$j\in\\{0,\ldots,n\\}$ we have the bound
$\left|\left|F^{(j)}(z)\right|\right|_{{\cal L}(V^{\otimes j};W)}\leq r_{0}$.
Fix $\delta\in[0,1]$ and define
$\Omega:=\Gamma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,\delta)$. For
$s\in\\{0,\ldots,n-1\\}$ inductively define

$E_{n-s}:=\left\\{\begin{array}[]{ll}\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A~{},~{}\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\right\\}&\mbox{if
}s=0\\\
\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-(s-1)}~{},~{}\min\left\\{E_{n-(s-1)}~{},~{}\delta
E_{n-(s-1)}+r_{0}e^{\delta}\right\\}\right\\}&\mbox{if }s\geq
1.\end{array}\right.$ (7.23)

We first prove that each $E_{n-s}$ is bounded from below by $r_{0}$. This is
the content of the following claim.

######  Claim 7.4.

For every $s\in\\{0,\ldots,n-1\\}$ we have

$E_{n-s}\geq r_{0}.$ (7.24)

###### Proof of Claim 7.4.

The claim is proven via induction on $s\in\\{0,\ldots,n-1\\}$. For $s=0$ we
have

$E_{n}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_Es}})}}}{{\geq}}\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A~{},~{}\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\right\\}\geq\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\geq
r_{0}$ (7.25)

where the last inequality uses that $r_{0}\leq A$. If (7.24) is valid for
$s\in\\{0,\ldots,n-2\\}$ then we compute that

$\displaystyle E_{n-(s+1)}$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_Es}})}}}{{=}}\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-s}~{},~{}\min\left\\{E_{n-s}~{},~{}\delta
E_{n-s}+r_{0}e^{\delta}\right\\}\right\\}$
$\displaystyle\geq\max\left\\{\sqrt{2\delta}E_{n-s}~{},~{}\min\left\\{E_{n-s}~{},~{}\delta
E_{n-s}+r_{0}e^{\delta}\right\\}\right\\}$
$\displaystyle\geq\min\left\\{E_{n-s}~{},~{}\delta
E_{n-s}+r_{0}e^{\delta}\right\\}\geq r_{0}$

where the last line uses that $E_{n-s}\geq r_{0}$ by the assumption that
(7.24) is valid for $s$, and that $\delta E_{n-s}+r_{0}e^{\delta}\geq r_{0}$
since $\delta\geq 0$. Thus we have established that the estimate (7.24) for
$s\in\\{0,\ldots,n-2\\}$ yields that the estimate (7.24) is true for $s+1$.
Since (7.25) establishes that (7.24) is true for $s=0$, we may use induction
to prove that (7.24) is in fact true for every $s\in\\{0,\ldots,n-1\\}$ as
claimed. This completes the proof of Claim 7.4. ∎

We now prove that, for each $s\in\\{0,\ldots,n-1\\}$, the
${\mathrm{Lip}}(n-s,\Omega,W)$-norm of
$F_{[n-s-1]}=\left(F^{(0)},\ldots,F^{(n-s-1)}\right)$ is bounded above by
$E_{n-s}$. This is the content of the following claim.

######  Claim 7.5.

For every $s\in\\{0,\ldots,n-1\\}$ we have that

$\left|\left|F_{[n-(s+1)]}\right|\right|_{{\mathrm{Lip}}(n-s,\Omega,W)}\leq
E_{n-s}.$ (7.26)

###### Proof of Claim 7.5.

We will prove (7.26) via induction on $s\in\\{0,\ldots,n-1\\}$. We begin with
the base case that $s=0$. In this case, consider
$\xi:=\frac{\rho-n}{2}\in(0,\rho-n)$ so that $n+\xi\in(n,\rho)$ with
$0<\xi\leq 1/2$. An initial application of Lemma 7.2, with $\Gamma$, $A$,
$r_{0}$, $\rho$ and $n$ here playing the same role and with the $\theta$ in
Lemma 7.2 being $n+\xi$ here, yields that

$||F||_{{\mathrm{Lip}}(n+\xi,\Omega,W)}\leq\max\left\\{(2\delta)^{\rho-n-\xi}A,\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\right\\}.$
(7.27)

We next apply Lemma 6.1, with the $\Gamma$, $\rho$ and $\theta$ of that result
as $\Omega$, $n+\xi$ and $n$ here respectively, to obtain that

$\left|\left|F_{[n-1]}\right|\right|_{{\mathrm{Lip}}(n,\Omega,W)}\leq\min\\{C_{1},C_{2}\\}||F||_{{\mathrm{Lip}}(n+\xi,\Omega,W)}$
(7.28)

where (cf. (6.3) and recalling both that $\operatorname{diam}(\Omega)\leq
2\delta\leq 2$ and that $0<\xi\leq 1$)

$C_{1}=\max\left\\{1~{},~{}\min\left\\{1+e~{},~{}\operatorname{diam}(\Omega)^{\xi}+\sum_{j=n}^{n}\frac{\operatorname{diam}(\Omega)^{j-n}}{(j-(n-1))!}\right\\}\right\\}=1+\operatorname{diam}(\Omega)^{\xi}\leq
1+(2\delta)^{\xi},$ (7.29)

and (cf. (6.4))

$C_{2}=\max\left\\{1,\min\left\\{1+e,\operatorname{diam}(\Omega)^{n-n}\right\\}\right\\}\left(1+\min\left\\{e,\operatorname{diam}(\Omega)\right\\}^{\xi}\right)\left(1+\min\left\\{e,\operatorname{diam}(\Omega)\right\\}\right)^{n-n}$
(7.30)

so that, since $\operatorname{diam}(\Omega)\leq 2\delta\leq 2$ and $0<\xi\leq
1$, we have

$C_{2}=1+\operatorname{diam}(\Omega)^{\xi}\leq 1+(2\delta)^{\xi}$ (7.31)

The combination of (7.27), (7.28), (7.29), and (7.31) yields that

$\displaystyle\left|\left|F_{[n-1]}\right|\right|_{{\mathrm{Lip}}(n,\Omega,W)}$
$\displaystyle\leq\left(1+(2\delta)^{\xi}\right)\max\left\\{(2\delta)^{\rho-n-\xi}A,\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\right\\}$
$\displaystyle=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A,\min\left\\{A~{},~{}A\delta^{\rho-n}+r_{0}e^{\delta}\right\\}\right\\}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_Es}})}}}{{=}}E_{n}.$

This completes the base case of our induction by verifying (7.26) when $s=0$.

Now assume that $n-1\geq 1$, that $s\in\\{1,\ldots,n-1\\}$, and that (7.26) is
true for $s-1$. Consider $\xi:=\frac{1}{2}\in(0,1)$ so that
$n-s+\xi\in(n-s,n-(s-1))$. An initial application of Lemma 7.2, with $\Gamma$,
$A$, $r_{0}$, $\rho$ and $\theta$ of that result as $\Omega$, $E_{n-(s-1)}$,
$r_{0}$, $n-(s-1)$ and $n-s+\xi$ here respectively, yields that

$\left|\left|F_{[n-s]}\right|\right|_{{\mathrm{Lip}}(n-s+\xi,\Omega,W)}\leq\max\left\\{(2\delta)^{1-\xi}E_{n-(s-1)},\min\left\\{E_{n-(s-1)}~{},~{}E_{n-(s-1)}\delta+r_{0}e^{\delta}\right\\}\right\\}.$
(7.32)

We next apply Lemma 6.1, with the $\Gamma$, $\rho$ and $\theta$ of that result
as $\Omega$, $n-s+\xi$ and $n-s$ here respectively, to obtain that

$\left|\left|F_{[n-(s+1)]}\right|\right|_{{\mathrm{Lip}}(n-s,\Omega,W)}\leq\min\\{D_{1},D_{2}\\}\left|\left|F_{[n-s]}\right|\right|_{{\mathrm{Lip}}(n-s+\xi,\Omega,W)}$
(7.33)

where (cf. (6.3) and recalling that $\operatorname{diam}(\Omega)\leq
2\delta\leq 2$ and $\xi:=\frac{1}{2}\leq 1$)

$D_{1}=\max\left\\{1~{},~{}\min\left\\{1+e,\operatorname{diam}(\Omega)^{\xi}+\sum_{j=n-s}^{n-s}\frac{\operatorname{diam}(\Omega)^{j-(n-s)}}{(j-(n-s-1))!}\right\\}\right\\}=1+\operatorname{diam}(\Omega)^{\xi}\leq
1+(2\delta)^{\xi},$ (7.34)

and (cf. (6.4))

$D_{2}=\max\left\\{1~{},~{}\min\left\\{1+e,\operatorname{diam}(\Omega)^{0}\right\\}\right\\}\left(1+\min\left\\{e,\operatorname{diam}(\Omega)\right\\}^{\xi}\right)\left(1+\min\left\\{e,\operatorname{diam}(\Omega)\right\\}\right)^{0}$
(7.35)

so that, since $\operatorname{diam}(\Omega)\leq 2\delta\leq 2$ and
$\xi:=\frac{1}{2}\leq 1$, we have

$D_{2}=1+\operatorname{diam}(\Omega)^{\xi}\leq 1+(2\delta)^{\xi}.$ (7.36)

The combination of (7.32), (7.33), (7.34), and (7.36) yields that

$\displaystyle\left|\left|F_{[n-(s+1)]}\right|\right|_{{\mathrm{Lip}}(n-s,\Omega,W)}$
$\displaystyle\leq\left(1+(2\delta)^{\xi}\right)\max\left\\{(2\delta)^{1-\xi}E_{n-(s-1)},\min\left\\{E_{n-(s-1)}~{},~{}\delta
E_{n-(s-1)}+r_{0}e^{\delta}\right\\}\right\\}$
$\displaystyle=\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-(s-1)},\min\left\\{E_{n-(s-1)}~{},~{}\delta
E_{n-(s-1)}+r_{0}e^{\delta}\right\\}\right\\}$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_Es}})}}}{{=}}E_{n-s}.$

This completes the proof of the inductive step by establishing that if (7.26)
is valid for $s-1$ with $s\in\\{1,\ldots,n-1\\}$, then (7.26) is in fact valid
for $s$.

