Proof. Define $\Phi \colon X\to Z$ by

$$ \begin{align*} \Phi(\mathbf{x})=\begin{cases} f_1(\mathbf{x}) & \text{ if }\left\|\mathbf{x}\right\|_{X}\le a ,\\ \displaystyle\frac{b-\left\|\mathbf{x}\right\|_{X}}{b-a}f_1(\mathbf{x})+ \frac{b\big(\left\|\mathbf{x}\right\|_{X}-a\big)}{\left\|\mathbf{x}\right\|_{X}(b-a)}f_2(\mathbf{x}) & \text{ if }a<\left\|\mathbf{x}\right\|_{X}<b,\\ f_2(\mathbf{x}) & \text{ if }\left\|\mathbf{x}\right\|_{X}\ge b. \end{cases} \end{align*} $$

Clearly, $\Phi $ satisfies (i) and (ii). Observe that $\Phi $ is a continuous mapping $X\to Z$ . Moreover, since $f_1,f_2\in \operatorname {Lip}_{1}$ , in order to show that $\Phi $ is Lipschitz and to check (iii), it is enough to verify

(3.1) $$ \begin{align} \left\|\Phi(\mathbf{y})-\Phi(\mathbf{x})\right\|_{Z}\le {\left(1+\frac{a}{b-a}\right)}\left\|\mathbf{y}-\mathbf{x}\right\|_{X} \end{align} $$

whenever $\left \|\mathbf {x}\right \|_{X},\left \|\mathbf {y}\right \|_{X}\in (a,b)$ . To show this, fix such $\mathbf {x},\mathbf {y}\in X$ and note first that

(3.2) $$ \begin{align} \begin{aligned} \left\|\Phi(\mathbf{y})-\Phi(\mathbf{x})\right\|_{Z}&\le \left\| \tfrac{b-\left\|\mathbf{y}\right\|_{X}}{b-a}f_1(\mathbf{y})- \tfrac{b-\left\|\mathbf{x}\right\|_{X}}{b-a}f_1(\mathbf{x}) \right\|_{Z} + \left\| \tfrac{b(\left\|\mathbf{y}\right\|_{X}-a)}{\left\|\mathbf{y}\right\|_{X}(b-a)}f_2(\mathbf{y}) - \tfrac{b(\left\|\mathbf{x}\right\|_{X}-a)}{\left\|\mathbf{x}\right\|_{X}(b-a)}f_2(\mathbf{x}) \right\|_{Z}. \end{aligned} \end{align} $$

In several estimates that follow we will use that

(3.3) $$ \begin{align} \left\|f_{i}(\mathbf{u})\right\|_{Z}\leq \operatorname{Lip}(f_{i})\left\|\mathbf{u}\right\|_{X} \qquad\text{ for all } \mathbf{u}\in X \text{ and } i=1,2. \end{align} $$

This holds due to the condition $f_{i}(\mathbf {0}_{X})=\mathbf {0}_{Z}$ for $i=1,2$ . Assuming without loss of generality that $\left \|\mathbf {y}\right \|_{X}\ge \left \|\mathbf {x}\right \|_{X}$ , the first term of (3.2) is bounded above by

$$ \begin{align*} &\frac{b-\left\|\mathbf{y}\right\|_{X}}{b-a} \left\| f_1(\mathbf{y})-f_1(\mathbf{x})\right\|_{Z} + \left|\frac{b-\left\|\mathbf{y}\right\|_{X}}{b-a}-\frac{b-\left\|\mathbf{x}\right\|_{X}}{b-a}\right| \left\| {f_1(\mathbf{x})}\right\|_{Z}\\ \\ \le\ &\frac{b-\left\|\mathbf{y}\right\|_{X}}{b-a} \operatorname{Lip}(f_1)\left\|\mathbf{y}-\mathbf{x}\right\|_{X} + \frac{\left\|\mathbf{y}-\mathbf{x}\right\|_{X}}{b-a}\operatorname{Lip}(f_1)\left\|\mathbf{x}\right\|_{X} \\ \\ =\ &\frac{b-\left\|\mathbf{y}\right\|_{X}+\left\|\mathbf{x}\right\|_{X}}{b-a}\operatorname{Lip}(f_1)\left\|\mathbf{y}-\mathbf{x}\right\|_{X} \le \left(1+\frac{a}{b-a}\right)\operatorname{Lip}(f_1)\left\|\mathbf{y}-\mathbf{x}\right\|_{X}. \end{align*} $$

The second term of (3.2) is bounded above by

$$ \begin{align*} &\frac{b\big(\left\|\mathbf{y}\right\|_{X}-a\big)}{\left\|\mathbf{y}\right\|_{X}(b-a)} \left\|f_2(\mathbf{y})-f_2(\mathbf{x})\right\|_{Z}+ \left|\frac{b\big(\left\|\mathbf{y}\right\|_{X}-a\big)}{\left\|\mathbf{y}\right\|_{X}(b-a)}- \frac{b\big(\left\|\mathbf{x}\right\|_{X}-a\big)}{\left\|\mathbf{x}\right\|_{X}(b-a)} \right|\left\|f_2(\mathbf{x})\right\|_{Z}\\ \le\ &\frac{b\big(\left\|\mathbf{y}\right\|_{X}-a\big)}{\left\|\mathbf{y}\right\|_{X}(b-a)} \operatorname{Lip}(f_2)\left\|\mathbf{y}-\mathbf{x}\right\|_{X} +\frac{ab}{\left\|\mathbf{x}\right\|_{X}\left\|\mathbf{y}\right\|_{X}(b-a)}\left\|\mathbf{x}-\mathbf{y}\right\|_{X}\operatorname{Lip}(f_2)\left\|\mathbf{x}\right\|_{X}\\ =\ &\left(1+\frac{a}{b-a}\right)\operatorname{Lip}(f_2)\left\|\mathbf{y}-\mathbf{x}\right\|_{X}. \end{align*} $$

Summing the derived upper bounds for the two terms of (3.2) establishes (3.1).

Finally, if $f_1=\mathbf {0}_{Z}$ , then

$$ \begin{align*} \left\|\Phi(\mathbf{x})-f_2(\mathbf{x})\right\|_{Z} = \begin{cases} \left\|f_2(\mathbf{x})\right\|_{Z} & \text{ if } \left\|\mathbf{x}\right\|_{X}\leq a,\\ \frac{a(b-\left\|\mathbf{x}\right\|_{X})}{(b-a)\left\|\mathbf{x}\right\|_{X}}\left\|f_2(\mathbf{x})\right\|_{Z} & \text{ if } a<\left\|\mathbf{x}\right\|_{X}<b,\\ 0 & \text{ if } \left\|\mathbf{x}\right\|_{X}\geq b, \end{cases} \end{align*} $$

and if $f_2=\mathbf {0}_{Z}$ , then

$$ \begin{align*} \left\|\Phi(\mathbf{x})\right\|_{Z}=\begin{cases} \left\|f_1(\mathbf{x})\right\|_{Z} & \text{ if } \left\|\mathbf{x}\right\|_{X}\leq a,\\ \frac{b-\left\|\mathbf{x}\right\|_{X}}{b-a}\left\|f_1(\mathbf{x})\right\|_{Z} & \text{ if } a<\left\|\mathbf{x}\right\|_{X}<b,\\ 0 & \text{ if } \left\|\mathbf{x}\right\|_{X}\geq b. \end{cases} \end{align*} $$

Applying (3.3) to these formulae, we verify (iv) and (v).

Proof. The approach we take to modify the mapping f to arrive at g is similar to that taken in [Reference Dymond11, Lemma 3.3].

The conclusion of this lemma is valid for f if and only if it is valid for any mapping of the form $f+p$ , where $p\colon Q\to Y$ is a constant mapping. Therefore, we may assume that $\mathbf {0}_{Y}\in f(\Gamma )$ . Lipschitz mappings $h\colon Q\to Y$ with the property $\mathbf {0}_{Y}\in h(Q)$ satisfy $\left \|h(\mathbf {x})\right \|_{Y}\leq \operatorname {Lip}(h)\operatorname {diam} Q$ for all $\mathbf {x}\in Q$ . This fact will be used later in the proof.

