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Mean-zero Poincare inequality on a rectangle

Statement

Let $d \geq 1$ , let $L_{1}, \dots, L_{d} > 0$ , and let $Ω = \prod_{i = 1}^{d} (0, L_{i})$ . If $w \in C^{1} (\overline{Ω}, R)$ satisfies $\int_{Ω} w (x) d x = 0$ , write $\nabla w = (\partial_{1} w, \dots, \partial_{d} w)$ and $∣ \nabla w ∣^{2} = \sum_{i = 1}^{d} ∣ \partial_{i} w ∣^{2}$ . Then $\int_{Ω} ∣ w (x) ∣^{2} d x \leq \frac{d}{2} \sum_{i = 1}^{d} L_{i}^{2} \int_{Ω} ∣ \partial_{i} w (x) ∣^{2} d x \leq C_{Ω} \int_{Ω} ∣ \nabla w (x) ∣^{2} d x,$ where $C_{Ω} = \frac{d}{2} \max_{i} L_{i}^{2}$ . All integrals are the multidimensional Riemann integrals of continuous functions on the indicated closed rectangles, with the usual extension to their interiors.

Facts & Assumptions

Given: A dimension $d \geq 1$ , positive side lengths $L_{i}$ , the open rectangle $Ω = \prod_{i} (0, L_{i})$ , and $w \in C^{1} (\overline{Ω}, R)$ with $\int_{Ω} w = 0$ .

[A1]

The zero-mean condition is $\int_{Ω} w (x) d x = 0$ .

[F1]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: if $G$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and its interior derivative has a Riemann-integrable extension $f$ , then $\int_{a}^{b} f = G (b) - G (a)$ .

[F2]

Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections: when every section of a Riemann-integrable function on a product of nondegenerate closed rectangles is integrable, the ordinary iterated integral equals the multiple integral. In particular, this applies to the continuous integrands used here.

Proof

1.1A1F2algebra

Expand the square in the product integral and use [A1]: $\int_{Ω} \int_{Ω} (w (x) - w (y))^{2} d y d x = 2 ∣ Ω ∣ \int_{Ω} w^{2} - 2 (\int_{Ω} w)^{2} = 2 ∣ Ω ∣ \int_{Ω} w^{2}$ . The integrand is continuous, so [F2] justifies the repeated integrals.

1.2givenalgebra

For $x, y \in Ω$ , define $z^{(0)} = x$ and $z^{(i)} = (y_{1}, \dots, y_{i}, x_{i + 1}, \dots, x_{d})$ for $1 \leq i \leq d$ , and set $δ_{i} = w (z^{(i - 1)}) - w (z^{(i)})$ . Then $w (x) - w (y) = \sum_{i} δ_{i}$ and $d \sum_{i} δ_{i}^{2} - (\sum_{i} δ_{i})^{2} = \sum_{i < j} (δ_{i} - δ_{j})^{2} \geq 0$ .

1.3F1algebra

Fix $i$ and the other coordinates. By [F1], $δ_{i}$ is the integral of $h = \partial_{i} w$ over the interval between $x_{i}$ and $y_{i}$ . If the endpoints differ, let $\overset{ˉ}{h}$ be the average of $h$ on that interval. Nonnegativity of $\int (h - \overset{ˉ}{h})^{2}$ gives $(\int h)^{2} \leq ∣ x_{i} - y_{i} ∣ \int h^{2}$ ; if they agree, $δ_{i} = 0$ . Since $∣ x_{i} - y_{i} ∣ \leq L_{i}$ , in both cases $δ_{i}^{2} \leq L_{i} \int_{0}^{L_{i}} ∣ \partial_{i} w (z_{1}, \dots, z_{i - 1}, s, z_{i + 1}, \dots, z_{d}) ∣^{2} d s$ .

2.1F2step 1.3algebra

Integrate the last bound over $x, y$ . The $x_{i}, y_{i}$ integrations contribute $L_{i}^{2}$ , the displayed segment bound contributes $L_{i}$ , each unused variable from a pair $x_{j}, y_{j}$ with $j \neq i$ contributes $L_{j}$ , and the remaining variables and $s$ give $\int_{Ω} ∣ \partial_{i} w ∣^{2}$ . By [F2], this yields $\int_{Ω} \int_{Ω} δ_{i}^{2} d y d x \leq ∣ Ω ∣ L_{i}^{2} \int_{Ω} ∣ \partial_{i} w ∣^{2}$ .

3.1step 1.1step 1.2step 2.1algebra∎

By step 1.2 and the finite squared-sum inequality, the double integral in step 1.1 is at most $d \sum_{i} \int_{Ω} \int_{Ω} δ_{i}^{2} d y d x$ . Apply step 2.1, divide by $2 ∣ Ω ∣ > 0$ , and use $\sum_{i} L_{i}^{2} ∣ \partial_{i} w ∣^{2} \leq (\max_{i} L_{i}^{2}) ∣ \nabla w ∣^{2}$ pointwise. This gives both stated inequalities with the displayed $C_{Ω}$ .

Depends on

Dependency tree · 2 levels

Sources

John K. Hunter, Notes on Partial Differential Equations, §4.4