Using the base case and the inductive step allows us to conclude that (7.26)
is valid for every $s\in\\{0,\ldots,n-1\\}$. This completes the proof of Claim
7.5. ∎

By appealing to Claim 7.5, we conclude that, for every
$s\in\\{0,\ldots,n-1\\}$ we have that

$\left|\left|F_{[n-(s+1)]}\right|\right|_{{\mathrm{Lip}}(n-s,\Omega,W)}\leq
E_{n-s}.$ (7.37)

Let $b_{q}:=n-(q+1)\in\\{0,\ldots,n-1\\}$ so that $q+1=n-b_{q}$. Then (7.37)
for $s:=b_{q}$ tells us that

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(q+1,\Omega,W)}\leq
E_{n-b_{q}}.$ (7.38)

A final application of Lemma 7.2, with $\Gamma$, $A$, $r_{0}$, $\rho$ and
$\theta$ of that result as $\Omega$, $E_{n-b_{q}}$, $r_{0}$, $q+1$ and
$\theta$ here, yields that

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\theta,\Omega,W)}\leq\max\left\\{(2\delta)^{q+1-\theta}E_{n-b_{q}},\min\left\\{E_{n-b_{q}}~{},~{}\delta
E_{n-b_{q}}+r_{0}e^{\delta}\right\\}\right\\}$ (7.39)

which is precisely the bound claimed in (7.17).

Now suppose that $r_{0}=0$. Then from (7.23), for $s\in\\{0,\ldots,n-1\\}$ we
have that

$E_{n-s}:=\left\\{\begin{array}[]{ll}\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A~{},~{}\min\left\\{A~{},~{}\delta^{\rho-n}A\right\\}\right\\}&\mbox{if
}s=0\\\
\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-(s-1)}~{},~{}\min\left\\{E_{n-(s-1)}~{},~{}\delta
E_{n-(s-1)}\right\\}\right\\}&\mbox{if }s\geq 1.\end{array}\right.$ (7.40)

Since $\delta\in[0,1]$ we have both that $\delta\leq\sqrt{\delta}\leq 1$ and
$\delta^{\rho-n}\leq\delta^{\frac{\rho-n}{2}}\leq 1$ Consequently, (7.40)
yields that

$E_{n-s}:=\left\\{\begin{array}[]{ll}\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)(2\delta)^{\frac{\rho-n}{2}}A&\mbox{if
}s=0\\\ \left(1+\sqrt{2\delta}\right)\sqrt{2\delta}E_{n-(s-1)}&\mbox{if }s\geq
1.\end{array}\right.$ (7.41)

Proceeding inductively via (7.41), we establish that for any
$s\in\\{0,\ldots,n-1\\}$ we have that

$E_{n-s}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{s}(2\delta)^{\frac{\rho-n+s}{2}}A.$
(7.42)

Observing that $\delta\in[0,1]$ means that $\delta\leq\delta^{q+1-\theta}\leq
1$, we may combine (7.39) and (7.42) for the choice $s:=b_{q}=n-(q+1)$ to
obtain that

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\theta,\Omega,W)}\leq(2\delta)^{q+1-\theta}E_{n-b_{q}}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_En-
s_r0=0}})}}}{{=}}\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{b_{q}}(2\delta)^{\frac{\rho-n+b_{q}}{2}+q+1-\theta}A.$
(7.43)

Since $\rho-n+b_{q}=\rho-(q+1)$ we see that (7.43) is precisely the estimate
claimed in (7.19).

Now assume that $r_{0}\in(0,A)$. We first let $\delta_{\ast}:=1$ To establish
(7.20) we must reduce $\delta_{\ast}$ to a smaller constant. With the benefit
of hindsight, it will suffice to reduce $\delta_{\ast}$, depending only on
$A$, $r_{0}$, and $\rho$, to ensure that whenever $\delta\in[0,\delta_{\ast}]$
we have the estimates

$\left\\{\begin{aligned} &({\bf
I})\quad\max\left\\{1+\sqrt{2\delta}~{},~{}1+\left(2\delta\right)^{\frac{\rho-n}{2}}\right\\}<2\quad\left(\text{in
particular }2\delta<1\right),\\\ &({\bf II})\quad r_{0}e^{\delta}\leq
A\left(1-\delta^{\rho-n}\right),\\\ &({\bf
III})\quad\left(2^{\frac{\rho-n}{2}}-\delta^{\frac{\rho-n}{2}}\right)\delta^{\frac{\rho-n}{2}}A\leq
r_{0}e^{\delta},\\\ &({\bf IV})\quad
2\sqrt{2\delta}\left(\delta^{\rho-n}A+r_{0}e^{\delta}\right)\leq
r_{0}e^{\delta},\quad\text{and}\\\ &({\bf
V})\quad\sqrt{2\delta}\left[2^{n}\left(\delta^{\rho-n}A+r_{0}\delta
e^{\delta}\right)+2r_{0}e^{\delta}\left(\frac{1-(2\delta)^{n}}{1-2\delta}\right)\right]\leq
r_{0}e^{\delta}.\end{aligned}\right.$ (7.44)

We now consider a fixed choice of $\delta\in[0,\delta_{\ast}]$ and establish
the estimate claimed in (7.20) for
$\Omega:={\overline{\mathbb{B}}}_{V}(p,\delta)~{}\cap~{}\Gamma$. We begin by
estimating the terms $E_{n-s}$ for $s\in\\{0,\ldots,n-1\\}$. We first prove,
for every $t\in\\{0,\ldots,n-1\\}$, that

$E_{n-t}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t}\left(\delta^{\rho-n+t}A+r_{0}\delta^{t}e^{\delta}\right)+{{\bf{X}}}_{t}(\delta)$
(7.45)

where ${{\bf{X}}}_{t}(\delta)$ is the quantity defined in (7.22). That is,

${{\bf{X}}}_{t}(\delta):=\left\\{\begin{array}[]{ll}0&\mbox{if }t=0\\\
\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}\sum_{j=0}^{t-1}\delta^{j}\left(1+\sqrt{2\delta}\right)^{j}&\mbox{if
}t\geq 1.\end{array}\right.$ (7.46)

We begin by considering $t:=0$. From (7.23) we have that

$E_{n}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A~{},~{}\min\left\\{A~{},~{}\delta^{\rho-n}A+r_{0}e^{\delta}\right\\}\right\\}.$
(7.47)

A consequence of (II) in (7.44) is that $\delta^{\rho-n}A+r_{0}e^{\delta}\leq
A$ so that from (7.47) we see that

$E_{n}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\max\left\\{(2\delta)^{\frac{\rho-n}{2}}A~{},~{}\delta^{\rho-n}A+r_{0}e^{\delta}\right\\}.$
(7.48)

A consequence of (III) in (7.44) is that
$(2\delta)^{\frac{\rho-n}{2}}A\leq\delta^{\rho-n}A+r_{0}e^{\delta}$ so that
from (7.48) we see that

$E_{n}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(\delta^{\rho-n}A+r_{0}e^{\delta}\right)$
(7.49)

which is the estimate claimed in (7.45) for $t=0$ since
${{\bf{X}}}_{0}(\delta):=0$.

Now consider $t\geq 1$ and assume that (7.45) is true for $t-1$. From (7.23)
we have that

$E_{n-t}=\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-(t-1)}~{},~{}\min\left\\{E_{n-(t-1)}~{},~{}\delta
E_{n-(t-1)}+r_{0}e^{\delta}\right\\}\right\\}.$ (7.50)

Since (7.45) is valid for $t-1$ we have that

$E_{n-(t-1)}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t-1}\left(\delta^{\rho-n+t-1}A+r_{0}\delta^{t-1}e^{\delta}\right)+{{\bf{X}}}_{t-1}(\delta)$
(7.51)

We claim that $\delta E_{n-(t-1)}+r_{0}e^{\delta}\leq E_{n-(t-1)}$.

If $t=1$, then (7.49) ensures that
$(1-\delta)E_{n}\geq(1-\delta)\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)r_{0}e^{\delta}$.
If we are able to conclude that
$(1-\delta)\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\geq 1$ then our desired
estimate $\delta E_{n}+r_{0}e^{\delta}\leq E_{n}$ is true. The required lower
bound $(1-\delta)\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\geq 1$ is
equivalent to
$(2\delta)^{\frac{\rho-n}{2}}-\delta-\delta(2\delta)^{\frac{\rho-n}{2}}\geq
0$. A consequence of (I) in (7.44) is that $(2\delta)^{\frac{\rho-n}{2}}<1$.
This tells us that $\delta+\delta(2\delta)^{\frac{\rho-n}{2}}\leq
2\delta\leq(2\delta)^{\frac{\rho-n}{2}}$ where the latter inequality is true
since $2\delta<1$ and $\frac{\rho-n}{2}<1$. Hence
$(1-\delta)\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\geq 1$ and so we have
that $\delta E_{n}+r_{0}e^{\delta}\leq E_{n}$ as required.

If $t>1$ then (7.51) ensures that
$(1-\delta)E_{n-(t-1)}\geq(1-\delta){{\bf{X}}}_{t-1}(\delta)\geq(1-\delta)\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}$.
If we are able to conclude that $(1-\delta)\left(1+\sqrt{2\delta}\right)\geq
1$ then our desired estimate $\delta E_{n-(t-1)}+r_{0}e^{\delta}\leq
E_{n-(t-1)}$ is true. The required lower bound
$(1-\delta)\left(1+\sqrt{2\delta}\right)\geq 1$ is equivalent to
$\sqrt{2\delta}-\delta-\delta\sqrt{2\delta}\geq 0$. A consequence of (I) in
(7.44) is that $\sqrt{2\delta}<1$. This tells us that
$\delta+\delta\sqrt{2\delta}\leq 2\delta\leq\sqrt{2\delta}$ where the latter
inequality is true since $2\delta<1$. Hence
$(1-\delta)\left(1+\sqrt{2\delta}\right)\geq 1$ and so we have that $\delta
E_{n-(t-1)}+r_{0}e^{\delta}\leq E_{n-(t-1)}$ as required.

