Measurable actions and inverse actions of parabolic forms

If $V=\{0\}$, all spaces and maps in the claim are zero and the result holds. Otherwise choose countable dense sets in $V$ and $V^{\ast }$, the latter available by [F3]. For fixed $v\in V$, the functional $A(t)v$ has norm at most $M\mathrm{\parallel }v{\mathrm{\parallel }}_V$. On a countable dense subset $(w_j)$ of the unit sphere, $\mathrm{\parallel }A(t)v-r{\mathrm{\parallel }}_{V^{\ast }}={\sup }_j\mathrm{\mid }a(t;v,w_j)-⟨r,w_j⟩\mathrm{\mid }$ for each fixed $r\in V^{\ast }$. Hence these norm distances are measurable. Approximating $A(t)v$ by the first point of a countable dense subset of $V^{\ast }$ within $1\mathrm{/}n$ gives a measurable countably-valued approximation.

For a.e. fixed $t$ outside the common null set, apply [F4] to $a(t)$ and to its transpose $a(t;w,v)$. Both forms have the same bounds and coercivity because their diagonal values agree. This gives the unique pointwise $x(t)$ and $x^{\ast }(t)$ solving the two equations.

Choose a countable linearly independent sequence $(e_j)$ whose linear span is dense in $V$; if $V$ is finite-dimensional, use a finite basis and set $V_n=V$ once it is exhausted. Let $V_n=\mathrm{span}(e_1,\dots ,e_n)$. The Galerkin systems for $x_n(t),x_n^{\ast }(t)\in V_n$ have measurable matrix entries and measurable right-hand sides. Their matrices are invertible: testing a homogeneous system by its trial vector and using coercivity forces that vector to be zero. Matrix inversion therefore makes the coefficient vectors, and hence $x_n,x_n^{\ast }$, strongly measurable.

Truncate the $n$th countably-valued approximation after its first $N_n$ values, replacing the tail by zero. By [F5], choose $N_n$ so the tail set has measure at most $2^{-n}$. These truncations are finite-valued measurable functions; [F6] shows that outside a null set only finitely many tails occur, so they converge pointwise to $A(t)v$. Thus $A(t)v$ is strongly measurable. Applying the same argument to the transposed form proves that $A^{\ast }(t)v$ is strongly measurable for each fixed $v$.

For any $z_n\in V_n$, Galerkin orthogonality and coercivity give $\alpha \mathrm{\parallel }x(t)-x_n(t){\mathrm{\parallel }}_V^2\le a(t;x-x_n,x-x_n)=a(t;x-x_n,x-z_n)\le M\mathrm{\parallel }x-x_n{\mathrm{\parallel }}_V\mathrm{\parallel }x-z_n{\mathrm{\parallel }}_V\mathrm{.}$ Thus $\mathrm{\parallel }x-x_n{\mathrm{\parallel }}_V\le (M\mathrm{/}\alpha ){\inf }_{z_n\in V_n}\mathrm{\parallel }x-z_n{\mathrm{\parallel }}_V$; the identical argument for the transposed form gives the same estimate for $x^{\ast }$. Density of ${\bigcup }_n{V}_n$ makes both Galerkin sequences converge pointwise to their unique solutions. Their limits are strongly measurable.

Approximate any strongly measurable $u\in L^2(0,T;V)$ pointwise a.e. by finite-valued simple functions $u_n$. For each $n$, $A(t)u_n(t)$ and $A^{\ast }(t)u_n(t)$ are strongly measurable by step 2.1. The uniform operator bounds give pointwise convergence to $A(t)u(t)$ and $A^{\ast }(t)u(t)$ off the common null set, and $\mathrm{\parallel }A(t)u(t){\mathrm{\parallel }}_{V^{\ast }},\mathrm{\parallel }{A}^{\ast }(t)u(t){\mathrm{\parallel }}_{V^{\ast }}\le M\mathrm{\parallel }u(t){\mathrm{\parallel }}_V$. This proves strong measurability and square integrability of both actions.

Testing each pointwise variational equation by its solution gives $\alpha \mathrm{\parallel }x(t){\mathrm{\parallel }}_V^2\le \mathrm{\parallel }q(t){\mathrm{\parallel }}_{V^{\ast }}\mathrm{\parallel }x(t){\mathrm{\parallel }}_V$ and the same inequality for $x^{\ast }(t)$. If a solution is zero the claimed bound is immediate; otherwise division by its norm yields the two pointwise estimates. Squaring and integrating proves $x,x^{\ast }\in L^2(0,T;V)$ with the asserted bounds.