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Zero-boundary Poincaré inequality on the unit interval

Statement

Assume Countable Choice. For every real $v \in H_{0}^{1} (0, 1)$ ,

$∥ v ∥_{L^{2} (0, 1)} \leq ∥ v^{'} ∥_{L^{2} (0, 1)} .$

Consequently $∥ v ∥_{H^{1} (0, 1)} \leq \sqrt{2} ∥ v^{'} ∥_{L^{2} (0, 1)}$ .

Facts & Assumptions

Given: Countable Choice and $v$ in the closure of $C_{c}^{\infty} (0, 1)$ in the $H^{1}$ norm.

[F1]

Zero-boundary Sobolev space as a norm closure: for every $ε > 0$ , some $φ \in C_{c}^{\infty} (0, 1)$ satisfies $∥ v - φ ∥_{H^{1}} < ε$ .

[F2]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : a smooth function's increment is the integral of its derivative.

[F3]

On $[0, 1]$ , Countable Choice gives $λ_{1} ([0, 1]) = λ_{1} ((0, 1)) = 1$ (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, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Hence Holder gives $\int_{0}^{1} ∣ g ∣ \leq ∥ g ∥_{L^{2}}$ and $∥ 1 ∥_{L^{2}} = 1$ (Holder's inequality for integrals, including the endpoint cases).

[F4]

Integer-order Sobolev spaces and their norms: for $k = 1, p = 2$ , $∥ v ∥_{H^{1}}^{2} = ∥ v ∥_{2}^{2} + ∥ v^{'} ∥_{2}^{2}$ .

[F5]

The nonnegative Lebesgue integral is monotone (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F6]

Continuous functions on closed intervals are bounded and Riemann integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). Under Countable Choice their Riemann and Lebesgue integrals agree (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

Proof

1.1F2F3F5F6givenalgebra

Extend $φ \in C_{c}^{\infty} (0, 1)$ smoothly by zero to $[0, 1]$ . Its derivative is continuous on every $[0, x]$ with $0 < x \leq 1$ , so [F6] makes it Riemann integrable and identifies its Riemann integral with its Lebesgue integral. Thus [F2] gives $φ (x) = \int_{0}^{x} φ^{'} (t) d t$ ; at $x = 0$ both sides are zero. Hence $∣ φ (x) ∣ \leq \int_{0}^{1} ∣ φ^{'} (t) ∣ d t \leq ∥ φ^{'} ∥_{2}$ by [F3]. Monotonicity [F5] and $λ_{1} ((0, 1)) = 1$ give $∥ φ ∥_{2}^{2} = \int_{0}^{1} ∣ φ ∣^{2} \leq ∥ φ^{'} ∥_{2}^{2}$ , so $∥ φ ∥_{2} \leq ∥ φ^{'} ∥_{2}$ .

2.1F1F4step 1.1algebra∎

Given $ε > 0$ , choose $φ$ by [F1]. The triangle inequality and step 1.1 give $∥ v ∥_{2} \leq ∥ φ ∥_{2} + ε \leq ∥ φ^{'} ∥_{2} + ε \leq ∥ v^{'} ∥_{2} + 2 ε$ . Since this holds for every positive $ε$ , $∥ v ∥_{2} \leq ∥ v^{'} ∥_{2}$ . By [F4], squaring and adding the derivative norm proves the $H^{1}$ estimate.

Depends on

Used by

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Sources

Richard S. Laugesen, Linear Analysis and Partial Differential Equations, Chapter 3, Sobolev spaces (standard reference, not scraped)
John K. Hunter, Notes on Partial Differential Equations, Chapter 4, Elliptic PDEs (standard reference, not scraped)