Fix $s\in (0,1)$ small enough so that $\Gamma $ is $4s$ -separated and the infimum of (3.4) is at least $4s$ . Choose a parameter

(3.6) $$ \begin{align} \beta=\frac{rs}{5(1+\operatorname{diam} Q)}\in (0,s/2) \end{align} $$

and define a mapping $g_{0}\colon Q\to Y$ by

$$ \begin{align*} g_{0}(\mathbf{z})=\begin{cases} f(\mathbf{z}) & \text{ if }\mathbf{z}\in Q\setminus \bigcup_{\mathbf{x}\in\Gamma}B_{X}(\mathbf{x},s),\\ f(\mathbf{x}+\Phi(\mathbf{z}-\mathbf{x})) & \text{ if }\mathbf{z}\in B_{X}(\mathbf{x},s)\text{ and }\mathbf{x}\in\Gamma, \end{cases} \end{align*} $$

where $\Phi :=\Phi (\beta ,s,\mathbf {0}_{X},\operatorname {Id}_X)\colon X\to X$ is the mapping given by Lemma 3.1 applied to X, $Z=X$ , $a=\beta $ , $b=s$ , $f_{1}=\mathbf {0}_{X}$ (the constant mapping $X\to X$ with value $\mathbf {0}_{X}$ ) and $f_{2}=\operatorname {Id}_{X}$ . Here we used Lemma 3.1(v) to conclude $\left \|\Phi (\mathbf {z}-\mathbf {x})\right \|_{X}\le \beta <s$ , so that $\mathbf {x}+\Phi (\mathbf {z}-\mathbf {x})\in Q$ whenever $\mathbf {x}\in \Gamma $ and $\mathbf {z}\in B_{X}(\mathbf {x},s)$ . Using again $\beta <s$ and Lemma 3.1(i), we note that

(3.7) $$ \begin{align} g_{0}(\mathbf{z})=g_{0}(\mathbf{x})=f(\mathbf{x})\qquad \text{whenever } \mathbf{x}\in \Gamma\text{ and }\mathbf{z}\in \overline{B}_{X}(\mathbf{x},\beta). \end{align} $$

It follows, in particular,

$$\begin{align*}\mathbf{0}_{Y}\in f(\Gamma)=g_{0}(\Gamma). \end{align*}$$

Further, we note that Lemma 3.1(iii) and (iv) imply $g_0$ is Lipschitz,

$$ \begin{align*} \operatorname{Lip}(g_{0})\leq 1+\frac{\beta}{s-\beta} \qquad \text{and} \qquad \left\|g_{0}-f\right\|_{\infty}\leq \beta. \end{align*} $$

Let

$$ \begin{align*} T=\frac{s}{s-\beta}L \qquad\text{and}\qquad \alpha=\frac{r^{2}s}{25(1+\operatorname{diam} Q)^{2}}. \end{align*} $$

Note $\left \|T\right \|_{\operatorname {op}}\le 1$ using $\beta <rs$ and $\left \|L\right \|_{\operatorname {op}}\le 1-r$ . Define $g_{1}\colon Q\to Y$ by

(3.8) $$ \begin{align} g_{1}(\mathbf{z})=\begin{cases} g_{0}(\mathbf{z}), & \text{ if }\mathbf{z}\in Q\setminus \bigcup_{\mathbf{x}\in\Gamma}B_{X}(\mathbf{x},\beta),\\ g_0({\mathbf{x}}) +T(\Psi(\mathbf{z}-\mathbf{x})), & \text{ if }\mathbf{z}\in B_{X}(\mathbf{x},\beta)\text{ and }\mathbf{x}\in \Gamma, \end{cases} \end{align} $$

where $\Psi :=\Phi (\alpha ,\beta ,\operatorname {Id}_{X},\mathbf {0}_{X})\colon X\to X$ is the mapping given by Lemma 3.1 applied to X, $Z=X$ , $a=\alpha $ , $b=\beta $ , $f_{1}=\operatorname {Id}_{X}$ and $f_{2}=\mathbf {0}_{X}$ . The properties of $g_0$ , $\Psi $ and (3.7) ensure that $g_{1}$ is Lipschitz. We may estimate its Lipschitz constant as

$$ \begin{align*} \operatorname{Lip}(g_{1})\leq \max\left\{\operatorname{Lip}(g_{0}),1+\frac{\alpha}{\beta-\alpha}\right\} \leq\max\left\{1+\frac{\beta}{s-\beta}, 1+\frac{\alpha}{\beta-\alpha}\right\}\leq 1+\frac{\beta}{s-\beta} =\frac{s}{s-\beta}, \end{align*} $$

where we used $\alpha \leq \frac {\beta ^{2}}{s}$ in the penultimate inequality. Moreover, we have

$$ \begin{align*} \left\|g_{1}-g_{0}\right\|_{\infty}\leq \beta\qquad\text{and}\qquad \mathbf{0}_{Y}\in g_{0}(\Gamma)=g_{1}(\Gamma). \end{align*} $$

To verify the former, it is enough to observe, using (3.7), $\left \|T\right \|_{\operatorname {op}}\leq 1$ , $\operatorname {Lip}(\operatorname {Id}_X)=1$ and Lemma 3.1(v), that for any $\mathbf {z}\in B_{X}(\mathbf {x},\beta )$ with $\mathbf {x}\in \Gamma $

$$ \begin{align*} \left\|g_1(\mathbf{z})-g_0(\mathbf{z})\right\|_{Y} =\left\|g_0(\mathbf{x})+T(\Psi(\mathbf{z}-\mathbf{x}))-g_0(\mathbf{z})\right\|_{Y} =\left\|T(\Psi(\mathbf{z}-\mathbf{x}))\right\|_{Y} \leq \left\|\Psi(\mathbf{z}-\mathbf{x})\right\|_{X} \leq \beta. \end{align*} $$

Finally we set

$$ \begin{align*} g=\frac{s-\beta}{s}\cdot g_{1}, \end{align*} $$

so that $g\in \operatorname {Lip}_{1} (Q,Y)$ and

$$ \begin{align*} \left\|g-g_{1}\right\|_{\infty}\leq \frac{\beta \operatorname{Lip}(g_{1})\operatorname{diam} Q}{s}\leq \frac{\beta \operatorname{diam} Q}{s-\beta}\leq \frac{2\beta \operatorname{diam} Q}{s}, \end{align*} $$

using $\beta <s/2$ from (3.6). Using the definition of $\beta $ , we conclude that

$$ \begin{align*} \left\|g-f\right\|_{\infty}\leq \left\|g-g_{1}\right\|_{\infty}+\left\|g_{1}-g_{0}\right\|_{\infty}+\left\|g_{0}-f\right\|_{\infty} \leq \frac{2\beta\operatorname{diam} Q}{s}+2\beta\leq \frac{4\beta(1+\operatorname{diam} Q)}{s} < r. \end{align*} $$

It is clear that (3.5) is satisfied as $g_1(\mathbf {x}+\mathbf {u})=g_1(\mathbf {x})+T\mathbf {u}$ whenever $\mathbf {x}\in \Gamma $ and $\left \|\mathbf {u}\right \|_{X}\le \alpha $ , from (3.8).

Proof. If $E=\emptyset $ , let $\Gamma _k=\emptyset $ for all $k\in \mathbb N$ .

Assume $E\ne \emptyset $ . Let $E_k=\left \{\mathbf {x}\in E\colon \operatorname {dist}_{X}(\mathbf {x},\partial Q)\geq 2^{-k}\right \}$ . Since $E\subseteq \operatorname {Int} Q$ , we have that ${\bigcup _{k\ge 1}E_k=E}$ . Let $n\ge 1$ be the smallest index such that $E_n\ne \emptyset $ . Set $\Gamma _k=\emptyset $ for any $0\le k\le n-1$ . For any $k\ge n$ , let us make an inductive choice of $\Gamma _k\supseteq \Gamma _{k-1}$ to be a non-empty maximal $2^{-k}$ -separated subset of $E_k$ . Since for any $k\ge n$ the set $\Gamma _k\ne \emptyset $ is a $2^{-k}$ -net of $E_k$ , we conclude that $\bigcup _{k\ge n}\Gamma _k=\bigcup _{k\ge 1}\Gamma _k$ is dense in E.

