A mesh-dependent grid norm hides common-space instability

Let $D$ be the finite-support sequences. By [F4], $X$ is Banach; the identity maps $U(t)=I_X$ form a strongly continuous semigroup with finite-time bound $M_T=1$. By [F5], $D$ is dense and it is invariant under $U$. The positive steps ${\tau }_m=1\mathrm{/}m$ tend to zero, so this is common-space scheme data.

The positive coordinate weights make the displayed maximum a norm: homogeneity and definiteness hold coordinatewise, and the triangle inequality follows by applying the scalar triangle inequality in every coordinate before taking the maximum. The truncation satisfies $\mathrm{\parallel }{P}_{m}x{\mathrm{\parallel }}_{G_m}\le m\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, and inclusion satisfies $\mathrm{\parallel }{R}_{m}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{G_m}$, so both are bounded. Also $P_m{R}_m=I_{G_m}$ and $\ker P_m$ is invariant under $U=I$. For $g={\sum }_{j<m}{g}_j{e}_j$, the zeroth output coordinate is $g_0+m{g}_{m-1}$, coordinates $1$ through $m-2$ if present remain $g_j$, and coordinate $m-1$ becomes $0$. Since $\mathrm{\mid }{g}_0\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{G_m}$, $m\mathrm{\mid }{g}_{m-1}\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{G_m}$, and $\mathrm{\mid }{g}_j\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{G_m}$ for $1\le j\le m-2$, it follows that $\mathrm{\parallel }{Q}_{m}g{\mathrm{\parallel }}_{G_m}\le 2\mathrm{\parallel }g{\mathrm{\parallel }}_{G_m}$, including $m=2$. Thus $Q_m$ is bounded and the grid lift [F2] is defined. Also $Q_m^2=Q_m$: it fixes each $e_j$ for $j<m-1$, and $Q_m^2{e}_{m-1}=Q_m(m{e}_0)=m{e}_0=Q_m{e}_{m-1}$. By induction [F7], $Q_m^n=Q_m$ for every $n\ge 1$: the base $n=1$ is immediate, and $Q_m^n=Q_m$ implies $Q_m^{n+1}=Q_m{Q}_m=Q_m^2=Q_m$. Meanwhile $Q_m^0=I$. By [F6], $\mathrm{\parallel }{Q}_m^n{\mathrm{\parallel }}_{G_m\to G_m}\le 2$ for every $n\ge 0$, so the grid updates are stable in their mesh norms.

Fix $v\in D$ and choose $N\in \mathbb{N}$ such that $v_j=0$ for every $j>N$ (take $N=0$ for the zero sequence). For $m\ge N+2$, one has $P_{m}v=v$, the high coordinate $v_{m-1}=0$ since $m-1>N$, and $Q_{m}v=v$. Since $U=I$, the lift gives $S_{m}v=v$. Thus the local defect is zero on every $v\in D$ for all sufficiently large $m$, which proves consistency.

For $e_{m-1}$, the lift gives $S_m{e}_{m-1}=m{e}_0$. Its common-space norm is therefore $m$, while $\mathrm{\parallel }{e}_{m-1}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. Testing this unit vector in [F6] gives $\mathrm{\parallel }{S}_m{\mathrm{\parallel }}_{X\to X}\ge m$. For every $T>0$, all sufficiently large $m$ have ${\tau }_m\le T$, so the common-space family is not stable.

The same vector satisfies $\mathrm{\parallel }{e}_{m-1}{\mathrm{\parallel }}_{G_m}=m$ but $\mathrm{\parallel }{R}_m{e}_{m-1}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$; thus no positive uniform lower comparison constant $a$ exists. Also $\mathrm{\parallel }{P}_m{\mathrm{\parallel }}_{X\to G_m}\ge m$ by testing the unit vector $e_{m-1}$ in [F6]. The hypotheses needed in [F3] fail, exactly where the grid norm ceases to control the reconstructed common-space norm and the restriction norm. The grid stability calculation therefore cannot certify common-space stability.