Having established that $\delta E_{n-(t-1)}+r_{0}e^{\delta}\leq E_{n-(t-1)}$,
(7.50) tells us that

$E_{n-t}=\left(1+\sqrt{2\delta}\right)\max\left\\{\sqrt{2\delta}E_{n-(t-1)}~{},~{}\delta
E_{n-(t-1)}+r_{0}e^{\delta}\right\\}.$ (7.52)

We claim that $\sqrt{2\delta}E_{n-(t-1)}\leq\delta
E_{n-(t-1)}+r_{0}e^{\delta}$. If $t=1$ then we compute, using (7.51) for
$t=1$, that

$\displaystyle\left(\sqrt{2\delta}-\delta\right)E_{n}$
$\displaystyle=\sqrt{\delta}\left(\sqrt{2}-\sqrt{\delta}\right)\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(\delta^{\rho-n}A+r_{0}e^{\delta}\right)$
$\displaystyle\stackrel{{\scriptstyle({\bf
I})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{\leq}}2\sqrt{2\delta}\left(\delta^{\rho-n}A+r_{0}e^{\delta}\right)\stackrel{{\scriptstyle({\bf
IV})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{\leq}}r_{0}e^{\delta}.$

Consequently we have $\sqrt{2\delta}E_{n}\leq\delta E_{n}+r_{0}e^{\delta}$ as
claimed. If $t>1$ then we compute, using (7.51) for $t>1$, that

$\displaystyle\left(\sqrt{2\delta}-\delta\right)E_{n-(t-1)}$
$\displaystyle=\left(\sqrt{2\delta}-\delta\right)\left(\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t-1}\left(\delta^{\rho-n+t-1}A+r_{0}\delta^{t-1}e^{\delta}\right)+{{\bf{X}}}_{t-1}(\delta)\right)$
$\displaystyle\stackrel{{\scriptstyle({\bf
I})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{\leq}}2^{t}\sqrt{2\delta}\left(\delta^{\rho-n+t-1}A+r_{0}\delta^{t-1}e^{\delta}\right)+\sqrt{2\delta}{{\bf{X}}}_{t-1}(\delta)$
$\displaystyle=2^{t}\sqrt{2\delta}\left(\delta^{\rho-n+t-1}A+r_{0}\delta^{t-1}e^{\delta}\right)+\sqrt{2\delta}\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}\sum_{j=0}^{t-1}\delta^{j}\left(1+\sqrt{2\delta}\right)^{j}$
$\displaystyle\stackrel{{\scriptstyle({\bf
I})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{\leq}}2^{t}\sqrt{2\delta}\left(\delta^{\rho-n}A+r_{0}\delta
e^{\delta}\right)+2\sqrt{2\delta}r_{0}e^{\delta}\sum_{j=0}^{t-1}(2\delta)^{j}$
$\displaystyle\stackrel{{\scriptstyle({\bf
I})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{=}}\sqrt{2\delta}\left[2^{t}\left(\delta^{\rho-n}A+r_{0}\delta
e^{\delta}\right)+2r_{0}e^{\delta}\left(\frac{1-(2\delta)^{t}}{1-2\delta}\right)\right]$
$\displaystyle\leq\sqrt{2\delta}\left[2^{n}\left(\delta^{\rho-n}A+r_{0}\delta
e^{\delta}\right)+2r_{0}e^{\delta}\left(\frac{1-(2\delta)^{n}}{1-2\delta}\right)\right]\stackrel{{\scriptstyle({\bf
IV})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_de<de_star_conds}})}}}{{\leq}}r_{0}e^{\delta}.$

Consequently we have that $\sqrt{2\delta}E_{n-(t-1)}\leq\delta
E_{n-(t-1)}+r_{0}e^{\delta}$ as claimed.

Returning our attention to (7.52), the inequality
$\sqrt{2\delta}E_{n-(t-1)}\leq\delta E_{n-(t-1)}+r_{0}e^{\delta}$ means that

$\displaystyle E_{n-t}$
$\displaystyle=\left(1+\sqrt{2\delta}\right)\left(\delta
E_{n-(t-1)}+r_{0}e^{\delta}\right)$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_r0>0_En-(t-1)}})}}}{{=}}\left(1+\sqrt{2\delta}\right)\left(\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t-1}\left(\delta^{\rho-n+t}A+r_{0}\delta^{t}e^{\delta}\right)+\delta{{\bf{X}}}_{t-1}(\delta)+r_{0}e^{\delta}\right)$
$\displaystyle=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t}\left(\delta^{\rho-n+t}A+r_{0}\delta^{t}e^{\delta}\right)+\left(1+\sqrt{2\delta}\right)\left(r_{0}e^{\delta}+\delta{{\bf{X}}}_{t-1}(\delta)\right).$

We observe that

$\delta\left(1+\sqrt{2\delta}\right){{\bf{X}}}_{t-1}(\delta)\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_norm_control_r0>0_bX_t}})}}}{{=}}\left\\{\begin{array}[]{ll}0&\mbox{if
}t-1=0\\\
\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}\sum_{j=1}^{t}\delta^{j}\left(1+\sqrt{2\delta}\right)^{j}&\mbox{if
}t-1\geq 1.\end{array}\right.$ (7.53)

Via (7.53) we see that

$\displaystyle\left(1+\sqrt{2\delta}\right)\left(r_{0}e^{\delta}+\delta{{\bf{X}}}_{t-1}(\delta)\right)$
$\displaystyle=\left\\{\begin{array}[]{ll}\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}&\mbox{if
}t-1=0\\\
\left(1+\sqrt{2\delta}\right)r_{0}e^{\delta}\sum_{j=0}^{t}\delta^{j}\left(1+\sqrt{2\delta}\right)^{j}&\mbox{if
}t-1\geq 1\end{array}\right.$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_norm_control_r0>0_bX_t}})}}}{{=}}\left\\{\begin{array}[]{ll}{{\bf{X}}}_{1}(\delta)&\mbox{if
}t=1\\\ {{\bf{X}}}_{t}(\delta)&\mbox{if }t\geq 2\end{array}\right.$
$\displaystyle={{\bf{X}}}_{t}(\delta).$

Therefore we have established that

$E_{n-t}=\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{t}\left(\delta^{\rho-n+t}A+r_{0}\delta^{t}e^{\delta}\right)+{{\bf{X}}}_{t}(\delta)$
(7.54)

which is the estimate claimed in (7.45) for $t$. Induction now allows us to
conclude that the estimate (7.45) is valid for every $t\in\\{0,\ldots,n-1\\}$
as claimed.

To conclude, recall that $\theta\in(q,q+1]$ and
$b_{q}:=n-(q+1)\in\\{0,\ldots,n-1\\}$. Then (7.39) yields that

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\theta,\Omega,W)}\leq\max\left\\{(2\delta)^{q+1-\theta}{\cal
E}~{},~{}\min\left\\{{\cal E}~{},~{}\delta{\cal
E}+r_{0}e^{\delta}\right\\}\right\\}$ (7.55)

where, via (7.45) for $t:=b_{q}$, ${\cal E}={\cal
E}(A,r_{0},\rho,\theta,\delta):=E_{n-b_{q}}$, i.e.

${\cal
E}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_control_r0>0_E_n-t}})}}}{{=}}\left(1+(2\delta)^{\frac{\rho-n}{2}}\right)\left(1+\sqrt{2\delta}\right)^{n-(q+1)}\left(\delta^{\rho-(q+1)}A+r_{0}\delta^{n-(q+1)}e^{\delta}\right)+{{\bf{X}}}_{n-(q+1)}(\delta)$
(7.56)

as claimed in (7.20) and (7.21). This completes the proof of Lemma 7.3. ∎

##  8 Proof of the Pointwise Lipschitz Sandwich Theorem

In this section we use the local pointwise Lipschitz estimates established in
Lemma 7.1 from Section 7 to establish the _Pointwise Lipschitz Sandwich_
Theorem 3.11

###### Proof of Theorem 3.11.

Assume that V and W are Banach spaces and that the tensor powers of V are all
equipped with admissible norms (cf. Definition 2.1). Assume $\Sigma\subset V$
is a closed subset. Let $\varepsilon,\gamma>0$ with $k\in{\mathbb{Z}}_{\geq
0}$ such that $\gamma\in(k,k+1]$, $(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq
0}\times{\mathbb{R}}_{\geq 0}\right)\setminus\\{(0,0)\\}$, and
$0\leq\varepsilon_{0}<\min\left\\{K_{1}+K_{2},\varepsilon\right\\}$. To ease
notation, we set $K_{0}:=K_{1}+K_{2}>0$ for the remainder of the proof. Let
$l\in\\{0,\ldots,k\\}$ and define
$\delta_{0}=\delta_{0}(\varepsilon,\varepsilon_{0},K_{0}=K_{1}+K_{2},\gamma,l)>0$
by

$\delta_{0}:=\sup\left\\{\theta>0~{}:~{}K_{0}\theta^{\gamma-l}+\varepsilon_{0}e^{\theta}\leq\min\left\\{K_{0},\varepsilon\right\\}\right\\}.$
(8.1)

A first consequence of (8.1) is that $\delta_{0}\leq 1$, and so for every
$s\in\\{0,\dots,l\\}$ we have that
$\delta_{0}^{\gamma-s}\leq\delta_{0}^{\gamma-l}$. A second consequence of
(8.1) is that

$K_{0}\delta_{0}^{\gamma-l}+\varepsilon_{0}e^{\delta_{0}}\leq\min\left\\{K_{0},\varepsilon\right\\}\leq\varepsilon.$
(8.2)

Now assume that $B\subset\Sigma$ is a $\delta_{0}$-cover of $\Sigma$ in the
sense that

$\Sigma\subset\bigcup_{x\in
B}{\overline{\mathbb{B}}}_{V}(x,\delta_{0})=B_{\delta_{0}}:=\left\\{v\in
V~{}:~{}\exists z\in B\text{ such that }||z-v||_{V}\leq\delta_{0}\right\\}.$
(8.3)

Suppose
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right),\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
satisfy the ${\mathrm{Lip}}(\gamma,\Sigma,W)$ norm estimates
$||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$ and
$||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Further suppose
that for every $j\in\\{0,\ldots,k\\}$ and every $x\in B$ the difference
$\psi^{(j)}(x)-\varphi^{(j)}(x)\in{\cal L}(V^{\otimes j};W)$ satisfies the
bound

$\left|\left|\psi^{(j)}(x)-\varphi^{(j)}(x)\right|\right|_{{\cal L}(V^{\otimes
j};W)}\leq\varepsilon_{0}.$ (8.4)

Define $F\in{\mathrm{Lip}}(\gamma,\Sigma,W)$ by $F:=\psi-\varphi$ so that for
every $j\in\\{0,\ldots,k\\}$ we have $F^{(j)}:=\psi^{(j)}-\varphi^{(j)}$. Then
$||F||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{0}:=K_{1}+K_{2}$ and, for
every integer $j\in\\{0,\ldots,k\\}$ and every point $x\in B$, (8.4) gives
that $\left|\left|F^{(j)}(x)\right|\right|_{{\cal L}(V^{\otimes
j};W)}\leq\varepsilon_{0}$.

