Parabolic finite-element methods

Finite-valued simple functions are dense in Bochner $L^2$. Choose simple $s_n\to q$ almost everywhere as in [F1]. Replace $s_n$ by zero where $\mathrm{\parallel }{s}_n\mathrm{\parallel }>2\mathrm{\parallel }q\mathrm{\parallel }$. The clipped functions are still simple, converge to $q$ almost everywhere, and have squared error bounded by $9\mathrm{\parallel }q{\mathrm{\parallel }}^2$. Scalar dominated convergence [F7] gives convergence in $L^2(X)$. If $q=0$ almost everywhere, the zero simple sequence suffices.

For fixed $t$, the integrand in the definition of $q_{ϵ}(t)$ is strongly measurable and has integrable norm on the compact support of the kernel, so [F2] defines the Bochner integral. Differentiating the scalar kernel under this integral gives $q_{ϵ}^{(k)}(t)=\int {\eta }_{ϵ}^{(k)}(t-s)q(s)ds(k\ge 0)\mathrm{.}$ Indeed, for $t$ in a fixed compact interval the difference quotients of each kernel derivative converge uniformly and are bounded on one compact set; the $L^1(X)$ norm of $q$ on that set and [F5] let the limit pass through the integral. The same bound proves continuity of every displayed derivative, so $q_{ϵ}$ is smooth.

For a finite-valued simple function $s={\sum }_{j=1}^m{x}_j{\mathbf{1}}_{E_j}$, omit zero values; each remaining $E_j$ has finite measure. Then $\mathrm{\parallel }s(\cdot -h)-s{\mathrm{\parallel }}_{L^2(X)}\le {\sum }_j\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_X\mathrm{\parallel }{\mathbf{1}}_{E_j}(\cdot -h)-{\mathbf{1}}_{E_j}{\mathrm{\parallel }}_{L^2}$, which tends to zero by [F3]. For general $q$, use the density from step 1.1 and translation's isometry [F8] to obtain $\mathrm{\parallel }q(\cdot -h)-q{\mathrm{\parallel }}_{L^2(X)}\to 0$.

Fix $R>0$ with $\mathrm{supp}\eta \subset [-R,R]$ for the generating bump, so $\mathrm{supp}{\eta }_{ϵ}\subset [-Rϵ,Rϵ]$ by [F4]. Since ${\eta }_{ϵ}(s)ds$ is a probability measure, Cauchy–Schwarz and [F5] give $\mathrm{\parallel }{q}_{ϵ}(t)-q(t){\mathrm{\parallel }}_X^2\le \int {\eta }_{ϵ}(s)\mathrm{\parallel }q(t-s)-q(t){\mathrm{\parallel }}_X^{2}ds\mathrm{.}$ Integrate in $t$ and apply [F6]. The result is bounded by ${\sup }_{\mathrm{\mid }s\mathrm{\mid }\le Rϵ}\mathrm{\parallel }q(\cdot -s)-q{\mathrm{\parallel }}_{L^2(X)}^2$, which tends to zero by step 2.1. This proves the asserted $L^2$ convergence.

Expand the last square using [F1]: $\mathrm{\parallel }x-y{\mathrm{\parallel }}_H^2=\mathrm{\parallel }x{\mathrm{\parallel }}_H^2-2(x,y{)}_H+\mathrm{\parallel }y{\mathrm{\parallel }}_H^2$.

Substituting this expression into the right-hand side gives $2\mathrm{\parallel }x{\mathrm{\parallel }}_H^2-2(x,y{)}_H=2(x-y,x{)}_H$ by bilinearity. This is the stated identity.

If $X=\{0\}$, the right-kernel hypothesis forces $Y=\{0\}$; the only $F\in Y^{\ast }$ is zero and $x=0$ is the unique solution. If $Y=\{0\}$ and $X\mathrm{\ne }\{0\}$, the inf-sup hypothesis fails for any nonzero $x$, so no such instance satisfies the assumptions. Assume henceforth both spaces are nonzero. For each $x\in X$, the functional $y\mapsto b(x,y)$ is bounded and linear. By [F1], let $Sx\in Y$ be the unique vector satisfying $(Sx,y{)}_Y=b(x,y)$ for every $y\in Y$. Bilinearity and uniqueness make $S:X\to Y$ linear, and the Riesz norm identity gives $\mathrm{\parallel }Sx{\mathrm{\parallel }}_Y={\sup }_{\mathrm{\parallel }y{\mathrm{\parallel }}_Y\le 1}\mathrm{\mid }b(x,y)\mathrm{\mid }\le C_b\mathrm{\parallel }x{\mathrm{\parallel }}_X$, where $C_b$ is a continuity bound for $b$.