Proof. We assume that $E\ne \emptyset $ . Let

$$ \begin{align*} \mathcal{H}_{T}:=\left\{f\in\operatorname{Lip}_{1}(Q,Y)\colon \text{ the set }\left\{\mathbf{x}\in E\colon T\in \mathcal{D}_{f}(\mathbf{x})\right\}\text{ is residual in } E\right\}. \end{align*} $$

We prove that the set $\mathcal {H}_{T}$ is residual in $\operatorname {Lip}_{1}(Q,Y)$ by describing a winning strategy for Player II in the relevant Banach-Mazur game in $(\operatorname {Lip}_{1}(Q,Y),\left \|\cdot \right \|_{\infty })$ with the target $\mathcal {H}_T$ , in which Player I’s choices are balls $B(f_k,r_k)$ and Player II’s choices are balls $B(g_k,s_k)$ ; see Section 2.5 for details on the Banach-Mazur game. Here and throughout the proof, given a mapping $\phi \in \operatorname {Lip}_{1}(Q,Y)$ and $\rho>0$ we abbreviate the notation $B_{\operatorname {Lip}_{1}(Q,Y)}(\phi ,\rho )$ , for the open ball in the metric space $(\operatorname {Lip}_{1}(Q,Y),\left \|\cdot \right \|_{\infty })$ with centre $\phi $ and radius $\rho $ , to $B(\phi ,\rho )$ .

Before the game starts, let Player II prepare by fixing a nested sequence $(\Gamma _{k})_{k\in \mathbb {N}}$ of sets ${\Gamma _k\subseteq \Gamma _{k+1}\subseteq E}$ , given by Lemma 3.3.

Let $k\in \mathbb {N}$ , assume that $k-1$ rounds of the game have already completed, giving $f_i$ , $r_i$ , $g_i$ and $s_i$ for $i\leq k-1$ and let $f_{k}\in \operatorname {Lip}_{1}(Q,Y)$ and $r_{k}>0$ denote the k-th move of Player I. Since nothing prevents Player II from acting as if the radius $r_{k}$ was replaced by a smaller radius $\tilde r_{k}>0$ , we may assume that

(3.9) $$ \begin{align} r_{k}\leq 2^{-k}(1-\left\|T\right\|_{\operatorname{op}}), \quad\text{in addition to}\quad \overline{B}(f_{k},r_{k})\subseteq B(g_{k-1},s_{k-1})\text{ if } k\geq 2. \end{align} $$

In order to define their response, Player II applies Lemma 3.2 to find a mapping $g_{k}\in \operatorname {Lip}_{1}(Q,Y)$ and $\alpha _{k}\in (0,r_k)$ satisfying $\left \|g_k-f_k\right \|_{\infty }<r_{k}$ and

(3.10) $$ \begin{align} g_{k}(\mathbf{x}+\mathbf{u})= g_k(\mathbf{x})+T\mathbf{u},\qquad \text{whenever } \mathbf{x}\in\Gamma_k\,\text{ and }\mathbf{u}\in\overline{B}(\mathbf{0}_{X},\alpha_{k}). \end{align} $$

Finally, Player II chooses

(3.11) $$ \begin{align} 0<s_{k}<\alpha_k/(4k) \end{align} $$

small enough so that

(3.12) $$ \begin{align} \overline{B}(g_{k},s_{k})\subseteq B(f_{k},r_{k}) \end{align} $$

and declares $g_{k}\in \operatorname {Lip}_{1}(Q,Y)$ and $s_{k}>0$ as their k-th move.

Due to the conditions (3.9) and (3.12), the intersection

$$ \begin{align*} \bigcap_{k=1}^{\infty}B(f_{k},r_{k})=\bigcap_{k=1}^{\infty}B(g_{k},s_{k}) \end{align*} $$

is a singleton set containing only the Lipschitz mapping $g:=\lim _{k\to \infty }g_{k}\in \operatorname {Lip}_{1}(Q,Y)$ .

To complete the proof, we show that Player II wins the game, that is, that $g\in \mathcal {H}_{T}$ , see Section 2.5. Consider the sequence $U_{k}:=\bigcup _{\mathbf {x}\in \Gamma _{k}} B_{X}(\mathbf {x},s_{k})$ of open sets in X and the set $J:=E\cap \bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }U_{k}\subseteq E$ . Clearly, J is a relatively $G_{\delta }$ subset of E. Moreover, for each $n\ge 1$ , $\bigcup _{k\ge n}U_{k}\supseteq \bigcup _{k\ge n}\Gamma _{k}=\bigcup _{k\ge 1}\Gamma _{k}$ , as $\Gamma _k$ are nested, and the latter is a dense subset of E by Lemma 3.3; thus $J\supseteq \bigcup _{k\ge 1}\Gamma _{k}$ is dense in E. We conclude that J is a relatively residual subset of E.

To prove $g\in \mathcal {H}_T$ , we verify $J\subseteq \{\mathbf {x}\in E\colon T\in \mathcal {D}_g(\mathbf {x})\}$ . Let $\mathbf {x}\in J$ and $\varepsilon>0$ . Choose $k\in \mathbb {N}$ with $k\geq {1}/{\varepsilon }$ such that $\mathbf {x}\in U_{k}$ and $\alpha _{k}<\varepsilon $ . Let $\mathbf {x}_{k}\in \Gamma _{k}$ be such that $\mathbf {x}\in B_{X}(\mathbf {x}_{k},s_{k})$ ; let $\mathbf {u}\in \overline {B}_{X}(\mathbf {0}_{X},\alpha _{k})$ be arbitrary. Then, applying (3.10), we get $g_{k}(\mathbf {x}_{k}+\mathbf {u})=g_{k}(\mathbf {x}_{k})+T\mathbf {u}$ . Using this identity, we derive

(3.13) $$ \begin{align} \left\|g(\mathbf{x}+\mathbf{u})-g(\mathbf{x})-T\mathbf{u}\right\|_{Y}&\leq\left\|g(\mathbf{x}+\mathbf{u})-g_{k}(\mathbf{x}+\mathbf{u})\right\|_{Y}+\left\|g_{k}(\mathbf{x}+\mathbf{u})-g_{k}(\mathbf{x}_{k}+\mathbf{u})\right\|_{Y}\nonumber\\ &+\left\|g_{k}(\mathbf{x}_{k}+\mathbf{u})-g_{k}(\mathbf{x}_{k})-T\mathbf{u}\right\|_{Y}+\left\|g_{k}(\mathbf{x}_{k})-g_{k}(\mathbf{x})\right\|_{Y}+\left\|g_{k}(\mathbf{x})-g(\mathbf{x})\right\|_{Y}\nonumber\\ &\leq 2\left\|g_{k}-g\right\|_{\infty}+2\left\|\mathbf{x}_{k}-\mathbf{x}\right\|_{X}+0\leq 4 s_k\leq \frac{\alpha_{k}}{k}, \end{align} $$

where the last inequality is due to Player II’s choice (3.11) of $s_{k}$ . This argument verifies

$$\begin{align*}\sup_{\mathbf{u}\in\overline{B}_{X}(\mathbf{0}_X,\alpha_{k})}\frac{\left\|g(\mathbf{x} + \mathbf{u})-g(\mathbf{x})-T\mathbf{u}\right\|_{Y}}{\alpha_k} \le\frac1k\leq \varepsilon. \end{align*}$$

and subsequently $T\in \mathcal {D}_{g}(\mathbf {x})$ .

Proof of Theorem 1.1.