Now fix $x\in\Sigma$ and $s\in\\{0,\ldots,l\\}$. From (8.3) we conclude that
there exists a point $z\in B$ with $||z-x||_{V}\leq\delta_{0}$. Then apply
Lemma 7.1, with $A:=K_{0}(=K_{1}+K_{2})$, $r_{0}:=\varepsilon_{0}$,
$\rho:=\gamma$, $p:=z$, $n:=k$ and $q:=k$, to conclude that (cf. (7.3))

$\left|\left|F^{(s)}(x)\right|\right|_{{\cal L}(V^{\otimes
s};W)}\leq\min\left\\{K_{0},K_{0}\delta_{0}^{\gamma-s}+\varepsilon_{0}\sum_{j=0}^{k-s}\frac{1}{j!}\delta_{0}^{j}\right\\}\leq\min\left\\{K_{0},K_{0}\delta_{0}^{\gamma-l}+\varepsilon_{0}e^{\delta_{0}}\right\\}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_C0_thm_de0_prop}})}}}{{\leq}}\varepsilon$
(8.5)

where we have used that $\delta_{0}^{\gamma-s}\leq\delta_{0}^{\gamma-l}$.
Since $F=\psi-\varphi$, the arbitrariness of $s\in\\{0,\ldots,l\\}$ and of
$x\in\Sigma$ ensure that (8.5) gives the bounds claimed in (3.9) This
completes the proof of Theorem 3.11. ∎

##  9 Proof of the Single-Point Lipschitz Sandwich Theorem

In this section we prove the _Single-Point Lipschitz Sandwich_ Theorem 3.8.
Our approach is to alter the constant $\delta_{0}$ appearing in the _Pointwise
Lipschitz Sandwich_ Theorem 3.11 in order to strengthen the conclusions to an
estimate on the full ${\mathrm{Lip}}(\eta)$-norm of the difference.

To be more precise, recall that $\Sigma\subset V$ is closed and $\gamma>0$
with $k\in{\mathbb{Z}}_{\geq 0}$ such that $\gamma\in(k,k+1]$. Let
$\eta\in(0,\gamma)$, $\varepsilon>0$,
$(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq 0}\times{\mathbb{R}}_{\geq
0}\right)\setminus\\{(0,0)\\}$, and
$0\leq\varepsilon_{0}<\min\left\\{K_{1}+K_{2},\varepsilon\right\\}$. Retrieve
the constant
$\delta_{0}=\delta_{0}(K_{1}+K_{2},\varepsilon,\varepsilon_{0},\gamma)>0$
arising in Theorem 3.11 for the choice $l:=k$. Given a point $p\in\Sigma$,
define $\Omega:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,\delta_{0})$.

Suppose
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W),\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
satisfy the norm bounds $||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$
and $||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Further suppose
that for every $j\in\\{0,\ldots,k\\}$ the difference
$\psi^{(j)}(p)-\varphi^{(j)}(p)\in{\cal L}(V^{\otimes j};W)$ satisfies
$\left|\left|\psi^{(j)}(p)-\varphi^{(j)}(p)\right|\right|_{{\cal L}(V^{\otimes
j};W)}\leq\varepsilon_{0}$. Then by applying Theorem 3.11, for the choices
$l$, $\Sigma$ and $B$ there as $k$, $\Omega$ and $\\{p\\}$ here respectively,
we may conclude that for every $s\in\\{0,\ldots,k\\}$ and every $x\in\Omega$
we have $\left|\left|\psi^{(s)}(x)-\varphi^{(s)}(x)\right|\right|_{{\cal
L}(V^{\otimes s};W)}\leq\varepsilon$.

We will prove the _Single-Point Lipschitz Sandwich Theorem_ 3.8, by
establishing that after reducing the constant $\delta_{0}$, allowing it to
additionally depend on $\eta$, we may strengthen these pointwise bounds into a
bound on the full ${\mathrm{Lip}}(\eta,\Omega,W)$ norm of $\psi-\varphi$. We
do so by appealing to the local Lipschitz estimates established in Lemmas 7.2
and 7.3 in Section 7.

There is a natural dichotomy within this strategy between the case that
$\eta\in(k,\gamma)$ and the case that $\eta\in(0,k]$. We first use Lemma 7.2
to establish the _Single-Point Lipschitz Sandwich_ Theorem 3.8 in the simpler
case that $\eta\in(k,\gamma)$.

###### Proof of Theorem 3.8 when $\eta\in(k,\gamma)$.

Assume that V and W are Banach spaces and that the tensor powers of V are all
equipped with admissible norms (cf. Definition 2.1). Let $\Sigma\subset V$ be
non-empty and closed. Let $\varepsilon,\gamma>0$ with $k\in{\mathbb{Z}}_{\geq
0}$ such that $\gamma\in(k,k+1]$, $(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq
0}\times{\mathbb{R}}_{\geq 0}\right)\setminus\\{(0,0)\\}$, and
$0\leq\varepsilon_{0}<\min\left\\{K_{1}+K_{2},\varepsilon\right\\}$, and
$\eta\in(k,\gamma)$. For notational convenience we let $K_{0}:=K_{1}+K_{2}$.
With a view to later applying Lemma 7.2, define
$\delta_{0}=\delta_{0}(K_{0},\varepsilon,\varepsilon_{0},\gamma)>0$ by

$\delta_{0}:=\sup\left\\{\theta\in(0,1]~{}:~{}K_{0}(2\theta)^{\gamma-\eta}\leq\min\left\\{K_{0},\varepsilon\right\\}\text{
and
}K_{0}\theta^{\gamma-k}+\varepsilon_{0}e^{\theta}\leq\min\left\\{K_{0},\varepsilon\right\\}\right\\}>0.$
(9.1)

It initially appears that $\delta_{0}$ additionally depends on $k$. However,
$k$ is determined by $\gamma$, thus any dependence on $k$ is really dependence
on $\gamma$. We now fix the value of $\delta_{0}>0$ for the remainder of the
proof. We record that (9.1) ensures that $\delta_{0}\leq 1$ and

$({\bf I})\quad
K_{0}(2\delta_{0})^{\gamma-\eta}\leq\min\left\\{K_{0},\varepsilon\right\\}\qquad\text{and}\qquad({\bf
II})\quad
K_{0}\delta_{0}^{\gamma-k}+\varepsilon_{0}e^{\delta_{0}}\leq\min\left\\{K_{0},\varepsilon\right\\}.$
(9.2)

Now assume that $p\in\Sigma$ and that
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right)$ and
$\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)$ are elements in
${\mathrm{Lip}}(\gamma,\Sigma,W)$ with
$||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$ and
$||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Further suppose
that for every $l\in\\{0,\ldots,k\\}$ the difference
$\psi^{(l)}(p)-\varphi^{(l)}(p)\in{\cal L}(V^{\otimes l};W)$ satisfies the
bound

$\left|\left|\psi^{(l)}(p)-\varphi^{(l)}(p)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\varepsilon_{0}.$ (9.3)

Define $\Omega:={\overline{\mathbb{B}}}_{V}(p,\delta_{0})~{}\cap~{}\Sigma$ and
$F\in{\mathrm{Lip}}(\gamma,\Sigma,W)$ by $F:=\psi-\varphi$, so that for every
$j\in\\{0,\ldots,k\\}$ we have $F^{(j)}=\psi^{(j)}-\varphi^{(j)}$. We apply
Lemma 7.2 to $F$, with $A:=K_{0}=K_{1}+K_{2}$, $r_{0}:=\varepsilon_{0}$,
$\rho:=\gamma$, $n:=k$, $z:=p$, $\theta:=\eta$ and $\delta:=\delta_{0}$, to
conclude both that $F\in{\mathrm{Lip}}(\eta,\Omega,W)$ and (cf. (7.11)) that

$\displaystyle||F||_{{\mathrm{Lip}}(\eta,\Omega,W)}\leq\max\left\\{K_{0}(2\delta_{0})^{\gamma-\eta}~{},~{}\min\left\\{K_{0}~{},~{}K_{0}\delta_{0}^{\gamma-k}+\varepsilon_{0}e^{\delta_{0}}\right\\}\right\\}$
$\displaystyle\stackrel{{\scriptstyle({\bf II})\text{ of
}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_easy_proof_de0_def_B}})}}}{{\leq}}\max\left\\{K_{0}(2\delta_{0})^{\gamma-\eta}~{},~{}\varepsilon\right\\}$
$\displaystyle\stackrel{{\scriptstyle({\bf I})\text{ of
}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_easy_proof_de0_def_B}})}}}{{\leq}}\max\left\\{\varepsilon~{},~{}\varepsilon\right\\}=\varepsilon.$

Since $F=\psi-\varphi$ and
$\Omega:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,\delta_{0})$, this is
precisely the estimate claimed in (3.5). This completes the proof of Theorem
3.8 for $\eta\in(k,\gamma)$. ∎

We now turn our attention to using Lemma 7.3 to establish the _Single-Point
Lipschitz Sandwich_ Theorem 3.8 in the more challenging case that
$\eta\in(0,k]$.