The inf-sup assumption gives $\mathrm{\parallel }Sx{\mathrm{\parallel }}_Y\ge \beta \mathrm{\parallel }x{\mathrm{\parallel }}_X$. Thus $S$ is injective. If $S{x}_n\to y$ in $Y$, then $\mathrm{\parallel }{x}_n-x_m{\mathrm{\parallel }}_X\le {\beta }^{-1}\mathrm{\parallel }S{x}_n-S{x}_m{\mathrm{\parallel }}_Y$, so completeness of $X$ gives $x_n\to x$ for some $x\in X$. The bound in step 1.1 gives $S{x}_n\to Sx$, hence $y=Sx$. Therefore $S(X)$ is closed.

If $y\in S(X{)}^{\perp }$, then $(Sx,y{)}_Y=b(x,y)=0$ for every $x\in X$; the right-kernel hypothesis gives $y=0$. Apply [F2] to the closed subspace $S(X)$. Its orthogonal complement is zero, so $Y=S(X)\oplus \{0\}=S(X)$; therefore $S$ is onto.

For the given $F\in Y^{\ast }$, let $g\in Y$ be its Riesz representative, so $F(y)=(y,g{)}_Y$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_Y=\mathrm{\parallel }F{\mathrm{\parallel }}_{Y^{\ast }}$. Surjectivity supplies $x\in X$ with $Sx=g$, whence $b(x,y)=(Sx,y{)}_Y=(g,y{)}_Y=(y,g{)}_Y=F(y)$ for all $y$, using symmetry of the real inner product. The lower bound $\beta \mathrm{\parallel }x{\mathrm{\parallel }}_X\le \mathrm{\parallel }Sx{\mathrm{\parallel }}_Y=\mathrm{\parallel }F{\mathrm{\parallel }}_{Y^{\ast }}$ proves the estimate and also uniqueness.

Let $g_1,g_2\in L^2(0,T;V^{\ast })$ both satisfy [F1], and put $h=g_1-g_2$. For every $v\in V$ and $\varphi \in C_c^{\mathrm{\infty }}(0,T)$, $\int ⟨h(t),v⟩\varphi (t)dt=0$. Therefore every Bochner convolution of the zero extension, $h_{ϵ}(t)={\int }_0^{T}h(s){\eta }_{ϵ}(t-s)ds$, vanishes on any compact subinterval whose distance from $\{0,T\}$ exceeds $ϵ$: its scalar pairing against each $v$ is the zero distribution tested by $s\mapsto {\eta }_{ϵ}(t-s)$. By [F2], $h_{ϵ}\to h$ in local $L^2(V^{\ast })$, so $h=0$ almost everywhere.

Regard $u$ as a $V^{\ast }$-valued function through the continuous embedding $V\hookrightarrow V^{\ast }$ and set $G(t)={\int }_0^{t}g(s)ds$. The Bochner integral exists by [F3], and [F5] gives $g\in L^1(0,T;V^{\ast })$. For each fixed $v\in V$, [F4] and scalar integration by parts show that $G$ has weak derivative $g$ in the sense of [F1]. Thus $w:=u-G$ has zero distribution derivative. For any $\psi \in C_c^{\mathrm{\infty }}(0,T)$ with integral zero, its primitive is compactly supported in $(0,T)$, hence $\int w(t)\psi (t)dt=0$ in $V^{\ast }$. Fix $\eta \in C_c^{\mathrm{\infty }}(0,T)$ with integral one and put $c=\int w(t)\eta (t)dt$. Subtracting $(\int \psi )\eta$ shows $\int (w-c)\psi =0$ for every test $\psi$. The convolutions of $w-c$ are therefore zero on compact subintervals as in step 1.1; [F2] makes them converge to $w-c$ in local $L^2(V^{\ast })$. Hence $w=c$ almost everywhere, and $\tilde{u}(t):=c+G(t)$ is a continuous $V^{\ast }$ representative of $u$.

Reflect the $V$-valued $L^2$ class evenly to $(-T,2T)$ across $0$ and $T$, and reflect $g$ oddly. The continuous $V^{\ast }$ representative from step 2.1 has matching values at each seam. Integration by parts on the three intervals therefore shows that the reflected function has weak $V^{\ast }$ derivative equal to the odd reflection of $g$; the endpoint terms cancel, so there are no point masses. Extend both functions by zero beyond $(-T,2T)$. For $0<ϵ<T\mathrm{/}2$, the convolution on the resulting neighborhood of $[0,T]$ does not see the extension jumps at $-T$ or $2T$.

