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A Poisson-type extension and its local and global Sobolev traces

Example

Assume the Axiom of Choice, let $d \geq 1$ and use the $2 π$ -normalised Fourier transform. For $g \in C_{c}^{\infty} (R^{d}; K)$ define $U (x, t) : = \int_{R^{d}} e^{2 π i x \cdot ξ} e^{- 2 π t ∣ ξ ∣} \hat{g} (ξ) d ξ, x \in R^{d}, t \geq 0.$ Then $U$ is smooth on $R^{d} \times [0, \infty)$ , bounded and harmonic on $H = R^{d} \times (0, \infty)$ , extends the datum with $U (\cdot, 0) = g$ , and $\nabla U \in L^{2} (H)$ with $∥ \nabla U ∥_{L^{2} (H)}^{2} = 2 π \int_{R^{d}} ∣ ξ ∣ ∣ \hat{g} (ξ) ∣^{2} d ξ < \infty .$ The classical boundary value is realised by the trace in two forms. (i) For every $R > 0$ , choose $κ_{R} \in C_{c}^{\infty} (R^{d})$ equal to one on $B_{R} (0)$ and a smooth compact normal cutoff $η$ equal to one near zero. Then $V_{R} (x, t) = κ_{R} (x) η (t) U (x, t)$ belongs to $W^{1, 2} (H)$ , is continuous with compact support in $\overline{H}$ , and $T_{+} V_{R} = κ_{R} g$ , giving the boundary value on $B_{R} (0)$ . (ii) If in addition $U \in L^{2} (H)$ — which holds for every $g$ when $d \geq 2$ , and for $d = 1$ exactly when $\int_{R} g = 0$ — then $U \in W^{1, 2} (H)$ and $T_{+} U = g$ for the flat trace $T_{+}$ of The half-space trace estimate and the half-space trace operator. For $d = 1$ with $\int_{R} g \neq 0$ one has $U \notin L^{2} (H)$ , so the half-space trace is not defined on $U$ and (i) is the correct local form of the identity. This exhibits a Poisson-type right inverse of the half-space trace for smooth data satisfying the stated $L^{2}$ condition; the cutoff form gives a local lift for every smooth datum. It is an illustration only: it does not prove the general- $p$ right inverse of A bounded right inverse of the trace, supported in a prescribed collar.

Facts & Assumptions

Given: The Axiom of Choice; $d \geq 1$ ; the $2 π$ -normalised transform of Fourier transform on complex L1 classes; a datum $g \in C_{c}^{\infty} (R^{d}; K)$ ; the extension $U$ defined by the displayed integral; the half-space $H = R^{d} \times (0, \infty)$ with its flat trace $T_{+}$ of The half-space trace estimate and the half-space trace operator.

[F1]

Fourier inversion on Schwartz space: for $f \in S (R^{d})$ and every $x$ , $f (x) = \int_{R^{d}} \hat{f} (ξ) e^{2 π i x \cdot ξ} d ξ$ , the integral converging absolutely. (Fourier inversion on Schwartz space)

[F2]

Plancherel gives a unitary Fourier transform on $L^{2}$ . For $h \in L^{1} \cap L^{2}$ , the inverse Fourier integral $\int h (ξ) e^{2 π i x \cdot ξ} d ξ$ represents $F_{2}^{- 1} h$ and has $L^{2}$ norm $∥ h ∥_{2}$ : apply integral/L2 agreement to $h$ , reflect $x \mapsto - x$ , and extend the Schwartz inversion identity by $L^{2}$ continuity. (Plancherel theorem, Agreement of the integral and L2 transforms, Fourier inversion on Schwartz space)

[F3]

The negative-sign, $2 π$ -normalised transform maps $S$ continuously to itself: for $g \in C_{c}^{\infty} (R^{d})$ the transform $\hat{g}$ is Schwartz and $F (\partial^{α} f) (ξ) = (2 π i ξ)^{α} \hat{f} (ξ)$ , $\partial^{β} \hat{f} = F ((- 2 π i x)^{β} f)$ ; in particular $∣ ξ ∣^{N} ∣ D^{β} \hat{g} (ξ) ∣$ is bounded on $R^{d}$ for all multi-indices and all $N$ . (Fourier transform acts continuously on Schwartz space, Fourier transform on complex L1 classes)

[F4]

Poisson kernel model in ambient dimension $n \geq 3$ : for bounded continuous $g$ on $\partial H = R^{n - 1}$ , the Poisson integral is bounded, smooth and harmonic on $H$ , continuous on $\overline{H}$ with boundary value $g$ , and it is the unique bounded harmonic function on $H$ with these properties. (Poisson kernel and bounded Dirichlet problem on a half-space)

[F5]

Tonelli for nonnegative measurable functions on a completed product measure: the iterated integral equals the product integral, finite or infinite. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

[F6]

Dominated convergence, in integral and in $L^{2}$ form: if $f_{k} \to f$ almost everywhere and $∣ f_{k} ∣ \leq G$ for a single integrable $G$ , then $\int f_{k} \to \int f$ ; if $∣ f_{k} ∣ \leq G$ for a single $G \in L^{2}$ , then $∥ f_{k} - f ∥_{2} \to 0$ . (Dominated convergence)

[F8]

The flat trace $T_{+} : W^{1, 2} (H) \to L^{2} (R^{d})$ is the unique bounded extension of classical restriction, and $T_{+} u = u (\cdot, 0)$ for every compactly supported $u \in C (\overline{H}) \cap W^{1, 2} (H)$ . (The half-space trace estimate and the half-space trace operator)

[F9]

Weak Leibniz rule with a smooth factor: if $η \in C^{\infty} (H)$ has bounded value and first derivatives and $u \in W^{1, 2} (H)$ , then $η u \in W^{1, 2} (H)$ with $\nabla (η u) = η \nabla u + u \nabla η$ as $L^{2}$ classes. (Weak Leibniz rule with a smooth factor)

Proof

technique · direct

1.1F1F2F3F4F6algebragiven

Smoothness, boundedness, harmonicity, boundary values and the Poisson identification. For every pair of multi-indices the differentiated integrand equals $(2 π i ξ)^{α} (- 2 π ∣ ξ ∣)^{k} e^{2 π i x \cdot ξ} e^{- 2 π t ∣ ξ ∣} \hat{g} (ξ)$ , which on $t \geq 0$ is dominated by $C_{α, k} ∣ ξ ∣^{∣ α ∣ + k} ∣ \hat{g} (ξ) ∣$ , an integrable function because $\hat{g}$ is Schwartz [F3]; differentiating under the integral sign is therefore legitimate, so $U \in C^{\infty} (R^{d} \times [0, \infty))$ with those derivative formulas, $∣ U ∣ \leq ∥ \hat{g} ∥_{1}$ , and $U (x, 0) = \int e^{2 π i x \cdot ξ} \hat{g} (ξ) d ξ = g (x)$ by inversion [F1]. For $t > 0$ the symbol identity $(- 2 π ∣ ξ ∣)^{2} + \sum_{j} (2 π i ξ_{j})^{2} = 4 π^{2} ∣ ξ ∣^{2} - 4 π^{2} ∣ ξ ∣^{2} = 0$ gives $(\partial_{t}^{2} + Δ_{x}) U = 0$ , so $U$ is harmonic on $H$ , and $U (\cdot, t) \to g$ in $L^{2} (R^{d})$ as $t ↓ 0$ by [F6] applied to $∣ (e^{- 2 π t ∣ ξ ∣} - 1) \hat{g} (ξ) ∣^{2} \leq 4 ∣ \hat{g} (ξ) ∣^{2}$ . For $d \geq 2$ the theorem [F4] applies with $n = d + 1 \geq 3$ to the bounded continuous datum $g$ and identifies $U$ with the bounded harmonic Poisson integral of $g$ .