Fix a dense sequence $(T_{n})_{n\in \mathbb {N}}$ in the closed unit ball $\mathbb {B}_{W}$ with $\left \|T_{n}\right \|_{\operatorname {op}}<1$ for all $n\in \mathbb {N}$ . Let the sets $\mathcal {H}_{T_{n}}\subseteq \operatorname {Lip}_{1}(Q,Y)$ be given by the conclusion of Lemma 3.4. Define a residual subset $\mathcal {F}=\bigcap _{n\in \mathbb {N}}\mathcal {H}_{T_{n}}$ of $\operatorname {Lip}_{1}(Q,Y)$ and let $f\in \mathcal {F}$ be arbitrary. Let the sets $\mathcal {P}_{T_{n},f}$ be given by the conclusion of Lemma 3.4, consider the residual subset $\mathcal {P}_f=\bigcap _{n\in \mathbb {N}}\mathcal {P}_{T_{n},f}$ of E and let $\mathbf {x}\in \mathcal {P}_f$ be arbitrary. Since $T_n\in \mathcal {D}_{f}(\mathbf {x})$ for all $n\ge 1$ and $\mathcal {D}_{f}(\mathbf {x})$ is closed in $(\mathcal {L}(X,Y),\left \|\cdot \right \|_{\operatorname {op}})$ , by Lemma 2.1, we have $\mathcal {D}_{f}(\mathbf {x})\supseteq \overline {\left \{T_n\colon n\in \mathbb {N}\right \}}^{\left \|\cdot \right \|_{\operatorname {op}}}= \mathbb {B}_{W}$ .

Proof. If $P=0$ we may take g as the constant zero function $X\to \mathbb {R}$ . So we may assume $P\neq 0$ .

Observe that if $g\colon X\to \mathbb {R}$ satisfies conditions (i)–(iv) for the functional $P=\frac {Q}{\left \|Q\right \|_{X^{\ast }}}$ , for some $Q\in X^{\ast }\setminus \left \{0\right \}$ , then the function $\left \|Q\right \|_{X^{\ast }}g\colon X\to \mathbb {R}$ satisfies the conditions (i)–(iv) for $P=Q$ . Hence, we may assume that $\left \|P\right \|_{X^{\ast }}=1$ .

Define $g\colon X\to \mathbb {R}$ by

(4.1) $$ \begin{align} \begin{aligned} g(\mathbf{x}) &= \sup\left\{ \mathcal{H}^{1} \left(G\cap \gamma((-\infty,b])\right)-s\colon \gamma\in \operatorname{Lip}(\mathbb{R},X),\, b\in\mathbb{R},\,s\geq 0,\,\right. \\ &\left. \gamma(b)=\mathbf{x}+s \mathbf{v}_{P},\, P\left(\gamma'(t)\right)\geq \alpha\left\|\gamma'(t)\right\|_{X}\text{ whenever } \gamma'(t) \text{ exists} \right\}. \end{aligned} \end{align} $$

We will now check that g satisfies (i)–(iv) and, in addition, the following two properties:

1. (A) $g(\mathbf {x})\leq g(\mathbf {x}+r\mathbf {v}_{P})\leq g(\mathbf {x})+r$ whenever $\mathbf {x}\in X$ and $r\geq 0$ .

2. (B) $g(\mathbf {x}')-g(\mathbf {x})=t$ whenever $\mathbf {x}'-\mathbf {x}=t\mathbf {v}_{P}$ , $t\in \mathbb {R}$ and $[\mathbf {x},\mathbf {x}']\subseteq G$ .

The fact that g is well-defined and the first inequality of(i) are witnessed by the triple $\gamma (t)=\mathbf {x}$ for all $t\in \mathbb {R}$ , $b=0$ and $s=0$ . The second inequality of (i) is immediate from $s\ge 0$ for any admissible triple $(\gamma ,b,s)$ in the definition of $g(\mathbf {x})$ and the definition of $(P,\alpha )$ -width of G in (2.2).

In the proof of the remaining parts we will use the following notation. If $\mathbf {z}\in X$ and $\eta>0$ then by $(\gamma _{\mathbf {z}},s_{\mathbf {z}},b_{\mathbf {z}})$ we denote an admissible triple in the sense of (4.1), that is, such that $\gamma _{\mathbf {z}}\in \operatorname {Lip}(\mathbb {R},X)$ , $b_{\mathbf {z}}\in \mathbb {R}$ , $s_{\mathbf {z}}\ge 0$ , $\gamma _{\mathbf {z}}(b_{\mathbf {z}})=\mathbf {z}+s_{\mathbf {z}}\mathbf {v}_{P}$ , $P(\gamma _{\mathbf {z}}'(t))\ge \alpha \left \|\gamma _{\mathbf {z}}'(t)\right \|_{X}$ whenever $\gamma _{\mathbf {z}}'(t)$ exists, with the additional property that

(4.2) $$ \begin{align} \mathcal{H}^{1}\left(G\cap \gamma_{\mathbf{z}}\big((-\infty,b_{\mathbf{z}}]\big)\right)-s_{\mathbf{z}}>g(\mathbf{z})-\eta. \end{align} $$

If, in addition, $\mathbf {u}\in \mathbb {S}_{X}$ and $P(\mathbf {u})\ge \alpha $ , we define

(4.3) $$ \begin{align} \gamma_{\mathbf{z},\mathbf{u}}(t) =\begin{cases} \gamma_{\mathbf{z}}(t) & \text{if }t\leq b_{\mathbf{z}},\\ \gamma_{\mathbf{z}}(b_{\mathbf{z}})+(t-b_{\mathbf{z}})\mathbf{u} & \text{if }t>b_{\mathbf{z}}. \end{cases} \end{align} $$

In both cases we suppress $\eta $ in the notation although the objects we define also depend on $\eta $ . Note that $\gamma _{\mathbf {z},\mathbf {u}}(t)\in \operatorname {Lip}(\mathbb {R},X)$ and $P(\gamma _{\mathbf {z},\mathbf {u}}'(t))\ge \alpha \left \|\gamma _{\mathbf {z},\mathbf {u}}'(t)\right \|_{X}$ whenever $\gamma _{\mathbf {z},\mathbf {u}}'(t)$ exists. In particular, we have that $P\circ \gamma _{\mathbf {z},\mathbf {u}}$ is monotone increasing and hence

(4.4) $$ \begin{align} \gamma_{\mathbf{z},\mathbf{u}}\left((-\infty,b_{\mathbf{z}}]\right)\cap\gamma_{\mathbf{z},\mathbf{u}}\left([b_{\mathbf{z}},+\infty)\right)=\{\gamma_{\mathbf{z},\mathbf{u}}(b_{\mathbf{z}})\}. \end{align} $$

To prove (ii), consider an arbitrary $\mathbf {x}\in X$ . Since (ii) is trivially satisfied for $\mathbf {y}=\mathbf {0}_X$ , assume $\mathbf {y}\in \ker P\setminus \{\mathbf {0}_X\}$ . Fix any $\eta \in (0,1-\alpha )$ , an admissible triple $(\gamma _{\mathbf {x}},s_{\mathbf {x}},b_{\mathbf {x}})$ and $\beta \in (0,\infty )$ so that $\mathbf {u}:={\frac {\alpha }{1-\eta } \mathbf {v}_{P}+\beta \frac {\mathbf {y}}{\left \|\mathbf {y}\right \|_{X}}}\in \mathbb {S}_{X}$ noting

(4.5) $$ \begin{align} \beta \geq 1-\frac{\alpha}{1-\eta}. \end{align} $$

Consider the mapping ${\gamma }_{1}=\gamma _{\mathbf {x},\mathbf {u}}$ , as defined in (4.3), and let

$$ \begin{align*} {s}_{1}:=s_{\mathbf{x}}+\frac{\alpha\left\|\mathbf{y}\right\|_{X}}{(1-\eta)\beta},\qquad {b}_{1}:=b_{\mathbf{x}}+\frac{\left\|\mathbf{y}\right\|_{X}}{\beta}. \end{align*} $$