###### Proof of Theorem 3.8 for $0<\eta\leq k$.

Assume that V and W are Banach spaces and that the tensor powers of V are all
equipped with admissible norms (cf. Definition 2.1). Let $\Sigma\subset V$ be
closed and non-empty. Let $\varepsilon,\gamma>0$ with $k\in{\mathbb{Z}}_{\geq
1}$ such that $\gamma\in(k,k+1]$, $(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq
0}\times{\mathbb{R}}_{\geq 0}\right)\setminus\\{(0,0)\\}$, and $\eta\in(0,k]$.
Observe that this requires $1\leq k<\gamma$. Let $q\in\\{0,\ldots,k-1\\}$ such
that $\eta\in(q,q+1]\subset(0,k]$. Finally let
$0\leq\varepsilon_{0}<\min\left\\{K_{1}+K_{2},\varepsilon\right\\}$. For
notational convenience, we set $K_{0}:=K_{1}+K_{2}>0$ for the remainder of the
proof

Our strategy is to establish the desired ${\mathrm{Lip}}(\eta)$-norm bounds
via an application of Lemma 7.3. For this purpose we retrieve the constant
$\delta_{\ast}$ arising in Lemma 7.3 for $A:=K_{0}=K_{1}+K_{2}$,
$r_{0}:=\varepsilon_{0}$, $\rho:=\gamma$ and $\theta:=\eta$. Note that we are
not actually applying Lemma 7.3, but simply retrieving a constant in
preparation for its future application.

Let $\delta_{0}:=\min\left\\{1~{},~{}\delta_{\ast}\right\\}>0$, which depends
only on $K_{0}=K_{1}+K_{2}$, $\varepsilon_{0}$, $\gamma$ and $\eta$. In order
to ensure that applying Lemma 7.3 yields the desired
${\mathrm{Lip}}(\eta)$-norm estimate, we will allow ourselves to (potentially)
further reduce $\delta_{0}$, additionally now depending on $\varepsilon$. With
the benefit of hindsight, it will suffice to alter $\delta_{0}$ to ensure that

$\left\\{\begin{aligned} &({\bf
A})\quad\max\left\\{1+\left(2\delta_{0}\right)^{\frac{\gamma-k}{2}}~{},~{}1+\sqrt{2\delta_{0}}\right\\}<2,\quad\left(\text{In
particular, }2\delta_{0}<1\right),\\\ &({\bf
B})\quad\left(2\delta_{0}\right)^{\frac{\gamma-\eta}{2}+\frac{q+1-\eta}{2}}\leq\frac{\varepsilon}{2^{k-q}K_{0}},\\\
&({\bf
C})\quad\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\left(\delta_{0}^{\gamma-k}K_{0}+\varepsilon_{0}e^{\delta_{0}}\right)\leq\varepsilon,\\\
&({\bf D})\quad
2^{k-q}\left(\delta_{0}^{\gamma-k}K_{0}+\varepsilon_{0}\delta_{0}e^{\delta_{0}}\right)+\left(\frac{1+\sqrt{2\delta_{0}}}{1-2\delta_{0}}\right)\varepsilon_{0}e^{\delta_{0}}\leq\varepsilon,\quad\text{and}\\\
&({\bf
E})\quad\varepsilon_{0}e^{\delta_{0}}\leq\left(1-\delta_{0}\right)\varepsilon.\end{aligned}\right.$
(9.4)

We now fix the value of
$\delta_{0}=\delta_{0}(K_{0},\gamma,\eta,\varepsilon_{0},\varepsilon)>0$ for
the remainder of the proof. Since $K_{0}:=K_{1}+K_{2}$ we note that
$\delta_{0}$ has the claimed dependencies.

Now let $p\in\Sigma$ and assume that
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right)$ and
$\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)$ are in
${\mathrm{Lip}}(\gamma,\Sigma,W)$ with
$||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$ and
$||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Suppose that for
every integer $l\in\\{0,\ldots,k\\}$ the difference
$\psi^{(l)}(p)-\varphi^{(l)}(p)\in{\cal L}(V^{\otimes l};W)$ satisfies the
bound

$\left|\left|\psi^{(l)}(p)-\varphi^{(l)}(p)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\varepsilon_{0}.$ (9.5)

Define $\Omega:={\overline{\mathbb{B}}}_{V}(p,\delta_{0})~{}\cap~{}\Sigma$ and
$F\in{\mathrm{Lip}}(\gamma,\Sigma,W)$ by $F:=\psi-\varphi$ so that for every
$j\in\\{0,\ldots,k\\}$ we have $F^{(j)}=\psi^{(j)}-\varphi^{(j)}$.

We begin with the case that $\varepsilon_{0}=0$. Since $\delta_{0}\leq 1$, the
bounds (9.5) allow us to apply Lemma 7.3 to $F$, with
$A:=K_{0}(=K_{1}+K_{2})$, $r_{0}:=\varepsilon_{0}$, $\rho:=\gamma$,
$\theta:=\eta$ and $\delta:=\delta_{0}$, to conclude that (cf. (7.19))

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}\leq\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\left(1+\sqrt{2\delta_{0}}\right)^{k-(q+1)}(2\delta_{0})^{\frac{\gamma-\eta}{2}+\frac{q+1-\eta}{2}}K_{0}.$
(9.6)

We compute that

$\displaystyle\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_ep0=0_est_A}})}}}{{\leq}}\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\left(1+\sqrt{2\delta_{0}}\right)^{k-(q+1)}(2\delta_{0})^{\frac{\gamma-\eta}{2}+\frac{q+1-\eta}{2}}K_{0}$
$\displaystyle\stackrel{{\scriptstyle({\bf
A})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_de0}})}}}{{\leq}}2^{k-q}(2\delta_{0})^{\frac{\gamma-\eta}{2}+\frac{q+1-\eta}{2}}K_{0}\stackrel{{\scriptstyle({\bf
B})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_de0}})}}}{{\leq}}2^{k-q}\frac{\varepsilon}{2^{k-q}K_{0}}K_{0}=\varepsilon.$

Recalling that
$\Omega:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,\delta_{0})$ and
$F:=\psi-\varphi$, this is precisely the estimate claimed in (3.5), and our
proof is complete for the case that $\varepsilon_{0}=0$.

Now consider the case that $\varepsilon_{0}>0$. Recalling how we chose
$\delta_{0}$, the bounds (9.5) allow us to apply Lemma 7.3 to $F$, with
$A:=K_{0}(=K_{1}+K_{2})$, $r_{0}:=\varepsilon_{0}$, $\rho:=\gamma$,
$\theta:=\eta$ and $\delta:=\delta_{0}$, to conclude via (7.20) that

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}\leq\max\left\\{(2\delta_{0})^{q+1-\eta}{\cal
E}~{},~{}\min\left\\{{\cal E}~{},~{}\delta_{0}{\cal
E}+\varepsilon_{0}e^{\delta_{0}}\right\\}\right\\}$ (9.7)

for ${\cal E}={\cal E}(K_{0},\gamma,\eta,\varepsilon_{0})>0$ defined by (cf.
(7.21))

${\cal
E}:=\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\left(1+\sqrt{2\delta_{0}}\right)^{k-(q+1)}\left(\delta_{0}^{\gamma-(q+1)}K_{0}+\varepsilon_{0}\delta_{0}^{k-(q+1)}e^{\delta_{0}}\right)+{{\bf{X}}}_{k-(q+1)}(\delta_{0})$
(9.8)

where, for $t\in\\{0,\ldots,k-1\\}$, the quantity ${{\bf{X}}}_{t}(\delta)$ is
defined by (cf. (7.22))

${{\bf{X}}}_{t}(\delta):=\left\\{\begin{array}[]{ll}0&\mbox{if }t=0\\\
\left(1+\sqrt{2\delta_{0}}\right)\varepsilon_{0}e^{\delta_{0}}\sum_{j=0}^{t-1}\delta_{0}^{j}\left(1+\sqrt{2\delta_{0}}\right)^{j}&\mbox{if
}t\geq 1.\end{array}\right.$ (9.9)

We first prove that $\left(1-\delta_{0}\right){\cal
E}\geq\varepsilon_{0}e^{\delta_{0}}$.

If $k=q+1$, then (9.8) ensures that $\left(1-\delta_{0}\right){\cal
E}\geq\left(1-\delta_{0}\right)\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\varepsilon_{0}e^{\delta_{0}}$.
If we are able to conclude that
$\left(1-\delta_{0}\right)\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\geq
1$ then our desired estimate $\left(1-\delta_{0}\right){\cal
E}\geq\varepsilon_{0}e^{\delta_{0}}$ is true. The required lower bound
$\left(1-\delta_{0}\right)\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\geq
1$ is equivalent to
$(2\delta_{0})^{\frac{\gamma-k}{2}}-\delta_{0}-\delta_{0}(2\delta_{0})^{\frac{\gamma-k}{2}}\geq
0$. A consequence of (A ) in (9.4) is that
$(2\delta_{0})^{\frac{\gamma-k}{2}}<1$. This tells us that
$\delta_{0}+\delta_{0}(2\delta_{0})^{\frac{\gamma-k}{2}}\leq
2\delta_{0}\leq(2\delta_{0})^{\frac{\gamma-k}{2}}$ where the latter inequality
is true since $2\delta_{0}<1$ and $\frac{\gamma-k}{2}<1$. Hence
$\left(1-\delta_{0}\right)\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\geq
1$ and so we have $\left(1-\delta_{0}\right){\cal
E}\geq\varepsilon_{0}e^{\delta_{0}}$ as required.

