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Completeness of the periodic Sobolev spaces H^s(T^3_{2 pi}; R^N)

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $s \geq 0$ be an integer and $N \geq 1$ , write $H^{s} = H^{s} (T_{2 π}^{3}; R^{N})$ for the periodic Sobolev space with its norm $∥ \cdot ∥_{H^{s}}$ and Fourier coefficients $\hat{u} (k) \in C^{N}$ of The 2 pi-periodic three-torus, its normalized measure, test functions and the Sobolev spaces H^s, and put $N_{s} (u)^{2} : = \sum_{k \in Z^{3}} (1 + ∣ k ∣^{2})^{s} ∣ \hat{u} (k) ∣^{2} \leq + \infty .$ Then:

(a) every Cauchy sequence in $(H^{s}, ∥ \cdot ∥_{H^{s}})$ converges in $(H^{s}, ∥ \cdot ∥_{H^{s}})$ ; that is, $H^{s}$ is a real Banach space;

(b) more precisely, for every Cauchy sequence $(u_{n})$ in $H^{s}$ there is $u \in H^{s}$ with $N_{s} (u_{n} - u) \to 0$ , and then $∥ u_{n} - u ∥_{H^{s}} \leq C_{s}^{1 / 2} N_{s} (u_{n} - u) \to 0$ , where $C_{s}$ is the constant of clause (b) of Periodic Fourier differentiation identity and equivalence of the derivative and Fourier-weight Sobolev norms; this $u$ is the unique limit of $(u_{n})$ .

Facts & Assumptions

Given: Countable Choice, integers $s \geq 0$ and $N \geq 1$ , the torus $T_{2 π}^{3}$ with its normalized measure and the spaces of The 2 pi-periodic three-torus, its normalized measure, test functions and the Sobolev spaces H^s, and a sequence $(u_{n})$ in $H^{s}$ that is Cauchy for $∥ \cdot ∥_{H^{s}}$ .

[L1]

Fourier differentiation and norm equivalence: there are constants $0 < c_{s} \leq C_{s} < \infty$ , depending only on $s$ , with $c_{s} N_{s} (u)^{2} \leq ∥ u ∥_{H^{s}}^{2} \leq C_{s} N_{s} (u)^{2}$ for every $u \in H^{s}$ ; and conversely, if a class $u \in L^{2} (T_{2 π}^{3}; R^{N})$ satisfies $N_{s} (u) < \infty$ , then $u \in H^{s}$ and the same two inequalities hold (Periodic Fourier differentiation identity and equivalence of the derivative and Fourier-weight Sobolev norms clause (b)).

[L2]

$L^{2} (T_{2 π}^{3}; R^{N})$ is a real Hilbert space, hence complete; for $u \in L^{2}$ and every $k$ one has $∣ \hat{u} (k) ∣_{C^{N}} \leq ∥ u ∥_{2}$ , because $∣ χ_{k} ∣ = 1$ and Cauchy--Schwarz applies ( $L^{2}$ with the integral pairing is a Hilbert space, The 2 pi-periodic three-torus, its normalized measure, test functions and the Sobolev spaces H^s, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[L3]

$(χ_{k})_{k \in Z^{3}}$ is an orthonormal family of $L^{2}$ , so for finitely supported coefficient families $∥ \sum_{k} a_{k} χ_{k} ∥_{2}^{2} = \sum_{k} ∣ a_{k} ∣^{2}$ (The trigonometric characters are orthonormal in $L^{2}$ of the torus, The Fourier basis and Parseval's identity on the finite torus, The 2 pi-periodic three-torus, its normalized measure, test functions and the Sobolev spaces H^s).

[L4]

$ℓ^{2} (Z^{3}; C^{N})$ is complete. Under the elementary identification $c \leftrightarrow (c_{k}^{j})_{(k, j)}$ it is the space $ℓ^{2} (Z^{3} \times {1, \dots, N}; C)$ of Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , and $ℓ^{2} (I; F)$ is complete for every index set $I$ and every $F \in {R, C}$ (A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set).

[L6]

A convergent sequence in a metric space has exactly one limit (A sequence in a metric space has at most one limit).

Proof

technique · direct

1.1L1algebra

(Weighted coefficient Cauchy sequences.) $H^{s}$ is a vector space, so $u_{n} - u_{m} \in H^{s}$ ; put $w_{k} : = (1 + ∣ k ∣^{2})^{s / 2}$ and $c^{(n)} : = (w_{k} \hat{u_{n}} (k))_{k}$ . By [L1], $∥ c^{(n)} - c^{(m)} ∥_{ℓ^{2}}^{2} = \sum_{k} ∣ w_{k} \hat{u_{n}} (k) - w_{k} \hat{u_{m}} (k) ∣^{2} = N_{s} (u_{n} - u_{m})^{2} \leq c_{s}^{- 1} ∥ u_{n} - u_{m} ∥_{H^{s}}^{2}$ , so $(c^{(n)})$ is a Cauchy sequence in $ℓ^{2} (Z^{3}; C^{N})$ .