Observe that ${s}_{1}\geq 0$ , ${\gamma }_{1}({b}_{1})=\gamma _{\mathbf {x}}(b_{\mathbf {x}})+\frac {\alpha \left \|\mathbf {y}\right \|_{X} }{\beta (1-\eta )}\mathbf {v}_{P}+\mathbf {y} = \mathbf {x}+\mathbf {y}+s_1\mathbf {v}_{P}$ and $P(\mathbf {u})=\frac {\alpha }{1-\eta }>\alpha $ , thus we conclude that

$$ \begin{align*} g(\mathbf{x}+\mathbf{y})&\geq \mathcal{H}^{1}\left(G\cap{\gamma}_{1}\left((-\infty,{b}_{1}]\right)\right)-{s}_{1}\\ &\geq \mathcal{H}^{1}\left(G\cap\gamma_{\mathbf{x}}\left((-\infty,b_{\mathbf{x}}]\right)\right)-s_{\mathbf{x}}-\frac{\alpha\left\|\mathbf{y}\right\|_{X}}{(1-\eta)\beta}>g(\mathbf{x})-\eta-\frac{\alpha\left\|\mathbf{y}\right\|_{X}}{1-\eta-\alpha}, \end{align*} $$

using (4.2) and (4.5) for the last inequality. Letting $\eta \to 0$ , we obtain $g(\mathbf {x}+\mathbf {y})\geq g(\mathbf {x})-\frac {\alpha \left \|\mathbf {y}\right \|_{X}}{1-\alpha }$ and applying the above argument to the pair $\mathbf {\widetilde {x}}:=\mathbf {x}+\mathbf {y}$ and $\mathbf {\widetilde {y}}:=-\mathbf {y}$ in place of $\mathbf {x}$ and $\mathbf {y}$ delivers the reverse inequality $g(\mathbf {x})\geq g(\mathbf {x}+\mathbf {y})-\frac {\alpha \left \|\mathbf {y}\right \|_{X}}{1-\alpha }$ . This finishes the proof of (ii).

We now turn our attention to the two additional properties. For (A), fix $\mathbf {x}\in X$ and $r\geq 0$ . Let $\eta>0$ be arbitrary, and consider again $(\gamma _{\mathbf {x}},b_{\mathbf {x}},s_{\mathbf {x}})$ ; let ${\gamma }_{2}=\gamma _{\mathbf {x},\mathbf {v}_{P}}$ be defined by (4.3).

If $s_{\mathbf {x}}>r\ge 0$ , we note $\gamma _{\mathbf {x}}(b_{\mathbf {x}})=(\mathbf {x}+r\mathbf {v}_{P})+ (s_{\mathbf {x}}-r)\mathbf {v}_{P}$ , so by (4.2)

(4.6) $$ \begin{align} g(\mathbf{x}+r\mathbf{v}_{P})\geq \mathcal{H}^{1}\left(G\cap\gamma_{\mathbf{x}}((-\infty,b_{\mathbf{x}}])\right)-(s_{\mathbf{x}}-r)>g(\mathbf{x})-\eta+r\ge g(\mathbf{x})-\eta. \end{align} $$

If $0\le s_{\mathbf {x}}\leq r$ , consider ${b}_{2}=b_{\mathbf {x}}+(r-s_{\mathbf {x}})\ge b_{\mathbf {x}}$ and ${s}_{2}=0$ to deduce

(4.7) $$ \begin{align} {\gamma}_{2}({b}_{2}) =\gamma_{\mathbf{x}}(b_{\mathbf{x}})+(r-s_{\mathbf{x}})\mathbf{v}_{P} =\mathbf{x}+s_{\mathbf{x}}\mathbf{v}_{P}+(r-s_{\mathbf{x}})\mathbf{v}_{P} =\mathbf{x}+r\mathbf{v}_{P}+{s}_{2}\mathbf{v}_{P}. \end{align} $$

Therefore by (4.2)

(4.8) $$ \begin{align} g(\mathbf{x}+r\mathbf{v}_{P})&\geq \mathcal{H}^{1}\left(G\cap {\gamma}_{2}((-\infty,{b}_{2}])\right)-{s}_{2}=\mathcal{H}^{1}\left(G\cap {\gamma}_{2}((-\infty,{b}_{2}])\right)\nonumber\\ &\geq \mathcal{H}^{1}\left(G\cap {\gamma}_{2}((-\infty,{b}_{\mathbf{x}}])\right)-s_{\mathbf{x}} = \mathcal{H}^{1}\left(G\cap \gamma_{\mathbf{x}}((-\infty,{b}_{\mathbf{x}}])\right)-s_{\mathbf{x}}\ge g(\mathbf{x})-\eta. \end{align} $$

In either case, letting $\eta \to 0$ , we establish the first inequality of (A). To prove the second inequality of (A), let $\eta>0$ be arbitrary and consider $({\gamma }_{3},{s}_{3},{b}_{3})= (\gamma _{\mathbf {x}+r\mathbf {v}_{P}},s_{\mathbf {x}+r\mathbf {v}_{P}},b_{\mathbf {x}+r\mathbf {v}_{P}})$ . We observe ${\gamma }_{3}({b}_{3})=\mathbf {x}+(r+{s}_{3})\mathbf {v}_{P}$ , so using (4.2) we get

$$ \begin{align*} g(\mathbf{x})\geq \mathcal{H}^{1}\left(G\cap {\gamma}_{3}((-\infty,{b}_{3}])\right)-(r+{s}_{3})>g(\mathbf{x}+r\mathbf{v}_{P})-\eta-r, \end{align*} $$

which implies the second inequality of (A) when we let $\eta \to 0$ .

Assume $\mathbf {x},\mathbf {x}'\in X$ satisfy the conditions of (B). We may assume without loss of generality that $\mathbf {x}'-\mathbf {x}=r\mathbf {v}_{P}$ , where $r\geq 0$ . In light of (A), it is now enough to show $g(\mathbf {x}+r\mathbf {v}_{P})\ge g(\mathbf {x})+r$ . Let us again fix an arbitrary $\eta>0$ and consider $(\gamma _{\mathbf {x}},b_{\mathbf {x}},s_{\mathbf {x}})$ , ${\gamma }_{2}=\gamma _{\mathbf {x},\mathbf {v}_{P}}$ and $b_2=b_{\mathbf {x}}+(r-s_{\mathbf {x}})$ . If $s_{\mathbf {x}}>r$ , then (4.6) applies to give the desired inequality in the limit $\eta \to 0$ . If $0\le s_{\mathbf {x}}\le r$ , we may improve (4.8). Indeed, using the first line of (4.8) for the first inequality, followed by ${\gamma }_{2}\left ((b_{\mathbf {x}},{b}_{2}]\right )=[\mathbf {x}+s_{\mathbf {x}}\mathbf {v}_{P},\mathbf {x}+r\mathbf {v}_{P}]\subseteq [\mathbf {x},\mathbf {x}+r\mathbf {v}_{P}]\subseteq G$ and (4.4) for the equality, we conclude that

$$ \begin{align*} g(\mathbf{x}+r\mathbf{v}_{P}) \geq \mathcal{H}^{1}\left(G\cap {\gamma}_{2}((-\infty,{b}_{2}])\right) =\mathcal{H}^{1}\left(G\cap \gamma_{\mathbf{x}}((-\infty,{b}_{\mathbf{x}}])\right) +(r-s_{\mathbf{x}})>g(\mathbf{x})-\eta+r, \end{align*} $$

which implies $g(\mathbf {x}+r\mathbf {v}_{P})\geq g(\mathbf {x})+r$ if we let $\eta \to 0$ .

To prove part (iii), assume that $\mathbf {x},\mathbf {w}\in X$ are given, consider $\mathbf {y}:=\mathbf {w}-P(\mathbf {w})\mathbf {v}_{P}\in \ker P$ and note that $\left \|\mathbf {y}\right \|_{X}\leq 2\left \|\mathbf {w}\right \|_{X}$ . If $P(\mathbf {w})=0$ , then $\mathbf {w}=\mathbf {y}$ and, setting $\lambda (\mathbf {x},\mathbf {w}):=1$ , we have that the inequality of (iii) immediately follows from (ii). We may therefore assume $P(\mathbf {w})\ne 0$ . In this case we let

$$ \begin{align*} \lambda(\mathbf{x},\mathbf{w}):= \frac{g(\mathbf{x}+P(\mathbf{w})\mathbf{v}_{P})-g(\mathbf{x})}{P(\mathbf{w})}, \end{align*} $$

and observe that $\lambda (\mathbf {x},\mathbf {w})\in [0,1]$ , as $g|_{\mathbf {x}+\mathbb {R}\mathbf {v}_{P}}$ is a $1$ -Lipschitz increasing function by (A), with $\lambda (\mathbf {x},\mathbf {w})=1$ if $\left [\mathbf {x},\mathbf {x}+\frac {P(\mathbf {w})}{\left \|P\right \|_{X^{\ast }}}\mathbf {v}_{P}\right ]\subseteq G$ , by (B). We now use (ii) to complete verification of (iii):

$$ \begin{align*} \left|g(\mathbf{x}+\mathbf{w})-g(\mathbf{x})-\lambda(\mathbf{x},\mathbf{w}) P(\mathbf{w})\right| = \left|g(\mathbf{x}+\mathbf{w})-g(\mathbf{x}+P(\mathbf{w})\mathbf{v}_{P})\right| \leq \frac{\alpha}{1-\alpha}\left\|\mathbf{y}\right\|_{X}\leq \frac{2\alpha}{1-\alpha}\left\|\mathbf{w}\right\|_{X}. \end{align*} $$

To establish (iv), we note that (iii) implies that for any $\mathbf {x},\mathbf {w}\in X$

$$ \begin{align*} \left|g(\mathbf{x}+\mathbf{w})-g(\mathbf{x})\right| \leq \frac{2\alpha}{1-\alpha}\left\|\mathbf{w}\right\|_{X}+\left|\lambda(\mathbf{x},\mathbf{w})\right| \left\|\mathbf{w}\right\|_{X} \leq \left(1+\frac{2\alpha}{1-\alpha}\right)\left\|\mathbf{w}\right\|_{X}.\\[-34pt] \end{align*} $$

Proof. The statement of the lemma is clear if $T=0$ , so assume, without loss of generality, that $T\ne 0$ . Also, although the proof below will work independently of whether E is an empty or a non-empty set, it might be worth mentioning that in case $E=\emptyset $ , it would be enough to take $H=\emptyset $ and $g\equiv 0$ .

Let $1\le l\leq \dim X$ denote the rank of T and $\mathbf {w}_{1},\ldots ,\mathbf {w}_{l}\in T(X)\subseteq Y$ be a basis of $T(X)$ for which

$$ \begin{align*}\max_{1\leq i\leq l}\left\|\sum_{i=1}^{j}\mathbf{w}_{i}^{\ast}\circ T(\cdot)\,\mathbf{w}_{i}\right\|_{\operatorname{op}}= \mathfrak{C}(T), \end{align*} $$

where we adopt the notation of Definition 4.2; let $T_{i}=\mathbf {w}_{i}^{\ast }\circ T\in X^*$ for each $1\le i\le l$ , so that

(4.9) $$ \begin{align} \sum_{i=1}^l T_i(\cdot)\mathbf{w}_i=T(\cdot). \end{align} $$

Let $U_0$ be an open set given by Remark C.4, satisfying $E\subseteq U_0\subseteq \overline {U_0}\subseteq U$ and $\partial U_0$ has Lebesgue measure zero. For each $i=1,\ldots ,l$ we will construct sequences of smooth functions $\varphi _{k}^{(i)}\colon X\to \mathbb {R}$ , positive numbers $\varepsilon _{k}^{(i)}$ , sets $G_{k}^{(i)}\subseteq X$ and Lipschitz functions $g_{k}^{(i)}\colon X\to \mathbb {R}$ respectively, as well as positive integers $K_{i}\in \mathbb {N}$ and open sets $U_{i}\subseteq X$ such that the following conditions hold for each $i=1,2,\ldots ,l$ :

1. (A) $(\varphi _{k}^{(i)})_{k\in \mathbb {N}}$ is a smooth, locally finite partition of unity with supports contained in $U_{i-1}$ .

2. (B) For each $k\ge 1$ , we have $\varepsilon _k^{(i)}\in (0,1)$ and

   $$ \begin{align*} \sum_{k\in\mathbb{N}}\frac{\varepsilon_{k}^{(i)}\left(1+\operatorname{Lip}(\varphi_{k}^{(i)})\right)}{1-\varepsilon_{k}^{(i)}} \leq \frac{\theta}{4l\left(1+{\left\|T_{i}\right\|_{X^*}} \right)\left(1+\left\|\mathbf{w}_{i}\right\|_{Y}\right)}. \end{align*} $$

3. (C) For each $k\ge 1$ the set $G_{k}^{(i)}$ is open and satisfies

   $$ \begin{align*} E\subseteq G_{k}^{(i)}\subseteq U_{i-1} \end{align*} $$
   and
   $$ \begin{align*} \begin{aligned} &\sup\left\{\mathcal{H}^{1}\left(G^{(i)}_{k}\cap \gamma(\mathbb{R})\right)\colon \gamma\in \operatorname{Lip}(\mathbb{R},X),\,\right.\\ &\qquad\qquad\qquad \left.T_{i}\left(\gamma'(t)\right)\geq \varepsilon_{k}^{(i)}\left\|\gamma'(t)\right\|_{X}\left\|T_{i}\right\|_{X^{\ast}}\text{ whenever } \gamma'(t) \text{ exists}\right\}\leq \varepsilon_{k}^{(i)}. \end{aligned} \end{align*} $$

4. (D) For each $k\geq 1$ the function $g_{k}^{(i)}\colon X\to \mathbb {R}$ satisfies the following conditions:

   1. (D1) $g^{(i)}_{k}$ is Lipschitz and $0\leq g^{(i)}_{k}(\mathbf {x})\leq \left \|T_i\right \|_{X^*}\varepsilon _{k}^{(i)}$ for all $\mathbf {x}\in X$ ,

   2. (D2) For every $\mathbf {x},\mathbf {w}\in X$ and every $k\ge 1$ there exists $\lambda _{k}^{(i)}=\lambda _{k}^{(i)}(\mathbf {x},\mathbf {w})\in [0,1]$ such that

      $$ \begin{align*} \left|g_{k}^{(i)}(\mathbf{x}+\mathbf{w})-g_{k}^{(i)}(\mathbf{x})-\lambda_{k}^{(i)} T_{i}(\mathbf{w})\right|\leq \frac{2\varepsilon_{k}^{(i)}\left\|T_{i}\right\|_{X^{\ast}}}{1-\varepsilon_{k}^{(i)}}\left\|\mathbf{w}\right\|_{X}. \end{align*} $$

   3. (D3) Whenever $B(\mathbf {x},r)\subseteq G_{k}^{(i)}$ and $\left \|\mathbf {w}\right \|_{X}<r$ , the inequality (D2) is satisfied with $\lambda _{k}^{(i)}(\mathbf {x},\mathbf {w})=1$ .

5. (E) $K_{i}\in \mathbb {N}$ and $U_{i}$ is an open subset of X such that $\partial U_{i}$ has Lebesgue measure zero,

   $$ \begin{align*} \operatorname{supp}(\varphi_{k}^{(i)})\cap U_i=\emptyset \quad\text{for all } k>K_i\qquad\text{and}\qquad E\subseteq U_{i}\subseteq\overline{U_{i}}\subseteq U_{i-1}\cap \bigcap_{k=1}^{K_{i}}G_{k}^{(i)}. \end{align*} $$

Suppose that $1\le j\le l$ and that we have constructed the above listed objects of levels $i=1,\ldots ,j-1$ such that conditions (A)–(E) are satisfied for $i=1,\ldots ,j-1$ . In the case $j=1$ no objects are yet constructed and all conditions are vacuous and therefore satisfied.

We then proceed to construct the objects of level j as follows: First we choose the sequences $(\varphi _{k}^{(j)})_{k\in \mathbb {N}}$ , $(\varepsilon _{k}^{(j)})_{k\in \mathbb {N}}$ and $(G_{k}^{(j)})_{k\in \mathbb {N}}$ , in that order, arbitrarily subject to the conditions (A), (B) and (C) respectively for $i=j$ . To choose $G_{k}^{(j)}$ as in (C) we are using that compact purely $1$ -unrectifiable sets satisfy a stronger property proved in Theorem C.2 in Appendix C. If $1\leq j\leq l$ with $T_{j}=0$ , we let $g_{k}^{(j)}\colon X\to \mathbb {R}$ be defined as the constant $0$ function for every $k\in \mathbb {N}$ . Observe that all parts of (D) are then trivially satisfied for such j. For the remaining $1\leq j\leq l$ , those with $T_{j}\neq 0$ , we let $g_{k}^{(j)}\colon X\to \mathbb {R}$ for each $k\in \mathbb {N}$ be given by the conclusion of Lemma 4.1 for $G=G_{k}^{(j)}$ , $P=T_{j}$ and $\mathbf {v}_{T_j}=\mathbf {v}_{j}\in \mathbb {S}_{X}$ is any vector satisfying the condition $T_{j}(\mathbf {v}_{j})=\left \|T_{j}\right \|_{X^{\ast }}$ , and $\alpha =\varepsilon _{k}^{(j)}$ . Then Lemma 4.1 provides all of the stated properties in (D) for $i=j$ . In particular $\lambda _{k}^{(j)}(\mathbf {x},\mathbf {w})$ in (D2) can be chosen as $\lambda (\mathbf {x},\mathbf {w})$ from Lemma 4.1(iii). Note that $\lambda _{k}^{(j)}(\mathbf {x},\mathbf {w})=1$ if the additional conditions of (D3) are satisfied, since $\left \|\frac {T_i(\mathbf {w})}{\left \|T_i\right \|_{X^*}}\mathbf {v}_{j}\right \|_{X}\le \left \|\mathbf {w}\right \|_{X}$ for all $\mathbf {w}\in X$ . Finally, we let $K_{j}$ and $U_{j}$ with $\partial U_{j}$ of Lebesgue measure $0$ be the pair given by the conclusion of Lemma C.5 applied to X, E, $V=U_{j-1}$ , $(G_{k})_{k\in \mathbb {N}}:=(G_{k}^{(j)})_{k\in \mathbb {N}}$ and $(\varphi _{k})_{k\in \mathbb {N}}:=(\varphi _{k}^{(j)})_{k\in \mathbb {N}}$ . Such choice of $K_{j}$ and $U_{j}$ ensures that (E) is satisfied for $i=j$ . This completes the construction of all above mentioned objects for levels $i=1,\ldots ,l$ so that conditions (A)–(E) are satisfied for $i=1,\ldots ,l$ .

We define the mapping $g\colon X\to T(X)$ by

(4.10) $$ \begin{align} g(\mathbf{x})=\sum_{i=1}^{l}\sum_{k\in\mathbb{N}}\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\mathbf{w}_{i} \end{align} $$

and put $H:=U_{l}$ . Then we have

(4.11) $$ \begin{align} \operatorname{supp} g \subseteq U_0\subseteq U\qquad\text{and}\qquad Dg(\mathbf{x})=0 \quad\text{for all } \mathbf{x}\in X\setminus \overline{U_{0}}\supseteq X\setminus U. \end{align} $$

due to (A) and the fact, coming from (E), that all $U_{i}$ ’s are contained in $U_{0}\subseteq \overline {U_{0}}\subseteq U$ . Moreover,

$$ \begin{align*} \left\|g(\mathbf{x})\right\|_{Y}\leq \sum_{i=1}^{l}\sum_{k\in\mathbb{N}}\left|g_{k}^{(i)}(\mathbf{x})\right|\left\|\mathbf{w}_{i}\right\|_{Y}\leq\sum_{i=1}^{l}\sum_{k\in\mathbb{N}}\left\|T_i\right\|_{X^*}\varepsilon_{k}^{(i)}\left\|\mathbf{w}_{i}\right\|_{Y}\leq \theta \end{align*} $$

by the inequalities (D1) and (B). This establishes (a) and (b). Part (c) is clear from the choice $H=U_{l}$ , (E) and $\overline {U_0}\subseteq U$ .

In order to show that g is Lipschitz, we argue first that g is a locally Lipschitz mapping. Fix an arbitrary $\mathbf {x}\in X$ . For each $i=1,\dots ,l$ the collection $(\varphi _{k}^{(i)})_{k\in \mathbb {N}}$ forms a locally finite partition of unity, hence there is an open ball $B=B(\mathbf {x},r)$ and an index $n\in \mathbb {N}$ such that $\sum _{k\in \mathbb {N}}\varphi _{k}^{(i)}(\mathbf {y})g_{k}^{(i)}(\mathbf {y})=\sum _{k=1}^{n}\varphi _{k}^{(i)}(\mathbf {y})g_{k}^{(i)}(\mathbf {y}) $ for all $\mathbf {y}\in B$ and all $1\le i\le l$ . Since by (A) all $\varphi _{k}^{(i)}$ are Lipschitz and bounded on B, and by (D1) each $g_{k}^{(i)}$ is a Lipschitz bounded function as well, we conclude that $g|_B$ is Lipschitz too. We have thus established that g is locally Lipschitz on X.

We now derive bounds on the norm in Y of vectors of the form $g(\mathbf {x}+\mathbf {z})-g(\mathbf {x})$ , aiming to get an estimate with the Lipschitz constant given in (e). We will approximate this vector closely with an appropriate linear mapping evaluated at $\mathbf {z}$ . The appropriate linear mapping to use will be determined by which sets in the nested sequence $U_{0}\supseteq U_{1}\supseteq \ldots \supseteq U_{l}=H$ contain the segment $[\mathbf {x},\mathbf {x}+\mathbf {z}]$ .

For any $\mathbf {x},\mathbf {z}\in X$ and any $\lambda \in \mathbb {R}$ we may use (D1) and (A) to write

(4.12) $$ \begin{align} &\left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)-\lambda\varphi_{k}^{(i)}(\mathbf{x})T_{i}(\mathbf{z})\right|\notag\\ &\leq \left|\varphi_{k}^{(i)}(\mathbf{x})\right|\left|\left(g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-g_{k}^{(i)}(\mathbf{x})\right)-\lambda T_{i}(\mathbf{z})\right|+\left|g_{k}^{(i)}(\mathbf{x}+\mathbf{z})\right|\left|\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})\right|\notag\\ &\leq \left|\left(g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-g_{k}^{(i)}(\mathbf{x})\right)-\lambda T_{i}(\mathbf{z})\right|+\left\|T_i\right\|_{X^*}\varepsilon^{(i)}_{k}\operatorname{Lip}(\varphi^{(i)}_{k})\left\|\mathbf{z}\right\|_{X}. \end{align} $$

Observe that the set $U_0\setminus \bigcup _{j=1}^{l}\partial U_{j}$ may be partitioned as a union of open sets

$$ \begin{align*} U_0\setminus \bigcup_{j=1}^{l}\partial U_{j}=H\cup\bigcup_{j=1}^{l}(U_{j-1}\setminus \overline{U_{j}}). \end{align*} $$

Therefore, to estimate the Lipschitz constant of g locally at every point of this set, we may consider an arbitrary $\mathbf {x}\in U_0\setminus \bigcup _{j=1}^{l}\partial U_{j}$ and distinguish two cases:

(4.13) $$ \begin{align} \mathbf{x}\in U(\mathbf{x}):=H=U_{l}\quad \text{ or }\quad \mathbf{x}\in U(\mathbf{x}):= U_{j-1}\setminus \overline{U_{j}}\text{ for some } j\in\left\{1,\ldots,l\right\}. \end{align} $$