Let ${\eta }_{ϵ}$ be a nonnegative smooth mollifier of mass one, supported in $(-ϵ,ϵ)$, and convolve the reflected functions in time. On a neighborhood of $[0,T]$, $u_{ϵ}={\eta }_{ϵ}\ast u$ is smooth with values in $V$ by [F2]. Testing the reflected weak derivative identity against a fixed $v\in V$ and integrating by parts in the scalar variable gives $⟨u_{ϵ}^{\mathrm{\prime }}(t),v⟩=\int ⟨g(s),v⟩{\eta }_{ϵ}(t-s)ds=⟨({\eta }_{ϵ}\ast g)(t),v⟩$. The bounded inclusion $V\to V^{\ast }$ commutes with the Bochner integral by [F3], so $u_{ϵ}^{\mathrm{\prime }}={\eta }_{ϵ}\ast g$ in $V^{\ast }$.

Apply the $L^2$ approximation in [F2] to the reflected $V$-valued function and its $V^{\ast }$ derivative, then restrict to $(0,T)$. This gives $u_{ϵ}\to u$ in $L^2(V)$ and $u_{ϵ}^{\mathrm{\prime }}\to g$ in $L^2(V^{\ast })$ as $ϵ\downarrow 0$.

Choose any sequence ${ϵ}_n\downarrow 0$ small enough that each convolution is defined on a neighborhood of $[0,T]$, and set $u_n=u_{{ϵ}_n}$. Step 5.1 gives both asserted convergences, and step 4.1 gives $u_n^{\mathrm{\prime }}={\eta }_{{ϵ}_n}\ast g$. The uniqueness in step 1.1 identifies $g=u^{\mathrm{\prime }}$.

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.

Let $g_1,g_2\in L^2(0,T;X)$ both satisfy the distributional derivative identity for $u$. Put $h=g_1-g_2$ and extend it by zero to $\mathbb{R}$. For every fixed $x^{\ast }\in X^{\ast }$ and every interior $t$ farther than $ϵ$ from the endpoints, the scalar pairing of ${\eta }_{ϵ}\ast h(t)$ with $x^{\ast }$ is zero: it is the zero distribution paired with $s\mapsto {\eta }_{ϵ}(t-s)$. Since continuous linear functionals separate points, ${\eta }_{ϵ}\ast h(t)=0$. By [F3] these convolutions converge to $h$ in $L^2$ on compact subintervals; exhausting $(0,T)$ proves $h=0$ a.e., so the weak derivative is unique.

The Bochner integral $G(t)={\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ exists by [F2] and [F5]. For each $x^{\ast }\in X^{\ast }$, scalar Fubini [F4] and integration by parts show that $G$ has weak derivative $u^{\mathrm{\prime }}$. Hence $w=u-G$ has zero distribution derivative. If $\psi \in C_c^{\mathrm{\infty }}(0,T)$ has integral zero, its primitive is compactly supported, so $\int w\psi =0$ in $X$. Choose $\eta \in C_c^{\mathrm{\infty }}(0,T)$ with integral one and put $c=\int w\eta$. Subtracting $(\int \psi )\eta$ shows that $w-c$ has zero distribution. As in step 1.1 its interior convolutions vanish, and [F3] makes them converge to $w-c$ locally in $L^2$. Thus $w=c$ a.e. and $u(t)=c+G(t)$ is a continuous representative. Any two continuous representatives equal a.e. agree everywhere, because their continuous difference vanishes on the dense set where they agree; hence this representative is unique.

For fixed $P$, test the weak derivative identity for $u$ against $y^{\ast }\circ P\in X^{\ast }$ for arbitrary $y^{\ast }\in Y^{\ast }$. Boundedness of $P$ gives ${\int }_0^T{y}^{\ast }(Pu(t)){\varphi }^{\mathrm{\prime }}(t)dt=-{\int }_0^T{y}^{\ast }(P{u}^{\mathrm{\prime }}(t))\varphi (t)dt\mathrm{.}$ The right side is represented by the $L^2(Y)$ function $P{u}^{\mathrm{\prime }}$, and uniqueness from step 1.1 gives $(Pu{)}^{\mathrm{\prime }}=P{u}^{\mathrm{\prime }}$.