If $k>q+1$, then (9.8) and (9.9) yield $\left(1-\delta_{0}\right){\cal
E}\geq\left(1-\delta_{0}\right){{\bf{X}}}_{k-(q+1)}(\delta_{0})\geq\left(1-\delta_{0}\right)\left(1+\sqrt{2\delta_{0}}\right)\varepsilon_{0}e^{\delta_{0}}$.
If we are able to conclude that
$\left(1-\delta_{0}\right)\left(1+\sqrt{2\delta_{0}}\right)\geq 1$ then our
desired estimate $\left(1-\delta_{0}\right){\cal
E}\geq\varepsilon_{0}e^{\delta_{0}}$ is true. The required lower bound
$\left(1-\delta_{0}\right)\left(1+\sqrt{2\delta_{0}}\right)\geq 1$ is
equivalent to $\sqrt{2\delta_{0}}-\delta_{0}-\delta_{0}\sqrt{2\delta_{0}}\geq
0$. A consequence of (A ) in (9.4) is that $\sqrt{2\delta_{0}}<1$. This tells
us that $\delta_{0}+\delta_{0}\sqrt{2\delta_{0}}\leq
2\delta_{0}\leq\sqrt{2\delta_{0}}$ where the latter inequality is true since
$2\delta_{0}<1$. Hence
$\left(1-\delta_{0}\right)\left(1+\sqrt{2\delta_{0}}\right)\geq 1$ and so we
have $\left(1-\delta_{0}\right){\cal E}\geq\varepsilon_{0}e^{\delta_{0}}$ as
required.

Having established that $\left(1-\delta_{0}\right){\cal
E}\geq\varepsilon_{0}e^{\delta_{0}}$ we observe that (9.7) becomes

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}\leq\max\left\\{(2\delta_{0})^{q+1-\eta}{\cal
E}~{},~{}\delta_{0}{\cal E}+\varepsilon_{0}e^{\delta_{0}}\right\\}$ (9.10)

We now prove the upper bound for ${\cal E}$ that ${\cal E}\leq\varepsilon$.
For this purpose note that when $k=q+1$ (9.8) yields that

${\cal
E}=\left(1+(2\delta_{0})^{\frac{\gamma-k}{2}}\right)\left(\delta_{0}^{\gamma-k}K_{0}+\varepsilon_{0}e^{\delta_{0}}\right)\stackrel{{\scriptstyle({\bf
C})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_de0}})}}}{{\leq}}\varepsilon$
(9.11)

since ${{\bf{X}}}_{0}(\delta_{0})=0$ from (9.9). If, however, $k>q+1$ then
$k-(q+1)\geq 1$ and so, recalling that (A ) in (9.4) ensures that that
$\delta_{0}(1+\sqrt{2\delta_{0}})<2\delta_{0}<1$, we have

${{\bf{X}}}_{k-(q+1)}(\delta_{0})\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_r0>0_bX_t}})}}}{{=}}\left(1+\sqrt{2\delta_{0}}\right)\varepsilon_{0}e^{\delta_{0}}\sum_{j=0}^{k-(q+1)}\delta_{0}^{j}\left(1+\sqrt{2\delta_{0}}\right)^{j}\leq\frac{\left(1+\sqrt{2\delta_{0}}\right)}{1-2\delta_{0}}\varepsilon_{0}e^{\delta_{0}}.$
(9.12)

Moreover, $2\delta_{0}<1$ ensures that
$\delta_{0}^{\gamma-(q+1)}\leq\delta_{0}^{\gamma-k}$ and
$\delta_{0}^{k-(q+1)}\leq\delta_{0}$. Hence we can combine (9.8) and (9.12) to
obtain that

${\cal E}\leq
2^{k-q}\left(\delta_{0}^{\gamma-k}K_{0}+\varepsilon_{0}\delta_{0}e^{\delta_{0}}\right)+\frac{\left(1+\sqrt{2\delta_{0}}\right)}{1-2\delta_{0}}\varepsilon_{0}e^{\delta_{0}}\stackrel{{\scriptstyle({\bf
D})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_de0}})}}}{{\leq}}\varepsilon.$

Therefore in both the case that $k=q+1$ and the case that $k>q+1$ we obtain
that

${\cal E}\leq\varepsilon.$ (9.13)

We complete the proof by using the upper bound in (9.13) to control
$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}$. Recalling
that (A ) in (9.4) means that $2\delta_{0}<1$, we have that

$({\bf I})\quad\left(2\delta_{0}\right)^{q+1-\eta}{\cal E}\leq{\cal
E}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_cal_e_ub}})}}}{{\leq}}\varepsilon\qquad\text{and}\qquad({\bf
II})\quad\delta_{0}{\cal
E}+\varepsilon_{0}e^{\delta_{0}}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_cal_e_ub}})}}}{{\leq}}\delta_{0}\varepsilon+\varepsilon_{0}e^{\delta_{0}}\stackrel{{\scriptstyle({\bf
E})~{}\text{in}~{}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimates_thm_de0}})}}}{{\leq}}\varepsilon.$
(9.14)

Thus

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega,W)}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_r0>0_est_B}})}}}{{\leq}}\max\left\\{(2\delta_{0})^{q+1-\eta}{\cal
E},\delta_{0}{\cal
E}+\varepsilon_{0}e^{\delta_{0}}\right\\}\stackrel{{\scriptstyle({\bf
I})\text{ in
}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_eta<q+1_A}})}}}{{\leq}}\max\left\\{\varepsilon,\delta_{0}{\cal
E}+\varepsilon_{0}e^{\delta_{0}}\right\\}\stackrel{{\scriptstyle({\bf
II})\text{ in
}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_ball_estimate_thm_eta<q+1_A}})}}}{{=}}\varepsilon.$
(9.15)

Recalling that
$\Omega:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,\delta_{0})$ and
$F:=\psi-\varphi$, (9.15) is precisely the estimate claimed in (3.5), and our
proof is complete for the case that $\varepsilon_{0}>0$. Having already
established the conclusion for the case that $\varepsilon_{0}=0$, this
completes the proof of Theorem 3.8 for the case that $\eta\in(0,k]$. ∎

##  10 Proof of the Lipschitz Sandwich Theorem 3.1

In this section we establish the full _Lipschitz Sandwich Theorem_ 3.1. Our
strategy to prove this result is to patch together the local Lipschitz bounds
achieved by the _Single-Point Lipschitz Sandwich_ Theorem 3.8 in a similar
spirit to the patching of local Lipschitz bounds in Lemma 1.16 in [Bou15]. We
do not necessarily have local Lipschitz bounds on a small ball centred at
_any_ point in $\Sigma$; we only have such estimates for points in the closed
subset $B\subset\Sigma$, and we do _not_ require that $B=\Sigma$.
Consequently, our patching is more complicated than the patching used in Lemma
1.16 in [Bou15].

To be more precise, recall that $\Sigma\subset V$ is closed and $\gamma>0$
with $k\in{\mathbb{Z}}_{\geq 0}$ such that $\gamma\in(k,k+1]$. Let
$\eta\in(0,\gamma)$, $\varepsilon>0$,
$(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq 0}\times{\mathbb{R}}_{\geq
0}\right)\setminus\\{(0,0)\\}$, and
$0\leq\varepsilon_{0}<\min\left\\{K_{1}+K_{2},\varepsilon\right\\}$. Retrieve
the constant
$\delta_{0}=\delta_{0}(K_{1}+K_{2},\varepsilon,\varepsilon_{0},\gamma,\eta)>0$
arising in the _Single-Point Lipschitz Sandwich Theorem_ 3.8. Assume that
$B\subset\Sigma$ is a $\delta_{0}$-cover of $\Sigma$ in the sense that the
$\delta_{0}$-fattening of $B$ contains $\Sigma$.

Suppose
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
and
$\varphi\in\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
both satisfy the norm bounds $||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq
K_{1}$ and $||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Further
suppose that for every $j\in\\{0,\ldots,k\\}$ and every $x\in B$ the
difference $\psi^{(j)}(x)-\varphi^{(j)}(x)\in{\cal L}(V^{\otimes j};W)$
satisfies $\left|\left|\psi^{(j)}(x)-\varphi^{(j)}(x)\right|\right|_{{\cal
L}(V^{\otimes j};W)}\leq\varepsilon_{0}$. Then given any point $p\in B$, we
can apply the _Single-Point Lipschitz Sandwich Theorem_ 3.8 to conclude that
$||\psi_{[q]}-\varphi_{[q]}||_{{\mathrm{Lip}}(\eta,\Omega_{p},W)}\leq\varepsilon$
for $\Omega_{p}:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,\delta_{0})$
and $q\in\\{0,\ldots,k\\}$ such that $\eta\in(q,q+1]$.

It may initially appear that since $\Sigma=\cup_{p\in B}\Omega_{p}$ these
local ${\mathrm{Lip}}(\eta)$-norm bounds should combine together to yield
$||\psi_{[q]}-\varphi_{[q]}||_{{\mathrm{Lip}}(\eta,\Sigma,W)}\leq\varepsilon$.
However, this is not necessarily true. For example, given any
$\alpha\in(0,1)$, consider the function
$F:[0,1]\cup[1+\alpha,2]\to{\mathbb{R}}$ defined by $F(x):=0$ if $x\in[0,1]$
and $F(x):=\alpha$ if $x\in[1+\alpha,2]$. Then
$F\in{\mathrm{Lip}}(1,[0,1]\cup[1+\alpha,2],{\mathbb{R}})$ and we have that
$||F||_{{\mathrm{Lip}}(1,[0,1],{\mathbb{R}})}=0$ and
$||F||_{{\mathrm{Lip}}(1,[1+\alpha,2],{\mathbb{R}})}=\alpha$. But
$|F(1+\alpha)-F(1)|=\alpha=|1+\alpha-1|$ and so
$||F||_{{\mathrm{Lip}}(1,[0,1]\cup[1+\alpha,2],{\mathbb{R}})}=1>\alpha$.

The main content of our proof of the _Lipschitz Sandwich Theorem_ 3.1 is to
overcome this problem. We prove that, by requiring the constant
$\varepsilon_{0}$ to be sufficiently small, depending only on
$\varepsilon,K_{1}+K_{2},\gamma$ and $\eta$, rather than an arbitrary real
number in the interval $[0,\min\left\\{K_{1}+K_{2},\varepsilon\right\\})$, we
_can_ patch together local Lipschitz estimates resulting from an application
of the _Single-Point Lipschitz Sandwich Theorem_ 3.8 to yield global Lipschitz
estimates throughout $\Sigma$. A key point is to ensure that the sets
$\Omega_{p}$ on which the _Single-Point Lipschitz Sandwich Theorem_ 3.8 yields
local Lipschitz estimates are not pairwie disjoint; that is, for each $p\in B$
there must be some $q\in B\setminus\\{p\\}$ such that the intersection
$\Omega_{p}\cap\Omega_{q}$ is non-empty.

###### Proof of Theorem 3.1.

Let $V$ and $W$ be Banach spaces, and assume that the tensor powers of $V$ are
all equipped with admissible norms (cf. Definition 2.1). Assume that
$\Sigma\subset V$ is non-empty and closed. Let $\varepsilon,\gamma>0$ with
$k\in{\mathbb{Z}}_{\geq 0}$ such that $\gamma\in(k,k+1]$, and
$(K_{1},K_{2})\in\left({\mathbb{R}}_{\geq 0}\times{\mathbb{R}}_{\geq
0}\right)\setminus\\{(0,0)\\}$. Further let $\eta\in(0,\gamma)$ with
$q\in\\{0,\ldots,k\\}$ such that $\eta\in(q,q+1]$. It suffices to prove the
theorem under the additional assumption that $\varepsilon\leq
K_{0}:=K_{1}+K_{2}$; the conclusion (3.3) being valid for $\varepsilon$
immediately means it is also valid for any constant
$\varepsilon^{\prime}\geq\varepsilon$.

Define $\theta:=\frac{1}{2(1+e)}>0$ and retrieve the constant $\delta_{0}>0$
arising from Theorem 3.8 for the same constants $K_{1}$, $K_{2}$, $\gamma$ and
$\eta$ as here respectively, and with the choices of $\theta\varepsilon$ and
$\frac{\theta}{2}\varepsilon$ here as the constants $\varepsilon$ and
$\varepsilon_{0}$ in Theorem 3.8 respectively. Note that we are not actually
applying Theorem 3.8, but simply retrieving a constant in preparation for its
future application. Examining the dependencies in Theorem 3.8 reveals that
$\delta_{0}>0$ depends only on $\varepsilon$, $K_{0}$, $\gamma$ and $\eta$. If
necessary, we reduce $\delta_{0}$, without additional dependencies, so that
$\delta_{0}\leq 1$. Finally, we replace the resulting constant $\delta_{0}$ by
$\delta_{0}/2$.

Our choice of $\delta_{0}>0$ means that if
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right)~{},~{}\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
with $||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$ and
$||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$, and if for a point
$p\in\Sigma$ and every $l\in\\{0,\ldots,k\\}$ we have the estimate
$\left|\left|\psi^{(l)}(p)-\varphi^{(l)}(p)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\frac{\theta}{2}\varepsilon$, then an application of Theorem 3.8
would allow us to conclude the estimate that
$\left|\left|\psi_{[q]}-\varphi_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega_{p},W)}\leq\theta\varepsilon$
for $\Omega_{p}:={\overline{\mathbb{B}}}_{V}(p,2\delta_{0})~{}\cap~{}\Sigma$.

We now fix the value of $\delta_{0}>0$ for the remainder of the proof. Having
done so, we define $\varepsilon_{0}>0$ by

$\varepsilon_{0}:=\min\left\\{\theta~{},~{}\frac{\delta_{0}^{\eta}}{e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)}\right\\}\frac{\varepsilon}{2}>0.$
(10.1)

Examining the dependencies in (10.1) reveals that $\varepsilon_{0}$ depends
only on $\varepsilon$, $K_{0}=K_{1}+K_{2}$, $\gamma$ and $\eta$. We may now
fix the value of $\varepsilon_{0}>0$ for the the remainder of the proof.

Let $B\subset\Sigma$ satisfy that

$\Sigma\subset\bigcup_{x\in B}{\overline{\mathbb{B}}}_{V}(x,\delta_{0}).$
(10.2)

Suppose
$\psi=\left(\psi^{(0)},\ldots,\psi^{(k)}\right),\varphi=\left(\varphi^{(0)},\ldots,\varphi^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
satisfy the ${\mathrm{Lip}}(\gamma,\Sigma,W)$ norm estimates
$||\psi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{1}$ and
$||\varphi||_{{\mathrm{Lip}}(\gamma,\Sigma,W)}\leq K_{2}$. Further assume that
whenever $l\in\\{0,\ldots,k\\}$ and $x\in B$ we have the estimate
$\left|\left|\psi^{(l)}(x)-\varphi^{(l)}(x)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\varepsilon_{0}$. Let $p\in B$. Recalling how we chose the constant
$\delta_{0}>0$ and that (10.1) means that
$\varepsilon_{0}\leq\frac{\theta}{2}\varepsilon$, we may appeal to Theorem 3.8
to conclude that

$\left|\left|\psi_{[q]}-\varphi_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega_{p},W)}\leq\theta\varepsilon$
(10.3)

where
$\Omega_{p}:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(p,2\delta_{0})$. The
arbitrariness of $p\in B$ allows us to conclude that the estimate (10.3) is
valid for every $p\in B$.

We complete the proof of Theorem 3.1 by establishing that having the bounds
(10.3) for every $p\in B$ allows us to conclude that
$\left|\left|\psi_{[q]}-\varphi_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Sigma,W)}\leq\varepsilon$.
This is proven in the following claim.

######  Claim 10.1.

If $F=\left(F^{(0)},\ldots,F^{(k)}\right)\in{\mathrm{Lip}}(\gamma,\Sigma,W)$
satisfies, for every $l\in\\{0,\ldots,k\\}$ and every $z\in B$, that
$\left|\left|F^{(l)}(z)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\varepsilon_{0}$ and
$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega_{z},W)}\leq\theta\varepsilon$,
where
$\Omega_{z}:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$, then
we have

$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Sigma,W)}\leq\varepsilon.$
(10.4)

###### Proof of Claim 10.1.

For each $l\in\\{0,\ldots,k\\}$ let $R^{F}_{l}:\Sigma\times\Sigma\to{\cal
L}(V^{\otimes l};W)$ denote the remainder term associated to $F^{(l)}$.
Therefore whenever $l\in\\{0,\ldots,k\\}$, $x,y\in\Sigma$ and $v\in V^{\otimes
l}$, we have that (cf. (2.9))

$R^{F}_{l}(x,y)[v]:=F^{(l)}(y)[v]-\sum_{s=0}^{k-l}\frac{1}{s!}F^{(l+s)}(x)\left[v\otimes(y-x)^{\otimes
j}\right].$ (10.5)

If $q=k$ then we may work with the unaltered remainder terms defined in
(10.5). But if $q<k$ then we must first appropriately alter the remainder
terms. For this purpose, for each $l\in\\{0,\ldots,q\\}$ we define
$\hat{R}^{F}_{l}:\Sigma\times\Sigma\to{\cal L}(V^{\otimes l};W)$ for
$x,y\in\Sigma$ and $v\in V^{\otimes l}$ by

$\hat{R}^{F}_{l}(x,y)[v]:=\left\\{\begin{array}[]{ll}R^{F}_{l}(x,y)[v]&\mbox{if
}q=k\\\
R^{F}_{l}(x,y)[v]+\sum_{s=q+1-l}^{k-l}\frac{1}{s!}F^{(l+s)}(x)\left[v\otimes(y-x)^{\otimes
s}\right]&\mbox{if }q<k.\end{array}\right.$ (10.6)

It follows from (10.5) and (10.6) that whenever $l\in\\{0,\ldots,q\\}$,
$x,y\in\Sigma$ and $v\in V^{\otimes l}$ we have

$F^{(l)}(y)[v]=\sum_{s=0}^{q-l}F^{(l+s)}(x)\left[v\otimes(y-x)^{\otimes
s}\right]+\hat{R}^{F}_{l}(x,y)[v].$ (10.7)

For each $z\in B$, the assumption that
$||F_{[q]}||_{{\mathrm{Lip}}(\eta,\Omega_{z},W)}\leq\theta\varepsilon$ for
$\Omega_{z}:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$ tells
us that for every $l\in\\{0,\ldots,q\\}$ and any
$x,y\in\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$ we have

$({\bf I})\quad\left|\left|F^{(l)}(x)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon\qquad\text{and}\qquad({\bf
II})\quad\left|\left|\hat{R}^{F}_{l}(x,y)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon||y-x||_{V}^{\eta-l}.$ (10.8)

Consider $p\in\Sigma$. From (10.2) we know that
$p\in\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,\delta_{0})$ for some $z\in
B$. Consequently the bound (I) in (10.8) holds for $x:=p$. Since $p\in\Sigma$
was arbitrary, we conclude that for any $p\in\Sigma$ we have

$\left|\left|F^{(l)}(p)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon.$ (10.9)

Consider $l\in\\{0,\ldots,q\\}$ and $p,w\in\Sigma$. If there exists $z\in B$
for which $p,w\in{\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$ then (II) in
(10.8) yields that

$\left|\left|\hat{R}^{F}_{l}(p,w)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon||w-p||_{V}^{\eta-l}.$ (10.10)

Now suppose that no single ball ${\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$
contains both $p$ and $w$. From (10.2) we know that
$p\in\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z_{i},\delta_{0})$ and
$w\in\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z_{j},\delta_{0})$ for some
$z_{i},z_{j}\in B$ which must be distinct. In fact, since
$w\notin\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z_{i},2\delta_{0})$, we
can conclude that

$||w-p||_{V}\geq\delta_{0}.$ (10.11)

Observe that from (10.7) we have, for any $v\in V^{\otimes l}$, that

$\hat{R}^{F}_{l}(p,w)[v]=F^{(l)}(w)[v]-\sum_{s=0}^{q-l}\frac{1}{s!}F^{(l+s)}(p)\left[v\otimes(w-p)^{\otimes
s}\right].$ (10.12)

We may further use (10.7) to compute that

$F^{(l)}(w)[v]=\sum_{u=0}^{q-l}\frac{1}{u!}F^{(l+u)}(z_{j})\left[v\otimes(w-z_{j})^{\otimes
u}\right]+\hat{R}^{F}_{l}(z_{j},w)[v].$ (10.13)