2.1step 1.1L4algebra

(Coefficientwise limit.) By [L4] there is $c \in ℓ^{2} (Z^{3}; C^{N})$ with $∥ c^{(n)} - c ∥_{ℓ^{2}} \to 0$ ; convergence in $ℓ^{2}$ gives coordinate convergence, so $w_{k} \hat{u_{n}} (k) \to c_{k}$ for every $k$ . Put $v_{k} : = w_{k}^{- 1} c_{k}$ ; then $\hat{u_{n}} (k) \to v_{k}$ and $N_{s} (v)^{2} : = \sum_{k} (1 + ∣ k ∣^{2})^{s} ∣ v_{k} ∣^{2} = \sum_{k} ∣ c_{k} ∣^{2} = ∥ c ∥_{ℓ^{2}}^{2} < \infty$ .

3.1step 2.1L3algebra

(Conjugate symmetry of the limit coefficients.) Each $u_{n}$ is real-valued, so $\hat{u_{n}} (- k) = \overline{\hat{u_{n}} (k)}$ for every $k$ , by conjugating the defining integral and using $χ_{- k} = \overline{χ_{k}}$ ; passing to the limit of the convergent coordinate sequences gives $v_{- k} = \overline{v_{k}}$ for every $k$ .

4.1step 2.1step 3.1L1L2L3algebra

(The limit class.) For $n \geq 0$ let $u^{(n)} : = \sum_{∣ k ∣_{\infty} \leq n} v_{k} χ_{k}$ , a finite sum defining a class in $L^{2} (T_{2 π}^{3}; R^{N})$ : conjugating the finite sum gives $\sum_{k} \overline{v_{k}} \overline{χ_{k}} = \sum_{k} v_{- k} χ_{- k} = \sum_{k} v_{k} χ_{k}$ by step 3.1, so it is real-valued, and its Fourier coefficients are $v_{k}$ for $∣ k ∣_{\infty} \leq n$ and $0$ otherwise. By [L3] and step 2.1, $∥ u^{(n)} - u^{(m)} ∥_{2}^{2} = \sum_{n < ∣ k ∣_{\infty} \leq m} ∣ v_{k} ∣^{2} \to 0$ as $n, m \to \infty$ , because $\sum_{k} ∣ v_{k} ∣^{2} \leq N_{s} (v)^{2} < \infty$ and $s \geq 0$ gives $(1 + ∣ k ∣^{2})^{s} \geq 1$ ; since $L^{2}$ is complete [L2], $(u^{(n)})$ converges in $L^{2}$ to a class $u \in L^{2} (T_{2 π}^{3}; R^{N})$ . For every $k$ , $∣ \hat{u} (k) - \hat{u^{(n)}} (k) ∣ = ∣ \hat{u - u^{(n)}} (k) ∣ \leq ∥ u - u^{(n)} ∥_{2} \to 0$ by [L2], so $\hat{u} (k) = v_{k}$ ; hence $N_{s} (u) = N_{s} (v) < \infty$ and [L1] gives $u \in H^{s}$ .

5.1step 1.1step 2.1step 4.1L1algebra

(Convergence in $H^{s}$ .) By steps 2.1 and 4.1, $N_{s} (u_{n} - u)^{2} = \sum_{k} (1 + ∣ k ∣^{2})^{s} ∣ \hat{u_{n}} (k) - v_{k} ∣^{2} = \sum_{k} ∣ w_{k} \hat{u_{n}} (k) - c_{k} ∣^{2} = ∥ c^{(n)} - c ∥_{ℓ^{2}}^{2} \to 0$ , and [L1] gives $∥ u_{n} - u ∥_{H^{s}} \leq C_{s}^{1 / 2} N_{s} (u_{n} - u) \to 0$ .

6.1step 4.1step 5.1L6∎

(Conclusion.) Steps 4.1 and 5.1 produce, for every Cauchy sequence in $H^{s}$ , a limit $u \in H^{s}$ with $N_{s} (u_{n} - u) \to 0$ and $∥ u_{n} - u ∥_{H^{s}} \to 0$ ; this is (a) and the first two assertions of (b), and the limit is unique by [L6].

Remarks

Role. This supplies the completeness of $H^{s - 1}$ used in Compactness of H^r-bounded families in H^{r-1} and locally uniform H^{s-1} subsequences of Galerkin solutions (the Cauchy diagonal subsequence converges) and of $H^{s - 2}$ used in A limit of Galerkin solutions is a Sobolev and then a classical solution of the symmetric hyperbolic system (the extracted limit of the time derivatives), and more generally the Banach-space property of the periodic Sobolev scale.
What is used. Only the Fourier characterisation of the norm and of the space, completeness of $L^{2}$ , completeness of the weighted coefficient space and the elementary conjugate-symmetry consequence of real-valuedness. No differentiability of the limit class is assumed; membership in $H^{s}$ follows from the finite weighted norm through clause (b) of Periodic Fourier differentiation identity and equivalence of the derivative and Fourier-weight Sobolev norms.

Completeness of the periodic Sobolev spaces H^s(T^3_{2 pi}; R^N)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources