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The coordinate inequality $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $j \in {1, \dots, n}$ and $f \in H^{1} (R^{n}; C)$ with $x f \in L^{2} (R^{n}; C^{n})$ . Here $H^{1} = W^{1, 2}$ under the regular-distribution identification of Integer-order W^{k,2} and H^k agree with equivalent norms, and $D_{j} f \in L^{2}$ denotes the weak partial derivative. Then $∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} \geq \frac{1}{2} ∥ f ∥_{2}^{2} .$ By the Fourier characterization of $H^{1}$ , this is the coordinate estimate on the natural domain where both $∣ x ∣ f$ and $∣ ξ ∣ \hat{f}$ lie in $L^{2}$ . Both sides vanish when $f = 0$ ; for nonzero $f$ the right side is positive.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $j \in {1, \dots, n}$ , and $f \in H^{1} (R^{n}; C)$ with $x f \in L^{2} (R^{n}; C^{n})$ .

[A1]

Countable Choice is the assumption carried by the Sobolev and compact-support integration-by-parts interfaces below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The Fourier characterization identifies $H^{1}$ with $W^{1, 2}$ under the regular-distribution embedding, so each weak derivative $D_{k} f$ belongs to $L^{2}$ ; its Fourier-side weight is $⟨ ξ ⟩ \hat{f}$ (Integer-order W^{k,2} and H^k agree with equivalent norms).

[F2]

There is a real smooth cutoff $χ$ with $0 \leq χ \leq 1$ , $χ = 1$ on $∣ x ∣ \leq 1$ , $χ = 0$ on $∣ x ∣ \geq 2$ , and $D_{j} (χ (x / R)) = R^{- 1} (D_{j} χ) (x / R)$ (Explicit compactly supported smooth cutoffs).

[F3]

For $u \in W^{1, 2}$ and a smooth multiplier $ρ$ with bounded derivatives, $ρ u \in W^{1, 2}$ and $D_{j} (ρ u) = (D_{j} ρ) u + ρ D_{j} u$ (Weak Leibniz rule with a smooth factor).

[F4]

If $u, v \in W^{1, 2} (R^{n})$ and one factor is compactly supported as an almost-everywhere class, then $\int u D_{j} v = - \int v D_{j} u$ (Integration by parts for dual-exponent Sobolev functions).

[F5]

Complex $L^{2}$ satisfies Cauchy–Schwarz, so the product of two $L^{2}$ functions is integrable (Complex completeness, density, and inner product: the consumer interface).

[F6]

If integrable functions converge almost everywhere under a common integrable majorant, their integrals converge (Dominated convergence).

[F7]

The Sobolev test identity is bilinear and does not conjugate its test function (Integer-order Sobolev spaces and their norms, Complex Lp classes and Euclidean test-function conventions).

Proof

technique · cutoff integration by parts, followed by vanishing $L^2$ tails

1.1A1F1F7given

Since $f \in H^{1} = W^{1, 2}$ by [F1], each weak derivative $D_{k} f$ lies in $L^{2}$ . Conjugating the bilinear weak-derivative test identity for $f$ shows that $\overline{f} \in W^{1, 2}$ and $D_{k} \overline{f} = \overline{D_{k} f}$ : for each $ϕ \in C_{c}^{\infty}$ , apply the identity to $\overline{ϕ}$ and conjugate it.

2.1F2F3F4step 1.1given

For $R \geq 1$ set $χ_{R} (x) = χ (x / R)$ and $ρ_{R} (x) = x_{j} χ_{R} (x)$ . By [F2], $ρ_{R} \in C_{c}^{\infty}$ ; by [F3] and step 1.1, $v_{R} : = ρ_{R} \overline{f}$ belongs to $W^{1, 2}$ and is compactly supported, with $D_{j} v_{R} = (χ_{R} + x_{j} D_{j} χ_{R}) \overline{f} + x_{j} χ_{R} \overline{D_{j} f} .$ Applying [F4] to $u = f$ and $v = v_{R}$ gives $\int f D_{j} v_{R} = - \int v_{R} D_{j} f,$ hence $\int χ_{R} ∣ f ∣^{2} + \int x_{j} D_{j} χ_{R} ∣ f ∣^{2} = - 2 Re \int x_{j} χ_{R} \overline{f} D_{j} f .$ The last identity uses that $χ_{R}$ and $x_{j}$ are real-valued, so the two derivative terms are complex conjugates.

3.1F2F5F6step 2.1

Let $R \to \infty$ through positive integers. We have $χ_{R} (x) \to 1$ pointwise and $0 \leq χ_{R} \leq 1$ , so dominated convergence [F6] gives $\int χ_{R} ∣ f ∣^{2} \to ∥ f ∥_{2}^{2}$ . The derivative $D_{j} χ_{R}$ vanishes unless $R \leq ∣ x ∣ \leq 2 R$ , and $∣ x_{j} D_{j} χ_{R} ∣ \leq 2 ∥ D_{j} χ ∥_{\infty}$ there by [F2]; for every fixed $x$ this factor is eventually zero. Since $∣ f ∣^{2} \in L^{1}$ , [F6] gives $\int x_{j} D_{j} χ_{R} ∣ f ∣^{2} \to 0$ . Finally, $x_{j} \overline{f} D_{j} f \in L^{1}$ by [F5], because $x_{j} f, D_{j} f \in L^{2}$ ; dominated convergence with $∣ χ_{R} ∣ \leq 1$ yields $∥ f ∥_{2}^{2} = - 2 Re \int x_{j} \overline{f} D_{j} f .$

4.1F5step 3.1∎

By [F5] and $∣ Re z ∣ \leq ∣ z ∣$ , $∥ f ∥_{2}^{2} \leq 2 \int ∣ x_{j} f ∣ ∣ D_{j} f ∣ \leq 2 ∥ x_{j} f ∥_{2} ∥ D_{j} f ∥_{2} .$ Dividing by $2$ proves the inequality. If $f = 0$ both sides vanish; if $f \neq 0$ the lower bound is positive.

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Sources

Calder Sheagren, Uncertainty Principles with Fourier Analysis (University of Chicago REU 2017, author PDF) (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)

The coordinate inequality ∥xjf∥2∥Djf∥2≥12∥f∥22∥xj​f∥2​∥Dj​f∥2​≥21​∥f∥22​