A normalized parabolic Bellman equation has a bounded weak solution

For each $a\in \mathcal{C}$, choose unit eigenvectors $e_i(a)$ and eigenvalues $s_i(a)\ge 0$, so $a={\sum }_{i=1}^d{s}_i(a)e_i(a)e_i(a{)}^{\mathsf{T}}$. For $0<h<1$, define $L_h^{a}u(t,x)={\sum }_i\frac{s_i(a)}{h^2}(u(t,x+h{e}_i)-2u(t,x)+u(t,x-h{e}_i))+\frac{r(a)}{h}(u(t+h,x)-u(t,x))\mathrm{.}$ All off-center coefficients are nonnegative. Their sum is $q_h(a)=2\mathrm{tr}(a)\mathrm{/}{h}^2+r(a)\mathrm{/}h$.

We will use this elementary representation fact on a bounded open ball $B$: if a linear functional $T$ on compact smooth tests satisfies $\mathrm{\mid }T\eta \mathrm{\mid }\le C\mathrm{\parallel }\eta {\mathrm{\parallel }}_1$, it is integration against a function of absolute value at most $C$. Indeed, $\mathrm{\parallel }\eta {\mathrm{\parallel }}_1\le \mathrm{\mid }B{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }\eta {\mathrm{\parallel }}_2$ extends $T$ continuously from the dense test subspace to $L^2(B)$. Real $L^2$ duality represents it by $g\in L^2(B)$. The original $L^1$ estimate extends to all $L^2$ tests, because their $L^2$ approximations also converge in $L^1$. Test with $\mathrm{sgn}(g){\mathbf{1}}_{\{\mathrm{\mid }g\mathrm{\mid }>C+ϵ\}}$ to obtain a contradiction unless that set is null. Hence $\mathrm{\mid }g\mathrm{\mid }\le C$. The unique local representatives agree on overlaps.

Write $L_h^{a}u=M_h^{a}u-q_h(a)u$, where $M_h^a$ is its off-center positive operator, and define $P_{h}u(z)={\sup }_{a\in \mathcal{C}}\frac{M_h^{a}u(z)+k(a)f(z)}{\lambda +q_h(a)}\mathrm{.}$ Put $Q_h=2h^{-2}+h^{-1}$ and ${\gamma }_h=Q_h\mathrm{/}(\lambda +Q_h)<1$. Positivity and $M_h^{a}1=q_h(a)\le Q_h$ give $\mathrm{\parallel }{P}_{h}u-P_{h}w{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\gamma }_h\mathrm{\parallel }u-w{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ The same argument bounds $\sup (P_{h}u-P_{h}w{)}^{+}$ by ${\gamma }_h\sup (u-w{)}^{+}$.

Starting with $u_0=0$, iterate $u_{j+1}=P_h{u}_j$. The contraction bounds successive differences by a summable geometric sequence. Their sums converge uniformly to a bounded function $v_h$, since a uniformly Cauchy sequence of bounded real functions has a bounded pointwise limit and converges uniformly. The contraction estimate gives $P_h{v}_h=v_h$ and uniqueness among bounded fixed points. Multiplying the individual residuals by their denominators, which lie in $[\lambda ,\lambda +Q_h]$, shows that this is equivalent to ${\sup }_{a\in \mathcal{C}}\{L_h^a{v}_h-\lambda v_h+k(a)f\}=0.\text{(1)}$.

Let $\psi$ be a supersolution as in the Statement. The centered second difference is an average of the corresponding directional second derivative, and the forward time difference is an average of ${\mathrm{\partial }}_t\psi$. Bounded uniform continuity gives $E_h\to 0$ such that, uniformly in $z$ and $a$, $\mid L_h^a\psi -r(a){\mathrm{\partial }}_t\psi -a:D_x^2\psi \mid \le E_h\mathrm{.}$ For example, the sum of the spatial-Hessian and time-derivative moduli of continuity at scale $h$ suffices, since $r(a)+\mathrm{tr}a=1$. Therefore ${\mathrm{\Psi }}_h=\psi +E_h\mathrm{/}\lambda$ satisfies $P_h{\mathrm{\Psi }}_h\le {\mathrm{\Psi }}_h$.

