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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.

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Sources

Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete 158-page graduate notes) (standard reference, not scraped)