]

The Sobolev conjugate satisfies $p^{*} > p$ , and $q < p$ implies $\frac{1}{q} > \frac{1}{p}$ (The Sobolev conjugate exponent and the scaling identity); the whole-space $p^{*}$ inequality holds for $1 < p < n$ by The Gagliardo-Nirenberg-Sobolev inequality for $1 < p < n$ and for compactly supported smooth functions at $p = 1$ by The p=1 Gagliardo-Nirenberg-Sobolev inequality.

[F4]

Such a mean-zero bump exists: take a nonzero nonnegative $η \in C_{c}^{\infty} (B (0, 1 / 4))$ from A Euclidean bump for a compact set inside an open set, and set $φ (x) = η (x - e_{1} / 2) - η (x + e_{1} / 2)$ . Its supports are disjoint inside $B (0, 1)$ , and its integral is zero by Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation.

[F5]

Classical smooth derivatives are weak derivatives (Classical derivatives agree with weak derivatives); real powers differentiate on positive bases (Continuity and derivatives of positive-base real powers); polar coordinates test radial integrability (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

Counterexample

technique · direct

1.1F1F2F4givenalgebra

The two norm scalings. Put $φ_{k} (x) : = φ (x / k)$ for $k \geq 1$ , so $supp φ_{k} \subseteq B (0, k)$ is compact and $φ_{k}$ is smooth with $φ_{k} \equiv̸ 0$ . Substituting $y = x / k$ and using [F1], $∥ φ_{k} ∥_{L^{q}}^{q} = k^{n} \int_{R^{n}} ∣ φ (y) ∣^{q} d y$ , so $∥ φ_{k} ∥_{L^{q}} = k^{n / q} ∥ φ ∥_{L^{q}}$ and similarly $∥ φ_{k} ∥_{L^{p}} = k^{n / p} ∥ φ ∥_{L^{p}}$ . Also $D φ_{k} (x) = k^{- 1} (D φ) (x / k)$ by the chain rule [F1], so $∥ D φ_{k} ∥_{L^{p}}^{p} = k^{- p} k^{n} ∥ D φ ∥_{L^{p}}^{p} = k^{n - p} ∥ D φ ∥_{L^{p}}^{p}$ , that is $∥ D φ_{k} ∥_{L^{p}} = k^{n / p - 1} ∥ D φ ∥_{L^{p}}$ ; moreover $∥ D φ ∥_{L^{p}} > 0$ , because a nonzero compactly supported smooth function is not constant. The support statement uses [F2].

2.1F1F3F4step 1.1givenalgebra

Failure of the inclusion and of the homogeneous inequality. By step 1.1, $∥ φ_{k} ∥_{L^{q}} = k^{n / q} ∥ φ ∥_{L^{q}}$ while $∥ φ_{k} ∥_{W^{1, p}}^{p} = k^{n} ∥ φ ∥_{L^{p}}^{p} + k^{n - p} \sum_{i} ∥ D_{i} φ ∥_{L^{p}}^{p} \leq k^{n} ∥ φ ∥_{W^{1, p}}^{p}$ , so the normalised functions $u_{k} : = φ_{k} / ∥ φ_{k} ∥_{W^{1, p}}$ satisfy $∥ u_{k} ∥_{W^{1, p}} = 1$ and $∥ u_{k} ∥_{L^{q}} \geq k^{n / q - n / p} ∥ φ ∥_{L^{q}} / ∥ φ ∥_{W^{1, p}} \to \infty$ because $q < p$ ; hence no constant $C$ can satisfy $∥ u ∥_{L^{q}} \leq C ∥ u ∥_{W^{1, p}}$ for all $u$ , so the whole-space inclusion fails for $0 < q < p$ . For the homogeneous inequality, dividing the two scalings of step 1.1 gives $∥ φ_{k} ∥_{L^{p}} / ∥ D φ_{k} ∥_{L^{p}} = k ∥ φ ∥_{L^{p}} / ∥ D φ ∥_{L^{p}} \to \infty$ (the denominator is positive by step 1.1); thus the homogeneous estimate $∥ u ∥_{L^{p}} \leq C ∥ D u ∥_{L^{p}}$ fails on $R^{n}$ for the same family. The dilates remain mean-zero since $\int φ_{k} = k^{n} \int φ = 0$ . The exponent $q = p^{*}$ inequality is unaffected by [F3].

3.1F5givenalgebra∎

Set inclusion fails as well. Choose $a : = \frac{1}{2} (n / p + n / q)$ , so $a p > n$ and $a q < n$ . The smooth function $w (x) : = (1 + ∣ x ∣^{2})^{- a / 2}$ and its classical gradient $D w = - a x (1 + ∣ x ∣^{2})^{- a / 2 - 1}$ are bounded near zero. For $∣ x ∣ = r \geq 1$ , $2^{- a / 2} r^{- a} \leq w (x) \leq r^{- a}$ and $∣ D w (x) ∣ \leq a r^{- a - 1}$ . Polar coordinates [F5] show $\int ∣ w ∣^{p} < \infty$ and $\int ∣ D w ∣^{p} < \infty$ , since $\int_{1}^{\infty} r^{n - 1 - a p} d r$ and $\int_{1}^{\infty} r^{n - 1 - (a + 1) p} d r$ converge. But $\int ∣ w ∣^{q} \geq c \int_{1}^{\infty} r^{n - 1 - a q} d r = \infty$ . By [F5], $w \in W^{1, p} (R^{n}) ∖ L^{q} (R^{n})$ , proving the stronger set-theoretic failure.

Source notes

The family is the outward dilation of a fixed bump; Kinnunen's dilation discussion (printed pp. 61-66) and Laugesen's necessity discussion (printed pp. 66-68) motivate the dilation calculation; the explicit ratio above proves that even the full Sobolev norm cannot give this subcritical inclusion, and that the homogeneous zero-trace estimate fails by dilation.

Sobolev Poincare and Morrey Inequalities — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Dilations force the Sobolev conjugate

Example

Facts & Assumptions

Verification

Source notes

Poincare on an interval: the length dependence is linear

Example

Facts & Assumptions

Verification

Source notes

Poincare-Wirtinger fails on disconnected bounded domains

Statement refuted

Facts & Assumptions

Counterexample

Source notes

A gradient-only Poincare estimate needs normalisation

Statement refuted

Facts & Assumptions

Counterexample

Source notes

The W1,p→LqW1,p→Lq bound fails for q>p∗q>p∗ by dilation

Statement refuted

Facts & Assumptions

Counterexample

Source notes

W1,nW1,n is not contained in L∞L∞

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Morrey's inequality has no p=np=n endpoint

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Radial powers approach the Morrey borderline exponent

Example

Facts & Assumptions

Verification

Source notes

Outward dilation defeats subcritical inclusion and homogeneous Poincare on Euclidean space

Statement refuted

Facts & Assumptions

Counterexample

Source notes

Sources

The $W^{1, p} \to L^{q}$ bound fails for $q > p^{*}$ by dilation

$W^{1, n}$ is not contained in $L^{\infty}$

Morrey's inequality has no $p = n$ endpoint