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Poincare-Wirtinger on bounded convex domains by the direct pairwise argument

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $Ω \subseteq R^{n}$ be open, bounded, convex and nonempty, and let $1 \leq p < \infty$ . Then $∥ u - u_{Ω} ∥_{L^{p} (Ω)} \leq C (n) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ for every $u \in W^{1, p} (Ω; K)$ .

Here $u_{Ω} = ∣ Ω ∣^{- 1} \int_{Ω} u d x$ is the mean of $u$ over $Ω$ , which is well defined because $Ω$ is nonempty open and bounded. The proof below gives the explicit choice $C (n) = \frac{2 (2^{n} - 1)}{n},$ which depends only on the dimension; no dependence on $p$ , on the shape of $Ω$ , on $∣ Ω ∣$ or on the regularity of $\partial Ω$ is used.

Facts & Assumptions

Given: The Axiom of Choice; an integer $n \geq 1$ ; an open, bounded, convex, nonempty set $Ω \subseteq R^{n}$ with $d : = diam (Ω)$ ; an exponent $1 \leq p < \infty$ ; a field $K \in {R, C}$ ; and a class $u \in W^{1, p} (Ω; K)$ .

[F1]

$W^{1, p} (Ω; K)$ consists of the $L^{p} (Ω; K)$ classes whose weak first derivatives exist as $L^{p}$ classes, and $∥ D u ∥_{L^{p} (Ω)}$ is the $L^{p}$ norm of the weak gradient (Integer-order Sobolev spaces and their norms); an element of $L^{p}$ is an almost-everywhere equivalence class of measurable representatives (The space $L^{p} (μ)$ as the quotient by null functions).

[F2]

The Axiom of Choice is the statement that every family of nonempty sets has a choice function (The Axiom of Choice); it implies the Axiom of Countable Choice, the statement that every at most countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

Under Countable Choice, for every open $U \subset \subset Ω$ the interior mollifications $u_{ε}$ of $u$ satisfy $u_{ε} \to u$ in $W^{1, p} (U; K)$ as $ε \to 0^{+}$ (Local smooth approximation in integer-order Sobolev spaces).

[F4]

On a completed sigma-finite product, nonnegative measurable functions may be integrated in either order and the iterated integrals agree, and integrable functions obey the same identity; this is Tonelli-Fubini (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability).

[F5]

Jensen's inequality: for a probability space, an integrable real function $f$ with values in an interval $I$ and a convex $φ$ on $I$ such that $φ \circ f \in L^{1} (P)$ , one has $φ (\int f d P) \leq \int φ (f) d P$ (Jensen's integral inequality for a probability measure).

[F6]

Holder's inequality: for conjugate exponents $p, q$ and measurable real $f, g$ in the corresponding $L$ -spaces, $\int ∣ f g ∣ d μ \leq ∥ f ∥_{p} ∥ g ∥_{q}$ (Holder's inequality for integrals, including the endpoint cases).

[F7]

For a $C^{1}$ diffeomorphism $T : U \to V$ between open subsets of $R^{m}$ and every nonnegative Borel $h : V \to [0, \infty]$ , $\int_{V} h (y) d y = \int_{U} h (T (x)) ∣ \det D T (x) ∣ d x$ , with equality in $[0, \infty]$ and the convention $0 \cdot \infty = 0$ (Borel change of variables from the compact-support formula and Radon uniqueness).

[F8]

If $f : [a, b] \to R^{m}$ is differentiable with integrable derivative then $\int_{a}^{b} f^{'} = f (b) - f (a)$ (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); if $g \circ f$ is formed from totally differentiable maps then $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F9]

Every Euclidean ball has positive finite Lebesgue measure (Euclidean balls have positive finite Lebesgue measure), a measure is monotone under inclusion (Measures are monotone), and every bounded subset of $R^{n}$ has finite outer measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F10]

Dominated convergence applies to integrable majorants (Dominated convergence).

