A consistent but unstable explicit spectral-grid heat step

Let $D$ be the finite-support sequences. It is dense by [F4] and invariant under the diagonal semigroup [F3]. Since $m=\nu +1$, ${\tau }_m=m^{-1}>0$ and ${\tau }_m\to 0$. For each $m$, take $P_m$ to be coordinate truncation and $R_m$ inclusion. The estimate $\mathrm{\parallel }{P}_{m}x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ and identity $\mathrm{\parallel }{R}_{m}g{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$, together with testing $e_0$ in [F6], give $\mathrm{\parallel }{P}_m\mathrm{\parallel }=\mathrm{\parallel }{R}_m\mathrm{\parallel }=1$. Then $P_m{R}_m=I$ and $\ker P_m$ is invariant under diagonal $U({\tau }_m)$. The finite maximum $L_m:={\max }_{0\le j<m}\mathrm{\mid }1-{\tau }_m{j}^2\mathrm{\mid }$ gives $\mathrm{\parallel }{Q}_{m}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le L_m\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $Q_m$ is bounded. Thus [F2] gives a bounded common-space step.

Fix $v\in D$ with support in coordinates $0,\dots ,N$. For $m>N$, every supported coordinate is updated by $1-{\tau }_m{j}^2$; since $0<\exp (-r{\tau }_m{j}^2)\le 1$, we have ${\max }_{\begin{array}{c}r\in \mathbb{N}\\ r{\tau }_m\le T\end{array}}\frac{\mathrm{\parallel }(S_m-U({\tau }_m))U(r{\tau }_m)v{\mathrm{\parallel }}_{\mathrm{\infty }}}{{\tau }_m}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}{\max }_{0\le j\le N}\frac{\mathrm{\mid }1-{\tau }_m{j}^2-\exp (-{\tau }_m{j}^2)\mathrm{\mid }}{{\tau }_m}\mathrm{.}$ For $j=0$ the quotient vanishes. For each fixed $1\le j\le N$, with $z={\tau }_m{j}^2$, the quotient equals $j^2\mathrm{\mid }(1-\exp (-z))\mathrm{/}z-1\mathrm{\mid }$, which tends to zero by [F5]. The finite maximum tends to zero: for any tolerance choose a threshold for each of the finitely many $j$ (or use threshold $1$ if there is no positive $j$), then take the largest threshold. Multiplication by fixed $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$ preserves the limit, uniformly in the permitted $r$. Thus the family is consistent.

On the highest grid mode $e_{m-1}$, the one-step multiplier is $1-\frac{(m-1{)}^2}{m}=3-m-\frac{1}{m}\mathrm{.}$ For $m\ge 4$ its magnitude is $m-3+1\mathrm{/}m$, which tends to infinity. Given any $T>0$, all sufficiently large $m$ also satisfy ${\tau }_m=1\mathrm{/}m\le T$; then the allowed power $n=1$ already has $\mathrm{\parallel }{S}_m{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\ge m-3+1\mathrm{/}m$ by testing the unit vector $e_{m-1}$ in [F6]. No uniform finite-time power bound can hold, so the family is unstable.