AM–GM applied to $r(a),s_1(a),\dots ,s_d(a)$, whose sum is $1$, gives $0\le k(a)\le 1\mathrm{/}n$. Hence $P_h$ preserves $0\le u\le V$: at the upper bound its numerator is at most $q_h(a)V+\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}n=(\lambda +q_h(a))V$. Thus $0\le v_h\le V$.

If $u$ has Lipschitz constant at most $L$, each affine control expression in $P_{h}u$ has Lipschitz constant at most $\frac{q_h(a)L+L_f\mathrm{/}n}{\lambda +q_h(a)}\le L\mathrm{.}$ A supremum preserves this common bound. Iteration and uniform convergence therefore make every $v_h$ globally $L$-Lipschitz.

If $u(z)+N\mathrm{\mid }z{\mathrm{\mid }}^2$ is convex, every translate of $u$ has the same semiconvexity constant $N$. The numerator in each affine control expression has semiconvexity constant at most $q_h(a)N+N_f\mathrm{/}n$. After division this is at most $N$, since $N_f\mathrm{/}n=\lambda N$. Adding $N\mathrm{\mid }z{\mathrm{\mid }}^2$ makes each expression convex; the translation corrections to its quadratic term are affine plus constants. Their supremum is convex. Starting from $0$ and passing to the uniform limit proves that $v_h+N\mathrm{\mid }z{\mathrm{\mid }}^2$ is convex.

Take a sequence $h_j\downarrow 0$. Diagonally select a subsequence along which $v_{h_j}$ converges at every rational point. Boundedness gives the successive subsequences, and finite rational nets plus the common Lipschitz constant give uniform convergence on every compact set. Call the limit $v$. It satisfies $0\le v\le V$, the global $L$-Lipschitz bound, and convexity of $v+N\mathrm{\mid }z{\mathrm{\mid }}^2$.

The one-sided contraction estimate gives $\sup (v_h-{\mathrm{\Psi }}_h{)}^{+}\le {\gamma }_h\sup (v_h-{\mathrm{\Psi }}_h{)}^{+}\mathrm{.}$ The supremum is finite, so it vanishes. Pass along the chosen subsequence to obtain $v\le \psi$. This proves the comparison assertion for the same constructed limit.

Fix $a\in \mathcal{C}$ and a nonnegative compactly supported smooth test $\eta$. Equation (1) implies $L_h^a{v}_h-\lambda v_h+k(a)f\le 0$. Integrate against $\eta$ and transfer translations to $\eta$. Its adjoint centered differences converge uniformly, on a common compact support, to the spatial second derivatives; its adjoint forward difference converges to $-{\mathrm{\partial }}_t\eta$. Local uniform convergence of $v_h$ gives $\int v(a:D_x^2\eta -r(a){\mathrm{\partial }}_t\eta -\lambda \eta )+\int k(a)f\eta \le 0.\text{(2)}$ Thus each fixed-control inequality holds as a distribution.

Choose a nonnegative smooth compactly supported unit-mass kernel $\rho$ and put ${\rho }_{ϵ}(z)={ϵ}^{-n}\rho (z\mathrm{/}ϵ)$, $v_{ϵ}=v\ast {\rho }_{ϵ}$. The common Lipschitz bound gives $\mathrm{\mid }{D}_z{v}_{ϵ}\mathrm{\mid }\le L$, and $v_{ϵ}\to v$ locally uniformly. Integration by parts bounds each distributional directional derivative by $L\mathrm{\parallel }\eta {\mathrm{\parallel }}_1$. Step 1.2 represents these by bounded weak derivatives. Using a countable dense set of unit directions gives $\mathrm{\mid }{D}_{z}v\mathrm{\mid }\le L$ almost everywhere.

