Energy control space with a time-integrable supremum norm

PDF p. 15: Proposition 3.9, definition of XT

Assume the Axiom of Choice. Let $T>0$ and let $\mathrm{\Omega }$ be the connected exterior of a nonempty compact obstacle with smooth boundary as in Exterior and truncated obstacle domains. The space $H_0^1(\mathrm{\Omega })$ is the closure of $C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{C})$ in the complex $H^1$ norm. A path $u$ belongs to $X_T$ if $u\in C([0,T];H_0^1(\mathrm{\Omega }))$ and it has a jointly measurable representative $U$ on $(0,T)\times \mathrm{\Omega }$ such that $U(t,\cdot )$ represents $u(t)$ for almost every $t$ and $\mathrm{\parallel }U{\mathrm{\parallel }}_{L_t^4{L}_x^{\mathrm{\infty }}}<\mathrm{\infty }$. The latter is the mixed norm of Continuous and mixed time-space function spaces, namely ${({\int }_0^T({\mathrm{ess\hspace{0.17em}sup}}_{x\in \mathrm{\Omega }}\mathrm{\mid }U(t,x)\mathrm{\mid }{)}^{4}dt)}^{1\mathrm{/}4}$.

Equip $X_T$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{X_T}={\sup }_{0\le t\le T}\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H_0^1(\mathrm{\Omega })}+\mathrm{\parallel }U{\mathrm{\parallel }}_{L_t^4{L}_x^{\mathrm{\infty }}}$. Here the $H_0^1$ norm is the inherited $H^1$ norm. The mixed term is independent of the chosen compatible representative. This is the Banach control space used in the article for $s=1$. The mixed condition does not require strong measurability of $t\mapsto U(t,\cdot )$ as an $L^{\mathrm{\infty }}(\mathrm{\Omega })$-valued map.

Exterior and truncated obstacle domains

Complex integer-order Sobolev notation

Continuous and mixed time-space function spaces

Dirichlet spectral Sobolev scale and duality

[F1] The integer Sobolev notation defines $H_0^1$ as the stated closure; the Dirichlet scale identifies it, with the displayed $H^1$ norm, with a Hilbert space. Complex integer-order Sobolev notation Dirichlet spectral Sobolev scale and duality

[F2] The time-space definition supplies continuous paths, compatible mixed norms and their norm properties; completed-product Tonelli identifies product-null sets with almost-everywhere null sections. Continuous and mixed time-space function spaces Tonelli and Fubini for the completed product, with only almost-everywhere section measurability

[F3] Scalar $L^4(0,T)$ is complete; Minkowski and monotone convergence control sums of nonnegative $L^4$ functions. Choice permits the countable representative and subsequence selections used below. Riesz-Fischer completeness of Lᵖ for 1 ≤ p ≤ ∞ Minkowski's inequality for integrals, including p = ∞ Monotone convergence for the integral The Axiom of Choice

[F4] For a nonempty topological space $S$ and a complete metric target $(Y,d)$, $C(S;Y)$ is complete for the bounded uniform metric $\bar{\rho }(v,w)={\sup }_{t\in S}\min \{d(v(t),w(t)),1\}$. If (Y,d) is complete then Y^X is complete in the uniform metric, and so is C(X,Y)

[F5] The interval $[0,T]$ is compact, as recorded in the time-space supplier. A continuous map from a compact metric space into any metric space is uniformly continuous. Continuous and mixed time-space function spaces Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous

[F6] Under Countable Choice, Complex Lp completeness and almost-everywhere subsequences gives completeness of complex $L^2$ on every measure space and an almost-everywhere convergent subsequence of measurable representatives for every norm-convergent sequence. Apply it on the completed product of time Lebesgue measure and restricted spatial Lebesgue measure; the stated AC assumption supplies Countable Choice. No Bochner measurability into spatial $L^{\mathrm{\infty }}$ is required.

Given: The geometry, $T$ and the spaces in the definition.

Goal: The definition and norm are independent of representatives and $X_T$ is complete.