Since $w\in{\overline{\mathbb{B}}}_{V}(z_{j},\delta_{0})$ we may use (10.10)
to conclude that

$\left|\left|\hat{R}^{F}_{l}(z_{j},w)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon||w-z_{j}||_{V}^{\eta-l}\leq\theta\varepsilon\delta_{0}^{\eta-l}.$
(10.14)

Additionally, since $z_{j}\in B$ we may compute that

$\sum_{u=0}^{q-l}\frac{1}{u!}\left|\left|F^{(l+u)}(z_{j})\left[v\otimes(w-z_{j})^{\otimes
u}\right]\right|\right|_{W}\leq\varepsilon_{0}\sum_{u=0}^{q-l}\frac{1}{u!}||w-z_{j}||_{V}^{u}||v||_{V^{\otimes
l}}\leq\varepsilon_{0}e^{\delta_{0}}||v||_{V^{\otimes l}}.$ (10.15)

Combining (10.13), (10.14), and (10.15) yields the estimate

$\left|\left|F^{(l)}(w)[v]\right|\right|_{W}\leq\left(\theta\varepsilon\delta_{0}^{\eta-l}+\varepsilon_{0}e^{\delta_{0}}\right)||v||_{V^{\otimes
l}}.$ (10.16)

Turning our attention to the second term in (10.12), note that for any
$s\in\\{0,\ldots,q-l\\}$ we have via (10.7), for
$v^{\prime}:=v\otimes(w-p)^{\otimes s}\in V^{\otimes(l+s)}$, that

$F^{(l+s)}(p)\left[v^{\prime}\right]=\sum_{u=0}^{q-l-s}\frac{1}{u!}F^{(l+s+u)}(z_{i})\left[v^{\prime}\otimes(p-z_{i})^{\otimes
u}\right]+\hat{R}^{F}_{l+s}(z_{i},p)\left[v^{\prime}\right].$ (10.17)

Since $p\in{\overline{\mathbb{B}}}_{V}(z_{i},\delta_{0})$ we may use (10.10)
to conclude that

$\left|\left|\hat{R}^{F}_{l+s}(z_{i},p)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\theta\varepsilon||p-z_{i}||_{V}^{\eta-(l+s)}\leq\theta\varepsilon\delta_{0}^{\eta-(l+s)}.$
(10.18)

Via similar computations to those used to establish (10.15), the fact that
$z_{i}\in B$ allows us to compute that

$\sum_{u=0}^{q-l-s}\frac{1}{u!}\left|\left|F^{(l+u+s)}(z_{j})\left[v^{\prime}\otimes(p-z_{i})^{\otimes
u}\right]\right|\right|_{W}\leq\varepsilon_{0}e^{\delta_{0}}||v^{\prime}||_{V^{\otimes(l+s)}}.$
(10.19)

Combining (10.17), (10.18), and (10.19) yields that

$\left|\left|F^{(l+s)}(p)\left[v^{\prime}\right]\right|\right|_{W}\leq\left(\varepsilon_{0}e^{\delta_{0}}+\theta\varepsilon\delta_{0}^{\eta-(l+s)}\right)||v^{\prime}||_{V^{\otimes(l+s)}}=\left(\varepsilon_{0}e^{\delta_{0}}+\theta\varepsilon\delta_{0}^{\eta-(l+s)}\right)||w-p||_{V}^{s}||v||_{V^{\otimes
l}}$ (10.20)

where the last equality uses that $v^{\prime}:=v\otimes(w-p)^{\otimes s}$ and
that the tensor powers of $V$ are equipped with admissible norms (cf.
Definition 2.1). A consequence of (10.20) is that

$\sum_{s=0}^{q-l}\frac{1}{s!}\left|\left|F^{(l+s)}(p)\left[v^{\prime}\right]\right|\right|_{W}\leq\sum_{s=0}^{q-l}\frac{1}{s!}\left(\varepsilon_{0}e^{\delta_{0}}+\theta\varepsilon\delta_{0}^{\eta-(l+s)}\right)||w-p||_{V}^{s}||v||_{V^{\otimes
l}}.$ (10.21)

Since from (10.11) we have that $\delta_{0}\leq||w-p||_{V}$, we may multiply
each term in the sum on the RHS of (10.21) by
$||w-p||_{V}^{\eta-(l+s)}\delta_{0}^{-(\eta-(l+s))}\geq 1$ to conclude that

$\sum_{s=0}^{q-l}\frac{1}{s!}\left|\left|F^{(l+s)}(p)\left[v^{\prime}\right]\right|\right|_{W}\leq\sum_{s=0}^{q-l}\frac{1}{s!}\left(\varepsilon_{0}e^{\delta_{0}}+\theta\varepsilon\delta_{0}^{\eta-(l+s)}\right)\delta_{0}^{-(\eta-
l-s)}||w-p||_{V}^{\eta-l}||v||_{V^{\otimes l}}.$ (10.22)

Combining (10.12), (10.16), and (10.22) allows us to deduce that

$\left|\left|\hat{R}^{F}_{l}(p,w)[v]\right|\right|_{W}\leq\left(\theta\varepsilon\delta_{0}^{\eta-l}+\varepsilon_{0}e^{\delta_{0}}+\sum_{s=0}^{q-l}\frac{1}{s!}\left(\varepsilon_{0}e^{\delta_{0}}\delta_{0}^{-(\eta-
l-s)}+\theta\varepsilon\right)||w-p||_{V}^{\eta-l}\right)||v||_{V^{\otimes
l}}.$ (10.23)

Observe that

$\theta\varepsilon\delta_{0}^{\eta-l}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_cover_thm_claim_de0_sep}})}}}{{\leq}}\theta\varepsilon||w-p||_{V}^{\eta-l},$
(10.24)
$\varepsilon_{0}e^{\delta_{0}}\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_cover_thm_claim_de0_sep}})}}}{{\leq}}\varepsilon_{0}\delta_{0}^{-(\eta-l)}e^{\delta_{0}}||w-p||_{V}^{\eta-l},$
(10.25)
$\varepsilon_{0}e^{\delta_{0}}\delta_{0}^{-(\eta-l)}\sum_{s=0}^{q-l}\frac{1}{s!}\delta_{0}^{s}\leq\varepsilon_{0}\delta_{0}^{-(\eta-l)}e^{2\delta_{0}},\quad\text{and}$
(10.26) $\theta\varepsilon\sum_{s=0}^{q-l}\frac{1}{s!}\leq\theta
e\varepsilon.$ (10.27)

Combining (10.24), (10.25), (10.26), and (10.27) with (10.23) yields

$\left|\left|\hat{R}^{\varphi}_{l}(p,w)[v]\right|\right|_{W}\leq\left(\theta\varepsilon(1+e)+\varepsilon_{0}\delta_{0}^{-(\eta-l)}e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)\right)||w-p||_{V}^{\eta-l}||v||_{V^{\otimes
l}}.$ (10.28)

Taking the supremum over $v\in V^{\otimes l}$ with unit $V^{\otimes l}$-norm
in (10.28) yields

$\left|\left|\hat{R}^{F}_{l}(p,w)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\left(\theta\varepsilon(1+e)+\varepsilon_{0}\delta_{0}^{-\eta}e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)\right)||w-p||_{V}^{\eta-l}$
(10.29)

since $\delta_{0}\leq 1$ means $\delta_{0}^{-(\eta-l)}\leq\delta_{0}^{-\eta}$
for every $l\in\\{0,\ldots,q\\}$.

Together (10.10), (10.29) and the inequality
$\theta\varepsilon<\theta\varepsilon(1+e)+\varepsilon_{0}\delta_{0}^{-\eta}e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)$
mean that for any $l\in\\{0,\ldots,q\\}$ and any $p,w\in\Sigma$ we have

$\left|\left|\hat{R}^{F}_{l}(p,w)\right|\right|_{{\cal L}(V^{\otimes
l};W)}\leq\left(\theta\varepsilon(1+e)+\varepsilon_{0}\frac{e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)}{\delta_{0}^{\eta}}\right)||w-p||_{V}^{\eta-l}.$
(10.30)

The definitions (10.7), the bounds (10.9), and the Hölder estimates (10.30)
tell us that

$\displaystyle\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Sigma,W)}$
$\displaystyle\leq\theta\varepsilon(1+e)+\varepsilon_{0}\frac{e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)}{\delta_{0}^{\eta}}$
$\displaystyle\stackrel{{\scriptstyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}({\ref{lip_k_cover_thm_ep0_def}})}}}{{\leq}}\frac{1}{2(1+e)}\varepsilon(1+e)+\frac{\varepsilon}{2}\frac{\delta_{0}^{\eta}}{e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)}\frac{e^{\delta_{0}}\left(1+e^{\delta_{0}}\right)}{\delta_{0}^{\eta}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

as claimed in (10.4). This completes the proof of Claim 10.1 ∎

Returning to the proof of Theorem 3.1 itself, we define
$F:=\psi-\varphi\in{\mathrm{Lip}}(\gamma,\Sigma,W)$ so that for every
$j\in\\{0,\ldots,k\\}$ we have $F^{(j)}=\psi^{(j)}-\varphi^{(j)}$. Then, by
assumption, we have for every $j\in\\{0,\ldots,k\\}$ and every $z\in B$ that
$\left|\left|F^{(j)}(z)\right|\right|_{{\cal L}(V^{\otimes
j};W)}\leq\varepsilon_{0}$. Moreover, (10.3) tells us that whenever $z\in B$
we have that
$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Omega_{z},W)}\leq\theta\varepsilon$
for $\Omega_{z}:=\Sigma~{}\cap~{}{\overline{\mathbb{B}}}_{V}(z,2\delta_{0})$.
Therefore we can apply Claim 10.1 to $F$ and conclude that
$\left|\left|F_{[q]}\right|\right|_{{\mathrm{Lip}}(\eta,\Sigma,W)}\leq\varepsilon$.
Since $F:=\psi-\varphi$ this gives the estimate claimed in (3.3) and completes
the proof of Theorem 3.1. ∎

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