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Mean-zero Poincare inequality on the normalized torus

Statement

Assume Countable Choice. Let $d \in {1, 2, 3}$ , $N \geq 1$ , and $Q_{d} = R^{d} / (2 π Z)^{d}$ with normalized Lebesgue measure $d μ = (2 π)^{- d} d x$ . For a smooth real vector field $f : Q_{d} \to R^{N}$ , define its mean by $\overset{ˉ}{f} = \int_{Q_{d}} f d μ$ and the Frobenius gradient norm by $∣ \nabla f ∣^{2} = \sum_{a = 1}^{N} \sum_{j = 1}^{d} ∣ \partial_{j} f_{a} ∣^{2}$ . If $\overset{ˉ}{f} = 0$ , then $∥ f ∥_{L^{2} (Q_{d})} \leq ∥ \nabla f ∥_{L^{2} (Q_{d})}$ .

Statement references

Periodic fluid Fourier Sobolev spaces

Kuksin and Shirikyan, Mathematics of Two-Dimensional Turbulence, §2.1.7

Facts & Assumptions

Given: Countable Choice, $d \in {1, 2, 3}$ , $N \geq 1$ , and a smooth real field $f : Q_{d} \to R^{N}$ with $\overset{ˉ}{f} = 0$ .

[A1]

Countable Choice: every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Periodic fluid Fourier Sobolev spaces: $\hat{f} (k) = \int_{Q_{d}} f (x) e^{- i k \cdot x} d μ (x)$ , the zero coefficient is the mean, and $∥ f ∥_{H^{0}}^{2} = \sum_{k \in Z^{d}} ∣ \hat{f} (k) ∣^{2}$ .

[F2]

The Fourier basis and Parseval's identity on the finite torus: under Countable Choice, Parseval identifies the $L^{2}$ norm on the normalized unit torus with the squared coefficient sum. Rescaling $x = 2 π y$ gives the normalized $2 π$ -periodic convention here. The derivative-coefficient identity is established locally below.

Proof

1.1givenF1algebra

By [F1] and $\overset{ˉ}{f} = 0$ , $\hat{f} (0) = 0$ . Every nonzero $k \in Z^{d}$ has $∣ k ∣^{2} \geq 1$ .

1.2givenF1algebra

For each component $f_{a}$ , mode $k$ , and coordinate $j$ , integrate $\partial_{j} (f_{a} (x) e^{- i k \cdot x})$ over $[0, 2 π]^{d}$ . The two faces perpendicular to the $j$ th direction cancel because both factors are periodic. Expanding the derivative and using the normalization in [F1] gives $\hat{\partial_{j} f_{a}} (k) = i k_{j} {\hat{f}}_{a} (k)$ . Smoothness makes all these integrands continuous and integrable.

2.1givenA1F2step 1.1step 1.2

By [F2] and step 1.2, componentwise Parseval and the locally proved derivative-coefficient formula give $∥ f ∥_{L^{2}}^{2} = \sum_{k \neq 0} ∣ \hat{f} (k) ∣^{2}$ and $∥ \nabla f ∥_{L^{2}}^{2} = \sum_{k \neq 0} ∣ k ∣^{2} ∣ \hat{f} (k) ∣^{2}$ . Termwise comparison using [step 1.1] yields $∥ f ∥_{L^{2}}^{2} \leq ∥ \nabla f ∥_{L^{2}}^{2}$ .

3.1step 2.1algebra∎

Both norms are nonnegative, so taking square roots proves the asserted inequality.

Depends on

Dependency tree · 2 levels

Sources

Sergei Kuksin and Armen Shirikyan, Mathematics of Two-Dimensional Turbulence, §2.1.7