Initial trace for the positive-interpolation and upwind HJB schemes

For a semi-Lagrangian row at a node $x_i$ and action $a$, view its interpolation weights as a finite probability vector on the cell nodes around $y=x_i+\tau b_a(x_i,t)$. In a periodic lift, the weighted mean of each coordinate is $(1-{\theta }_j)x_{i_j}+{\theta }_j(x_{i_j}+h_j)=y_j$, by the product formula in [F1]. The centered interpolation error $\xi$ therefore has weighted mean zero and satisfies $\mathrm{\mid }\xi {\mathrm{\mid }}^2\le {\sum }_j{h}_j^2$.

For an upwind row, the weights in [F2] are nonnegative and sum to one by CFL. The weighted mean displacement to the previous-time node is $\tau b_a$: in coordinate $j$ it is $\tau (b_j^{+}\mathrm{/}{h}_j)h_j-\tau (b_j^{-}\mathrm{/}{h}_j)h_j=\tau b_j$. Its centered error $\xi$ has mean zero and second moment at most $\tau {\sum }_j\mathrm{\mid }{b}_j\mathrm{\mid }{h}_j\le \tau B{\sum }_j{h}_j$, since subtracting the squared mean cannot increase the second moment.

Unroll either update for the finite number $n$ of steps. At each stage the minimum chooses one of finitely many control rows, and each row is a finite weighted average; induction therefore represents the value as the minimum of finite weighted path costs $u_0(X_n)+{\sum }_{r<n}\tau {\mathrm{\ell }}_r$. Along any such controlled path the sum of the drift increments has norm at most $Bt$. The centered increments have conditional mean zero. For $r<s$, summing the increment ${\xi }_s$ over all next-branch weights from each fixed path prefix through time $s-1$ gives zero; hence the finite weighted cross term $\mathbb{E}({\xi }_r\cdot {\xi }_s)$ is zero. Expanding the square therefore shows that their total second moment is at most $n{\sum }_j{h}_j^2$ for the semi-Lagrangian rows and $tB{\sum }_j{h}_j$ for the upwind rows. Thus the endpoint displacement obeys $\mathbb{E}\mathrm{\mid }{X}_n-x_i{\mathrm{\mid }}^2\le 2B^2{t}^2+2n{\sum }_j{h}_j^2$ for the semi-Lagrangian case and $\mathbb{E}\mathrm{\mid }{X}_n-x_i{\mathrm{\mid }}^2\le 2B^2{t}^2+2tB{\sum }_j{h}_j$ for the upwind case. Every path's accumulated reward has magnitude at most $Lt$.

Let $M_0=\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ and let ${\omega }_0(\delta )\to 0$ be the uniform modulus from [F4]. For any $0<\delta <1$, split the finite weighted endpoint average into paths with $\mathrm{\mid }{X}_n-x_i\mathrm{\mid }\le \delta$ and its complement. On the first set, $\mathrm{\mid }{u}_0(X_n)-u_0(x_i)\mathrm{\mid }\le {\omega }_0(\delta )$; on the complement, it is at most $2M_0$. The complement weight is at most $\mathbb{E}\mathrm{\mid }{X}_n-x_i{\mathrm{\mid }}^2\mathrm{/}{\delta }^2$. By step 1.3 this tends to zero as $t_n\to 0$ under $h_{\max }^{}\mathrm{/}\tau \to 0$ for the semi-Lagrangian scheme, and under $h_{\max }\to 0$ for the upwind scheme. The reward contributes at most $L{t}_n$; the same two-sided bound holds after taking a minimum over policies. First let the mesh and $t_n$ tend to zero, then let $\delta \downarrow 0$. This proves the asserted uniform initial trace.