2.1F2F3F5step 1.1algebra

The gradient energy. Fix $t > 0$ . The functions $h_{t} (ξ) : = (- 2 π ∣ ξ ∣) e^{- 2 π t ∣ ξ ∣} \hat{g} (ξ)$ and $h_{t, j} (ξ) : = (2 π i ξ_{j}) e^{- 2 π t ∣ ξ ∣} \hat{g} (ξ)$ belong to $L^{1} \cap L^{2}$ by the rapid decay of $\hat{g}$ (they need not be Schwartz at $ξ = 0$ ), and by step 1.1 the functions $\partial_{t} U (\cdot, t)$ and $\partial_{x_{j}} U (\cdot, t)$ are their inverse transforms. By Plancherel [F2], $\int_{R^{d}} ∣ \partial_{t} U (x, t) ∣^{2} d x = 4 π^{2} \int ∣ ξ ∣^{2} e^{- 4 π t ∣ ξ ∣} ∣ \hat{g} (ξ) ∣^{2} d ξ$ and $\int_{R^{d}} ∣ \partial_{x_{j}} U (x, t) ∣^{2} d x = 4 π^{2} \int ξ_{j}^{2} e^{- 4 π t ∣ ξ ∣} ∣ \hat{g} (ξ) ∣^{2} d ξ$ , so $\int_{R^{d}} ∣ \nabla U (x, t) ∣^{2} d x = 8 π^{2} \int ∣ ξ ∣^{2} e^{- 4 π t ∣ ξ ∣} ∣ \hat{g} (ξ) ∣^{2} d ξ$ because $∣ ξ ∣^{2} + \sum_{j} ξ_{j}^{2} = 2 ∣ ξ ∣^{2}$ . Integrating in $t$ over $(0, \infty)$ with Tonelli [F5] and using $\int_{0}^{\infty} e^{- 4 π t ∣ ξ ∣} d t = 1 / (4 π ∣ ξ ∣)$ for $ξ \neq 0$ gives $∥ \nabla U ∥_{L^{2} (H)}^{2} = 2 π \int ∣ ξ ∣ ∣ \hat{g} (ξ) ∣^{2} d ξ$ , finite because the integrand is bounded near zero and $∣ ξ ∣ ∣ \hat{g} (ξ) ∣^{2} \leq C (1 + ∣ ξ ∣)^{- d - 1}$ at infinity for a constant $C$ by the Schwartz bounds of [F3].

2.2F8step 1.1algebra

Local trace by compact cutoffs. Fix $R > 0$ and the cutoffs $κ_{R}, η$ of (i). Step 1.1 bounds $U$ and all its first derivatives on the compact support of these cutoffs; the Leibniz formula therefore gives $V_{R} \in W^{1, 2} (H)$ with compact support in $\overline{H}$ . Classical derivatives are weak derivatives by integration against interior tests. Its continuous boundary value is $κ_{R} g$ , so [F8] gives $T_{+} V_{R} = κ_{R} g$ , equal to $g$ on $B_{R} (0)$ . This local construction applies even when $U \notin L^{2} (H)$ .