Semiconvexity gives $D_z^2{v}_{ϵ}\ge -2NI$. For a unit spatial vector $e$, insert the rank-one control $a=e{e}^{\mathsf{T}}$ into (2). It has $r(a)=k(a)=0$, so $D_{ee}v-\lambda v\le 0$ as distributions. Convolution makes this $D_{ee}{v}_{ϵ}\le \lambda v_{ϵ}$ everywhere. Taking a countable dense set of $e$ and then continuity yields $-2NI\le D_x^2{v}_{ϵ}\le \lambda v_{ϵ}I\le \lambda VI\mathrm{.}\text{(3)}$.

Set $C=\max (2N,\lambda V)$. Every spatial Hessian entry in (3) has absolute value at most $C$. For a compact smooth test $\eta$, $\int v{\mathrm{\partial }}_{x_i{x}_j}\eta ={\lim }_{ϵ\downarrow 0}\int (D_{ij}{v}_{ϵ})\eta \mathrm{.}$ Step 1.2 therefore represents this distribution by an actual weak derivative $D_{ij}v\in L^{\mathrm{\infty }}$, with norm at most $C$. Passing the directional bounds through tests gives $-2NI\le D_x^{2}v\le \lambda vI$ almost everywhere. The control $a=0$ in (2) similarly gives ${\mathrm{\partial }}_{t}v\le \lambda v$.

Each individually existing derivative commutes with convolution by its weak-test identity: if $g=D^{\beta }v$, where $\beta$ is a first index or two spatial indices, test against $y\mapsto {\rho }_{ϵ}(z-y)$ to obtain $D^{\beta }(v\ast {\rho }_{ϵ})(z)=\int g(y){\rho }_{ϵ}(z-y)dy\mathrm{.}$ The factors $(-1{)}^{\mathrm{\mid }\beta \mathrm{\mid }}$ cancel. This argument uses only that particular weak derivative. All expressions in (2) now have bounded-function representatives. Discard the null sets for a countable dense subset of $\mathcal{C}$; continuity in $a$, including continuity of the determinant at singular matrices, extends the inequalities to every $a$ on one common full-measure set.

By semiconvexity and the local Alexandrov theorem, almost every $z_0$ has a direction-uniform expansion $v(z)=q(z)+o(\mathrm{\mid }z-z_0{\mathrm{\mid }}^2)$ for a quadratic polynomial $q$ with $q(z_0)=v(z_0)$. Require also that ball averages recover each bounded weak first and spatial second derivative $g$, and $g^2$, at $z_0$. These finitely many requirements remove only a null set. Expanding the square gives $\frac{1}{\mathrm{\mid }B(z_0,s)\mathrm{\mid }}{\int }_{B(z_0,s)}\mathrm{\mid }g-g(z_0){\mathrm{\mid }}^2⟶0,$ so Cauchy–Schwarz gives mean absolute convergence as well.

The first-order blowups $[v(z_0+sy)-v(z_0)]\mathrm{/}s$ converge locally uniformly to $Dq(z_0)\cdot y$. Their weak derivatives are $D_{z}v(z_0+sy)$, which converge locally in $L^1$ to the weak derivative values at $z_0$ by step 10.2. Distributional limits identify those values with $Dq(z_0)$. Similarly the second-order blowups $\frac{v(z_0+sy)-v(z_0)-sDq(z_0)\cdot y}{s^2}$ converge locally uniformly to $\frac{1}{2}{y}^{\mathsf{T}}{D}^{2}qy$, while their spatial second weak derivatives are $D_x^{2}v(z_0+sy)$ and converge locally in $L^1$ to $D_x^{2}v(z_0)$. Thus the spatial block of $D^{2}q$ equals that actual weak Hessian. These scaled weak identities follow directly by changing variables in the compact-test identities.