1.1 If $U$ and $V$ are compatible representatives of the same path, their spatial sections agree almost everywhere for almost every time. Completed-product Tonelli applied to their measurable disagreement set shows that $U=V$ product-almost everywhere. Thus F2 makes the mixed norm independent of the representative. F1 and F2 give the vector-space and norm axioms; in particular, a zero supremum term makes the continuous path identically zero. [given, F1, F2]

2.1 Let $(u_j)$ be Cauchy in $X_T$. Apply F4 with $S=[0,T]$, $Y=H_0^1(\mathrm{\Omega })$ and $d(v,w)=\mathrm{\parallel }v-w{\mathrm{\parallel }}_{H_0^1}$; $S$ is nonempty and F1 makes $Y$ complete. The sequence is Cauchy for $\bar{\rho }$, since this metric is bounded above by the supremum norm distance, so F4 gives a continuous limit $u$. When $\bar{\rho }(u_j,u)<1$, every pointwise norm distance equals its truncation and their supremum equals $\bar{\rho }(u_j,u)$. Thus $u_j\to u$ uniformly in the $H_0^1$ norm. Choose a subsequence $(u_{j_k})$ whose successive $X_T$ distances are at most $2^{-k}$ and choose compatible jointly measurable representatives $U_k$. Put $a_k(t)={\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\mid }{U}_{k+1}(t,x)-U_k(t,x)\mathrm{\mid }$. Minkowski and monotone convergence give $a={\sum }_k{a}_k\in L^4(0,T)$. For almost every $t$, the sum is finite and, outside a product-null set, every successive difference is bounded by its corresponding $a_k(t)$. Hence $U_k$ converges product-almost everywhere to a measurable $U$, and ${\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\mid }U(t,x)-U_k(t,x)\mathrm{\mid }\le {\sum }_{\mathrm{\ell }\ge k}{a}_{\mathrm{\ell }}(t)$ for almost every $t$. The $L^4$ norm of this tail is at most ${\sum }_{\mathrm{\ell }\ge k}{2}^{-\mathrm{\ell }}$, so $U_k\to U$ in the mixed norm. [given, F1, F2, F3, F4, step 1.1]

3.1 The inequality $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H_0^1(\mathrm{\Omega })}$ makes $u$ continuous into $L^2(\mathrm{\Omega })$. By F5 it is uniformly continuous on $[0,T]$. For each $m\ge 1$, choose a finite time-step path $s_m$ with values in $L^2(\mathrm{\Omega })$ and ${\sup }_t\mathrm{\parallel }{s}_m(t)-u(t){\mathrm{\parallel }}_2<2^{-m}$, by a partition whose mesh is smaller than a uniform-continuity radius. Choose measurable spatial representatives for the finitely many values and denote the resulting jointly measurable time-step function by $V_m(t,x)$. Tonelli gives $\mathrm{\parallel }{V}_m-V_l{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}\le T^{1\mathrm{/}2}(2^{-m}+2^{-l})$. F6 therefore gives a product $L^2$ limit $V$ with a measurable representative. Tonelli also makes $V(t,\cdot )$ an $L^2(\mathrm{\Omega })$ class for almost every $t$. Choose a subsequence with ${\sum }_k\mathrm{\parallel }{V}_{m_k}-V{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2<\mathrm{\infty }$. Tonelli and monotone convergence show that ${\sum }_k\mathrm{\parallel }{V}_{m_k}(t,\cdot )-V(t,\cdot ){\mathrm{\parallel }}_2^2<\mathrm{\infty }$ for almost every $t$. Hence $s_{m_k}(t)\to [V(t,\cdot )]$ in spatial $L^2$ there. Since $s_m(t)\to u(t)$ uniformly in that norm, uniqueness of the norm limit gives $[V(t,\cdot )]=u(t)$ for almost every $t$. This establishes the representative identification after constructing the product limit. [given, F1, F2, F3, F5, F6, step 2.1]

4.1 By the row identification in step 3.1 and Tonelli, the compatible representatives $U_k$ from step 2.1 satisfy $\mathrm{\parallel }{U}_k-V{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2={\int }_0^T\mathrm{\parallel }{u}_{j_k}(t)-u(t){\mathrm{\parallel }}_2^{2}dt\to 0$, because the paths converge uniformly in $H_0^1$. F6 gives a further subsequence of these complex product $L^2$ classes converging to $V$ product-almost everywhere. Their previously chosen representatives may be retained, since the countable union of their disagreements with the representatives supplied by F6 is null. But step 2.1 already gives $U_k\to U$ product-almost everywhere. Thus $U=V$ product-almost everywhere; Tonelli makes $U(t,\cdot )$ a representative of $u(t)$ for almost every $t$. Therefore $u\in X_T$, and step 2.1 gives convergence of the chosen subsequence in both terms of the $X_T$ norm. The triangle inequality and the original Cauchy property give convergence of the whole sequence. Hence $X_T$ is Banach. [F1, F2, F6, step 2.1, step 3.1] ∎

The original article is cited for the definition and notation in Proposition 3.9, not as proof of completeness. The mixed-norm convention above makes the measurable representative explicit and does not silently replace it by a Bochner $L^4(0,T;L^{\mathrm{\infty }}(\mathrm{\Omega }))$ condition.