Proof

technique · direct

1.1F1F6F9givenalgebra

Means and integrability. Since $Ω$ is nonempty and open it contains a Euclidean ball $B \subseteq Ω$ , and [F9] gives $0 < λ (B) \leq λ (Ω)$ by monotonicity; since $Ω$ is bounded, [F9] gives $∣ Ω ∣ < \infty$ . Thus $0 < ∣ Ω ∣ < \infty$ and $u_{Ω}$ is defined once $u$ is integrable. For $1 < p < \infty$ Holder's inequality [F6] with $g = 1_{Ω}$ gives $∥ u ∥_{1} \leq ∣ Ω ∣^{1 - 1 / p} ∥ u ∥_{p} < \infty$ , and for $p = 1$ integrability is immediate; in both cases $u \in L^{1} (Ω; K)$ and $u_{Ω}$ is an element of $K$ , understood componentwise when $K = C ≅ R^{2}$ .

1.2F5F8givenalgebra

The smooth segment inequality. Let $V \subseteq Ω$ be open and convex and let $v \in C^{\infty} (V; K) \cap W^{1, p} (V; K)$ . Fix $x, y \in V$ and put $γ (t) = x + t (y - x)$ for $t \in [0, 1]$ ; convexity gives $γ ([0, 1]) \subseteq V$ . By the chain rule [F8], $v \circ γ$ is differentiable with $(v \circ γ)^{'} (t) = D v (γ (t)) (y - x)$ , and the vector-valued fundamental theorem [F8] gives $v (x) - v (y) = \int_{0}^{1} D v (γ (t)) (x - y) d t$ . Hence $∣ v (x) - v (y) ∣ \leq ∣ x - y ∣ \int_{0}^{1} ∣ D v (γ (t)) ∣ d t$ , and, since $D v$ is continuous on the compact segment, $∣ D v \circ γ ∣^{p}$ is integrable. Jensen's inequality [F5] applied to the probability measure $d t$ on $[0, 1]$ and the convex function $φ (s) = s^{p}$ on $[0, \infty)$ yields $∣ v (x) - v (y) ∣^{p} \leq ∣ x - y ∣^{p} \int_{0}^{1} ∣ D v (γ (t)) ∣^{p} d t$ .

2.1F4F7step 1.2algebra

The first half of the substitution. For $t \in [1 / 2, 1]$ and fixed $x \in V$ , the affine diffeomorphism $y \mapsto z = (1 - t) x + t y$ has image $(1 - t) x + t V \subseteq V$ and inverse Jacobian $t^{- n}$ . Bound $∣ x - y ∣ \leq d$ before substituting in [F7]; this gives $\int_{V} ∣ x - y ∣^{p} ∣ D v ((1 - t) x + t y) ∣^{p} d y \leq d^{p} t^{- n} \int_{V} ∣ D v ∣^{p}$ . Integrating over $x \in V$ and $t \in [1 / 2, 1]$ yields $c_{n} d^{p} ∣ V ∣ \int_{V} ∣ D v ∣^{p}$ , where $c_{n} : = \int_{1 / 2}^{1} t^{- n} d t$ .

2.2F4F7step 1.2algebra

The second half. For $t \in [0, 1 / 2]$ and fixed $y \in V$ , substitute $z = (1 - t) x + t y \in V$ in the $x$ integral. The bound $∣ x - y ∣ \leq d$ and [F7] give $\int_{V} ∣ x - y ∣^{p} ∣ D v ((1 - t) x + t y) ∣^{p} d x \leq d^{p} (1 - t)^{- n} \int_{V} ∣ D v ∣^{p}$ . Integration over $y$ and $t$ , with $s = 1 - t$ , gives the same $c_{n} d^{p} ∣ V ∣ \int_{V} ∣ D v ∣^{p}$ .

3.1F4F5step 2.1step 2.2algebra

The mean-zero bound for smooth functions. Adding steps 2.1 and 2.2 and using [F4] to identify the iterated integral over $V \times V \times [0, 1]$ of the nonnegative integrand with the sum of its two halves, $\int_{V} \int_{V} ∣ v (x) - v (y) ∣^{p} d x d y \leq 2 c_{n} d^{p} ∣ V ∣ \int_{V} ∣ D v ∣^{p} .$ The normalized Lebesgue measure $d y / ∣ V ∣$ is a probability measure, so using $∣ \int h ∣ \leq \int ∣ h ∣$ and scalar Jensen [F5] for $s \mapsto s^{p}$ , with $v_{V} = ∣ V ∣^{- 1} \int_{V} v$ , the required integrability holds because $v \in L^{p} (V)$ implies $∣ v (x) - v (\cdot) ∣^{p} \in L^{1} (V)$ for every fixed $x$ ; hence one has $∣ v (x) - v_{V} ∣^{p} \leq ∣ V ∣^{- 1} \int_{V} ∣ v (x) - v (y) ∣^{p} d y$ for every $x \in V$ , and integrating in $x$ yields $\int_{V} ∣ v - v_{V} ∣^{p} \leq 2 c_{n} d^{p} \int_{V} ∣ D v ∣^{p} .$

4.1F1F2F3step 3.1givenalgebra

Passage to $W^{1, p}$ and to the whole of $Ω$ . Choose open convex sets $V_{1} \subseteq V_{2} \subseteq \dots \subset \subset Ω$ with $⋃_{j} V_{j} = Ω$ , for instance $V_{j} = {x \in Ω : dist (x, R^{n} ∖ Ω) > 1 / j} \cap B (0, j)$ . Fix $j$ ; by [F3] the mollifications $u_{ε}$ of $u$ converge to $u$ in $W^{1, p} (V_{j}; K)$ as $ε \to 0^{+}$ , and each $u_{ε}$ is smooth on a neighbourhood of $V_{j}$ . Step 3.1 applied to $v = u_{ε}$ on the convex set $V_{j}$ , followed by the limits $∥ u_{ε} - u ∥_{L^{p} (V_{j})} \to 0$ , $∥ D u_{ε} - D u ∥_{L^{p} (V_{j})} \to 0$ and $(u_{ε})_{V_{j}} \to u_{V_{j}}$ as $ε \to 0^{+}$ , gives $\int_{V_{j}} ∣ u - u_{V_{j}} ∣^{p} \leq 2 c_{n} d^{p} \int_{V_{j}} ∣ D u ∣^{p} \leq 2 c_{n} d^{p} \int_{Ω} ∣ D u ∣^{p} .$

5.1F10step 4.1algebra∎

Exhaustion and the constant. Discard the finitely many empty $V_{j}$ . Since $V_{j} ↑ Ω$ and $u \in L^{1}$ , dominated convergence [F10] gives $∣ V_{j} ∣ \to ∣ Ω ∣$ and $u_{V_{j}} \to u_{Ω}$ . The means are bounded; hence $1_{V_{j}} ∣ u - u_{V_{j}} ∣^{p}$ is dominated by $2^{p - 1} (∣ u ∣^{p} + \sup_{j} ∣ u_{V_{j}} ∣^{p})$ on the finite-measure set $Ω$ . By [F10] it converges in integral to $∣ u - u_{Ω} ∣^{p}$ , and similarly $\int_{V_{j}} ∣ D u ∣^{p} \to \int_{Ω} ∣ D u ∣^{p}$ . Step 4.1 yields $∥ u - u_{Ω} ∥_{p} \leq (2 c_{n})^{1 / p} d ∥ D u ∥_{p}$ . For $1 / 2 \leq t \leq 1$ , $t^{- n} \geq 1$ and $t^{- n} \leq t^{- n - 1}$ , so $1 \leq 2 c_{n} \leq 2 \int_{1 / 2}^{1} t^{- n - 1} d t = 2 (2^{n} - 1) / n$ . Thus $(2 c_{n})^{1 / p} \leq 2 c_{n} \leq 2 (2^{n} - 1) / n$ for every $p \geq 1$ , proving the stated dimension-only constant.

Source notes

The computation follows Kinnunen's ball proof of the pointwise oscillation estimate and the Poincare inequality, printed pp. 133-136, with the segment argument of Lemma 5.22: the ball is replaced by the convex set, polar coordinates and the maximal function are not needed, and the two halves of the parameter interval carry the substitution from the moving interior point to a fixed one. The constant is not claimed to be sharp; the dimension-only bound is the conclusion used here.

Depends on

Used by

Domain classes covered by the mean-zero Poincare inequality Remark
Poincare inequality on a ball Theorem

Dependency tree · two levels

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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)