Lax equivalence for a common Banach-space scheme

Fix $T>0$ and suppose the family is stable, with a bound $C_T$ from [F1]. For $x\in D$, set ${\eta }_{\nu }(x,T):={\max }_{\begin{array}{c}r\in \mathbb{N}\\ r{\tau }_{\nu }\le T\end{array}}\frac{\mathrm{\parallel }(S_{\nu }-U({\tau }_{\nu }))U(r{\tau }_{\nu })x\mathrm{\parallel }}{{\tau }_{\nu }}\mathrm{.}$ Consistency says ${\eta }_{\nu }(x,T)\to 0$ for each $x\in D$.

For each integer $n\ge 0$, the finite telescoping identity is $S_{\nu }^n-U(n{\tau }_{\nu })={\sum }_{r=0}^{n-1}{S}_{\nu }^{n-1-r}(S_{\nu }-U({\tau }_{\nu }))U(r{\tau }_{\nu })\mathrm{.}$ For $n=0$ both sides are zero. For $n\ge 1$, expand each summand as $S_{\nu }^{n-r}U(r{\tau }_{\nu })-S_{\nu }^{n-1-r}U((r+1){\tau }_{\nu })$; adjacent terms cancel, leaving $S_{\nu }^n-U(n{\tau }_{\nu })$.

Conversely, suppose the family is convergent. Fix $T>0$ and $x\in X$. By convergence, for all sufficiently large $\nu$, ${\max }_{n{\tau }_{\nu }\le T}\mathrm{\parallel }{S}_{\nu }^{n}x-U(n{\tau }_{\nu })x\mathrm{\parallel }\le 1.$ The finite-time solution bound in [F1] then gives ${\max }_{n{\tau }_{\nu }\le T}\mathrm{\parallel }{S}_{\nu }^{n}x\mathrm{\parallel }\le M_T\mathrm{\parallel }x\mathrm{\parallel }+1$ for those $\nu$.

If $x\in D$ and $n{\tau }_{\nu }\le T$, [F1], step 1.2, and the definition of ${\eta }_{\nu }$ yield $\mathrm{\parallel }{S}_{\nu }^{n}x-U(n{\tau }_{\nu })x\mathrm{\parallel }\le C_T{\sum }_{r=0}^{n-1}\mathrm{\parallel }(S_{\nu }-U({\tau }_{\nu }))U(r{\tau }_{\nu })x\mathrm{\parallel }\le C_{T}n{\tau }_{\nu }{\eta }_{\nu }(x,T)\le C_{T}T{\eta }_{\nu }(x,T)\mathrm{.}$ The last quantity tends to zero independently of the allowed $n$.

Only finitely many indices $\nu$ were omitted in step 1.3. For each fixed omitted $\nu$, there are only finitely many integers $n$ with $n{\tau }_{\nu }\le T$, and each $S_{\nu }^{n}x$ is a vector of finite norm. Thus the powers are pointwise bounded on every $x\in X$. The admissible index set $I_T:=\{(\nu ,n)\in \mathbb{N}\times \mathbb{N}:n{\tau }_{\nu }\le T\}$ exists by Separation; each pair determines a unique bounded linear operator $S_{\nu }^n$, so Replacement gives the set-family ${\mathcal{F}}_T:=\{S_{\nu }^n:(\nu ,n)\in I_T\}$. For fixed $\nu$, boundedness supplies a finite $K_{\nu }\ge 1$ with $\mathrm{\parallel }{S}_{\nu }z\mathrm{\parallel }\le K_{\nu }\mathrm{\parallel }z\mathrm{\parallel }$. The identity $S_{\nu }^0=I$ is linear with bound $1=K_{\nu }^0$; if $S_{\nu }^n$ is linear with bound $K_{\nu }^n$, then $S_{\nu }^{n+1}=S_{\nu }{S}_{\nu }^n$ is linear and $\mathrm{\parallel }{S}_{\nu }^{n+1}z\mathrm{\parallel }\le K_{\nu }\mathrm{\parallel }{S}_{\nu }^{n}z\mathrm{\parallel }\le K_{\nu }^{n+1}\mathrm{\parallel }z\mathrm{\parallel }$. Finite induction therefore proves every member of ${\mathcal{F}}_T$ bounded and linear. This argument fixes one $\nu$ at a time and chooses no sequence of bounds. Also $X$ is Banach by [F1].

Let $x\in X$ and $\epsilon >0$. By density in [F1], choose $y\in D$ with $\mathrm{\parallel }x-y\mathrm{\parallel }<\epsilon \mathrm{/}[2(1+C_T+M_T)]$. For every $n{\tau }_{\nu }\le T$, $\mathrm{\parallel }{S}_{\nu }^{n}x-U(n{\tau }_{\nu })x\mathrm{\parallel }\le (C_T+M_T)\mathrm{\parallel }x-y\mathrm{\parallel }+\mathrm{\parallel }{S}_{\nu }^{n}y-U(n{\tau }_{\nu })y\mathrm{\parallel }\mathrm{.}$ By step 2.1, the second term is below $\epsilon \mathrm{/}2$ for all such $n$ once $\nu$ is sufficiently large. This proves convergence for $x$, uniformly over the grid times in $[0,T]$. Since $x,T$ were arbitrary, stability implies convergence.

Apply the uniform boundedness principle [F2] to ${\mathcal{F}}_T$, using the explicit DC assumption [A1]. It gives ${\sup }_{\begin{array}{c}\nu ,n\in \mathbb{N}\\ n{\tau }_{\nu }\le T\end{array}}\mathrm{\parallel }{S}_{\nu }^n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<\mathrm{\infty },$ which is stability on $[0,T]$. Since $T>0$ was arbitrary, convergence implies stability; together with step 3.1 this proves the equivalence.