For the operator-norm $C^1$ family, fix $y^{\ast }\in Y^{\ast }$ and put $\mathrm{\ell }(t)=y^{\ast }\circ P(t)\in X^{\ast }$, so ${\mathrm{\ell }}^{\mathrm{\prime }}(t)=y^{\ast }\circ P^{\mathrm{\prime }}(t)$. Use the representative $u(t)=c+{\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ from step 2.1. For $\varphi \in C_c^{\mathrm{\infty }}(0,T)$, scalar Fubini and integration by parts give ${\int }_0^T\mathrm{\ell }(t)u(t){\varphi }^{\mathrm{\prime }}(t)dt=-{\int }_0^T({\mathrm{\ell }}^{\mathrm{\prime }}(t)u(t)+\mathrm{\ell }(t)u^{\mathrm{\prime }}(t))\varphi (t)dt\mathrm{.}$ Indeed, substitute $c+{\int }_0^t{u}^{\mathrm{\prime }}(s)ds$ on the left, interchange the integrals over $0\le s\le t\le T$, and use ${\int }_s^T\mathrm{\ell }(t){\varphi }^{\mathrm{\prime }}(t)dt=-\mathrm{\ell }(s)\varphi (s)-{\int }_s^T{\mathrm{\ell }}^{\mathrm{\prime }}(t)\varphi (t)dt$; a second Fubini interchange combines the ${\mathrm{\ell }}^{\mathrm{\prime }}$ terms into ${\mathrm{\ell }}^{\mathrm{\prime }}(t)u(t)$. Absolute integrability follows from boundedness of $\mathrm{\ell },{\mathrm{\ell }}^{\mathrm{\prime }},\varphi ,{\varphi }^{\mathrm{\prime }}$ and [F5]. Since this identity holds for all $y^{\ast }$ and $P^{\mathrm{\prime }}u,P{u}^{\mathrm{\prime }}\in L^2(0,T;Y)$, the vector-valued weak derivative is $P^{\mathrm{\prime }}u+P{u}^{\mathrm{\prime }}$.

The displayed derivative belongs to $L^2(0,T;Y)$ because $P$ and $P^{\mathrm{\prime }}$ are bounded on $[0,T]$ and $u,u^{\mathrm{\prime }}\in L^2(0,T;X)$. Hence $P(t)u(t)\in H^1(0,T;Y)$ with the asserted formula.

Apply Gram–Schmidt to a countable dense sequence in $X$, omitting zero remainders, to obtain a finite or countable orthonormal family $(e_j)$ with dense span. Use the following deterministic extraction for each bounded real sequence $(b_n)$ with $\mathrm{\mid }{b}_n\mathrm{\mid }\le C$. Start with $[-C,C]$; at each bisection retain the left closed half if it contains infinitely many terms, and otherwise the right closed half, which then contains infinitely many terms. From the $k$th retained interval select the least index larger than the preceding selected index whose term lies in that interval. The selected terms lie in nested intervals of diameters tending to zero, hence are Cauchy; [F9] gives a convergent subsequence, so the entire selected sequence converges. This rule determines the extracted indices uniquely, including when $C=0$, and can therefore be recursively applied to successive coordinates $(x_n,e_j{)}_X$ without dependent choice. For finitely many coordinates use the last extracted subsequence; for infinitely many take the diagonal of the nested subsequences, whose indices are strictly increasing and whose tail lies in every earlier subsequence. If the family is empty retain the original sequence. The resulting $(x_{n_k})$ has a limit $a_j={\lim }_k(x_{n_k},e_j{)}_X$ for each coordinate present.

Let $E$ be any separable real Hilbert space. The formula $(f,g{)}_{L^2(E)}={\int }_0^T(f(t),g(t){)}_{E}dt$ is well-defined by pointwise Cauchy–Schwarz and Holder [F4], and its induced norm is the square-integral norm. To prove completeness, let $(f_n)$ be Cauchy. By [A1], choose a subsequence $(f_{n_k})$ with summable successive $L^2(E)$ distances and representatives for that countable subsequence. Put $h_k(t)=\mathrm{\parallel }{f}_{n_{k+1}}(t)-f_{n_k}(t){\mathrm{\parallel }}_E$. Minkowski bounds partial-sum $L^2$ norms by the sums of the norms; monotone convergence [F4] gives $h={\sum }_k{h}_k\in L^2(0,T)$. Its infinite-value set is null, since positive measure would force $\int h^2\ge N^2\mathrm{\mid }\{h=\mathrm{\infty }\}\mathrm{\mid }$ for every $N$. Thus $f_{n_k}(t)$ converges in complete $E$ for a.e. $t$ to $f(t)$. The limit is strongly measurable: because $E$ is separable, distances to a countable dense set are measurable; first-dense-point approximants are countably valued, and finite truncations with summable exceptional-set measures converge a.e. by [F5]–[F6]. The pointwise tail estimate and Minkowski give $\mathrm{\parallel }f-f_{n_m}{\mathrm{\parallel }}_{L^2(E)}\le {\sum }_{k\ge m}\mathrm{\parallel }{f}_{n_{k+1}}-f_{n_k}{\mathrm{\parallel }}_{L^2(E)}\to 0.$ The original Cauchy sequence then converges to $f$. This proves that $L^2(0,T;E)$ is Hilbert.