3.1F2F3F5F6F8F9step 1.1step 2.1step 2.2algebra∎

The global trace under the $L^{2} (H)$ condition, and the exact condition. First compute $\int_{H} ∣ U ∣^{2}$ : by Plancherel in $x$ [F2] and Tonelli [F5], $\int_{H} ∣ U (x, t) ∣^{2} d x d t = \int_{R^{d}} ∣ \hat{g} (ξ) ∣^{2} \int_{0}^{\infty} e^{- 4 π t ∣ ξ ∣} d t d ξ = \frac{1}{4 π} \int_{R^{d}} \frac{∣ \hat{g} (ξ) ∣^{2}}{∣ ξ ∣} d ξ$ with the value $+ \infty$ allowed. This is finite exactly when $d \geq 2$ , or $d = 1$ and $\hat{g} (0) = \int_{R} g = 0$ : for $d \geq 2$ one has $\int_{B_{1}} ∣ ξ ∣^{- 1} d ξ < \infty$ ; for $d = 1$ and $\hat{g} (0) \neq 0$ continuity of $\hat{g}$ gives $∣ \hat{g} (ξ) ∣^{2} / ∣ ξ ∣ \geq c / ∣ ξ ∣$ near $ξ = 0$ , which is not integrable; and for $d = 1$ with $\hat{g} (0) = 0$ the mean value bound $∣ \hat{g} (ξ) ∣ \leq C ∣ ξ ∣$ on $B_{1}$ [F3] makes the integrand bounded by $C^{2} ∣ ξ ∣$ there, with Schwartz decay at infinity. Assume now $U \in L^{2} (H)$ ; then $U \in W^{1, 2} (H)$ by step 2.1. Choose $ψ \in C_{c}^{\infty} (R)$ with $0 \leq ψ \leq 1$ , $ψ = 1$ on $[- 1, 1]$ and $ψ = 0$ outside $[- 2, 2]$ , and set $η_{k} (x, t) : = ψ (∣ x ∣ / k) ψ (t / k)$ on $H$ , a smooth multiplier with $∣ \nabla η_{k} ∣ \leq 2 ∥ ψ^{'} ∥_{\infty} / k$ . By the weak Leibniz rule [F9], $η_{k} U \in W^{1, 2} (H)$ , it is compactly supported and continuous on $\overline{H}$ , and $(η_{k} U) (\cdot, 0) = ψ (∣ \cdot ∣ / k) g$ . Moreover $η_{k} U \to U$ in $W^{1, 2} (H)$ : both $∥ η_{k} U - U ∥_{2}$ and $∥ (η_{k} - 1) \nabla U ∥_{2}$ tend to $0$ since $η_{k} \to 1$ pointwise with $∣ η_{k} ∣ \leq 1$ , and $∥ U \nabla η_{k} ∥_{2} \leq 2 ∥ ψ^{'} ∥_{\infty} ∥ U ∥_{L^{2} (H)} / k \to 0$ . Hence, by continuity of $T_{+}$ and its agreement with classical restriction on compactly supported continuous elements [F8], $T_{+} U = \lim_{k} T_{+} (η_{k} U) = \lim_{k} ψ (∣ \cdot ∣ / k) g = g$ in $L^{2} (R^{d})$ , the last limit by [F6]. Finally, for $d = 1$ with $\hat{g} (0) \neq 0$ the first computation gives $U \notin L^{2} (H)$ , hence $U \notin W^{1, 2} (H)$ , so the half-space trace is not defined on $U$ and the local identity of step 2.2 is the correct form.

Source notes

Schikorra's Section V.2 (printed pp. 98-100) computes exactly this harmonic extension and the identity $∥ D U ∥_{L^{2}}^{2} = ∥ (- Δ)^{1 / 4} u ∥_{L^{2}}^{2}$ via Plancherel; Kampanou's Theorem 3.3 (printed pp. 23-26) constructs a scaled-kernel right inverse whose $p = 2$ smooth model is the Poisson kernel; Mironescu's Section 1 (printed pp. 99-101) explains why the endpoint lift is a scaled convolution rather than a pointwise formula. The example verifies all properties directly from the Fourier integral representation: for $d \geq 2$ the function coincides with the bounded harmonic Poisson integral by the uniqueness in [F4], while for $d = 1$ with nonzero boundary mean it lies in $L_{l o c}^{2} (H)$ but not in $L^{2} (H)$ , so the trace identity is stated locally using compact cutoffs and globally only when $U \in L^{2} (H)$ .

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Sources

Armin Schikorra, Partial Differential Equations (University of Pittsburgh, version 4 December 2019) (standard reference, not scraped)
Maria Kampanou, Trace Theorems for Sobolev Spaces (master's thesis, National and Kapodistrian University of Athens, July 2018) (standard reference, not scraped)
Petru Mironescu, Note on Gagliardo's theorem (fetch-verified Internet Archive capture of HAL hal-01131162v1), Annals of the University of Bucharest (Mathematical Series) 6 (LXIV) (2015), no. 1, 99-103 (standard reference, not scraped)