Fix $ϵ>0$ and set $q_{\pm }(z)=q(z)\pm ϵ\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$. On a sufficiently small closed ball centered at $z_0$, the expansion makes $q_{+}$ a strict upper contact and $q_{-}$ a strict lower contact. Let $z_j$ maximize $v_{h_j}-q_{+}$ there. Local uniform convergence implies $z_j\to z_0$ and $m_j=(v_{h_j}-q_{+})(z_j)\to 0$. For large $j$, every stencil neighbor of $z_j$ remains in that ball. Nonnegative off-center weights then give $0\le {\sup }_a\{L_{h_j}^a{q}_{+}-\lambda q_{+}+k(a)f\}(z_j)-\lambda m_j$.

Centered spatial differences are exact on a quadratic polynomial; its forward time difference differs from its time derivative by $h_j{\mathrm{\partial }}_{tt}{q}_{+}\mathrm{/}2$. This error is uniform in the controls, since $0\le r(a)\le 1$. Bounded coefficients and uniform Lipschitz continuity of $f$ therefore allow passage in step 11.2 to ${\max }_a\{r(a){\mathrm{\partial }}_t{q}_{+}+a:D_x^2{q}_{+}-\lambda q_{+}+k(a)f\}(z_0)\ge 0$.

Minimize $v_{h_j}-q_{-}$ on the same ball. The identical positive-weight calculation reverses the inequality and gives the corresponding maximum for $q_{-}$ nonpositive in the limit. Let $ϵ\downarrow 0$ in both conclusions. Continuity of the finite-dimensional maximum gives zero for $q$. By step 11.1 its first and spatial second coefficients are the actual weak derivatives of $v$ at $z_0$. This proves the asserted Bellman equation outside a null set.

At a point where the equation and derivative bounds hold, put $S=\lambda v-{\mathrm{\partial }}_{t}v\ge 0$ and $H=\lambda vI-D_x^{2}v\ge 0$. The equation becomes ${\max }_{\begin{array}{c}r\ge 0,a\ge 0\\ r+\mathrm{tr}a=1\end{array}}\{-rS-a:H+(r\det a{)}^{1\mathrm{/}n}f\}=0.\text{(4)}$ In particular every normalized control satisfies $rS+a:H\ge (r\det a{)}^{1\mathrm{/}n}f$.

For $\delta >0$, let $S_{\delta }=S+\delta$, $H_{\delta }=H+\delta I$, and $c_{\delta }=(S_{\delta }^{-1}+\mathrm{tr}{H}_{\delta }^{-1}{)}^{-1},r_{\delta }=c_{\delta }\mathrm{/}{S}_{\delta },a_{\delta }=c_{\delta }{H}_{\delta }^{-1}\mathrm{.}$ This is a normalized control, and $r_{\delta }S+a_{\delta }:H\le n{c}_{\delta },(r_{\delta }\det a_{\delta }{)}^{1\mathrm{/}n}=\frac{c_{\delta }}{(S_{\delta }\det H_{\delta }{)}^{1\mathrm{/}n}}\mathrm{.}$ Insert it into the inequality from step 14.1 and cancel $c_{\delta }>0$ to get $f\le n(S_{\delta }\det H_{\delta }{)}^{1\mathrm{/}n}$. Sending $\delta \downarrow 0$ gives $f\le n(S\det H{)}^{1\mathrm{/}n}$, including $S=0$ or singular $H$.

If $S>0$ and $H$ is positive definite, AM–GM for $rS$ and the eigenvalues of $a^{1\mathrm{/}2}H{a}^{1\mathrm{/}2}$ gives $rS+a:H\ge n(r\det aS\det H{)}^{1\mathrm{/}n}\mathrm{.}$ If $f<n(S\det H{)}^{1\mathrm{/}n}$, the expression in (4) is strictly negative for interior controls. It is also strictly negative for boundary controls: there $r\det a=0$, whereas $rS+a:H>0$ because $r+\mathrm{tr}a=1$. Compactness of the normalized control set and continuity make its maximum strictly negative, contradicting (4). Thus equality holds. If $S=0$ or $H$ is singular, step 15.1 already gives $f=0$ and $S\det H=0$. This proves the product identity in all cases, and completes every assertion.