Let $r=r(\mathbf {x})>0$ be sufficiently small so that $B_{X}(\mathbf {x},r)\subseteq U(\mathbf {x})$ and let $\mathbf {z}\in B_{X}(\mathbf {0}_X,r)$ be arbitrary. In the former case of (4.13), we have $\mathbf {x}\in B_X(\mathbf {x},r)\subseteq H\subseteq U_i\subseteq G_k^{(i)}$ , for every $i=1,\ldots ,l$ and $1\le k\le K_i$ , by (E), so we may apply (D3) and (4.12) with $\lambda =1$ to derive

$$ \begin{align*} &\left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)-\varphi_{k}^{(i)}(\mathbf{x})T_{i}(\mathbf{z})\right|\nonumber\\ &\qquad\qquad \leq\left(\frac{2\varepsilon_{k}^{(i)}\left\|T_{i}\right\|_{X^{\ast}}}{1-\varepsilon_{k}^{(i)}}+\left\|T_i\right\|_{X^*}\varepsilon_{k}^{(i)}\operatorname{Lip}(\varphi_{k}^{(i)})\right)\left\|\mathbf{z}\right\|_{X} \text{ for every }1\le i\le l\text{ and }1\le k\le K_i. \end{align*} $$

We also recall that for every $1\le i\le l$ and $k>K_i$ , (E) implies

$$ \begin{align*} \left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)-\varphi_{k}^{(i)}(\mathbf{x})T_{i}(\mathbf{z})\right|=0 , \end{align*} $$

since both $\mathbf {x},\mathbf {x}+\mathbf {z}\in H\subseteq U_i$ . Multiplying the expressions under the modulus sign on the left-hand side by $\mathbf {w}_i$ and summing over respective ranges for $k\ge 1$ and then over $i=1,\ldots ,l$ we obtain, according to (4.10), (4.9) and $\sum _{k=1}^\infty \varphi _{k}^{(i)}(\mathbf {x})=1$ ,

$$ \begin{align*} \left\|g(\mathbf{x}+\mathbf{z})-g(\mathbf{x})-T(\mathbf{z})\right\|_{Y}&\leq\sum_{i=1}^{l}\sum_{k=1}^{K_i}\left(\frac{2\varepsilon_{k}^{(i)}\left\|T_{i}\right\|_{X^{\ast}}}{1-\varepsilon_{k}^{(i)}}+\left\|T_i\right\|_{X^*}\varepsilon_{k}^{(i)}\operatorname{Lip}(\varphi_{k}^{(i)})\right)\left\|\mathbf{w}_{i}\right\|_{Y}\left\|\mathbf{z}\right\|_{X}\\ &\leq \theta\left\|\mathbf{z}\right\|_{X}, \end{align*} $$

where the last inequality holds due to (B). Using that $g\colon X\to T(X)$ is a locally Lipschitz mapping between finite-dimensional spaces, and hence $Dg(\mathbf {u})$ exists for almost all $\mathbf {u}\in X$ , and the fact that the inequality above has been obtained for an arbitrary pair of $\mathbf {x}\in H=U_{l}$ and $\mathbf {z}\in B_{X}(\mathbf {0}_X,r)$ , we establish (d). We also note that the last inequality implies

(4.14) $$ \begin{align} \left\|g(\mathbf{x}+\mathbf{z})-g(\mathbf{x})\right\|_{Y}\leq \left(\left\|T\right\|_{\operatorname{op}}+\theta\right)\left\|\mathbf{z}\right\|_{X} \end{align} $$

in the first case from (4.13).

In the remaining case from (4.13), $\mathbf {x}\in U_{j-1}\setminus \overline {U_{j}}$ , $1\le j\le l$ , note that $\mathbf {x},\mathbf {x}+\mathbf {z}\in B_{X}(\mathbf {x},r)\subseteq U_{j-1}\setminus \overline {U_{j}}$ . Then, by (A) and (E), we have that $\varphi ^{(i)}_{k}(\mathbf {x})=\varphi ^{(i)}_{k}(\mathbf {x}+\mathbf {z})=0$ for all $i\geq j+1$ and $k\ge 1$ . Therefore,

(4.15) $$ \begin{align} \left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)\right|=0 \qquad\text{for all } i\geq j+1 \text{ and } k\ge1. \end{align} $$

For $i=j$ and each $k\in \mathbb {N}$ we may consider the quantity $\lambda ^{(j)}_{k}=\lambda ^{(j)}_{k}(\mathbf {x},\mathbf {z})$ from (D2) and apply inequality (D2) in (4.12) with $\lambda =\lambda ^{(j)}_{k}$ to get, for all $k\ge 1$ ,

(4.16) $$ \begin{align} \left|\left(\varphi_{k}^{(j)}(\mathbf{x}+\mathbf{z})g_{k}^{(j)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(j)}(\mathbf{x})g_{k}^{(j)}(\mathbf{x})\right)\right.&-\left.\lambda^{(j)}_{k}\varphi_{k}^{(j)}(\mathbf{x})T_{j}(\mathbf{z})\right|\nonumber\\ &\leq \left(\frac{2\varepsilon^{(j)}_{k}\left\|T_{j}\right\|_{X^{\ast}}}{1-\varepsilon^{(j)}_{k}}+\left\|T_j\right\|_{X^*}\varepsilon_{k}^{(j)}\operatorname{Lip}(\varphi^{(j)}_{k})\right)\left\|\mathbf{z}\right\|_{X}. \end{align} $$

For $i=1,\ldots ,j-1$ we again use $\left \|\mathbf {z}\right \|_{X}<r$ and $[\mathbf {x},\mathbf {x}+\mathbf {z}]\subseteq B_{X}(\mathbf {x},r)\subseteq U_{j-1}\subseteq U_i\subseteq \bigcap _{k=1}^{K_{i}}G^{(i)}_{k}$ by (E), to conclude, by (D3) and (4.12) with $\lambda =1$ , that

(4.17) $$ \begin{align} &\left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)-\varphi_{k}^{(i)}(\mathbf{x})T_{i}(\mathbf{z})\right|\nonumber\\ &\qquad\leq\left(\frac{2\varepsilon_{k}^{(i)}\left\|T_{i}\right\|_{X^{\ast}}}{1-\varepsilon_{k}^{(i)}}+\left\|T_i\right\|_{X^*}\varepsilon_{k}^{(i)}\operatorname{Lip}(\varphi_{k}^{(i)})\right)\left\|\mathbf{z}\right\|_{X} \text{ for } i=1,\ldots,j-1 \text{ and } k=1,\ldots,K_{i}. \end{align} $$

Moreover, from (E) it follows that $\varphi ^{(i)}_{k}|_{U_{j-1}}$ is constant $0$ for each $i=1,\ldots ,j-1$ and $k>K_{i}$ . Therefore

(4.18) $$ \begin{align} \left|\left(\varphi_{k}^{(i)}(\mathbf{x}+\mathbf{z})g_{k}^{(i)}(\mathbf{x}+\mathbf{z})-\varphi_{k}^{(i)}(\mathbf{x})g_{k}^{(i)}(\mathbf{x})\right)-\varphi_{k}^{(i)}(\mathbf{x})T_{i}(\mathbf{z})\right|&=0 \nonumber\\ &\quad \text{for } i=1,\ldots,j-1 \text{ and } k>K_{i}. \end{align} $$

Setting $\lambda ^{(j)}(\mathbf {x},\mathbf {z}):=\sum _{k\in \mathbb {N}}\lambda ^{(j)}_{k}(\mathbf {x},\mathbf {z})\varphi _{k}^{(j)}(\mathbf {x})\in [0,1]$ and using $\sum _{k=1}^{\infty }\varphi _k^{(i)}(\mathbf {x})=1$ for each $1\le i\le j-1$ , from (A), we may now multiply the expressions under the modulus sign on the left-hand sides of (4.15), (4.16), (4.17) and (4.18) by $\mathbf {w}_i$ and sum over respective ranges of $k\ge 1$ and then over $i=1,\ldots ,l$ , to obtain

$$ \begin{align*} \left\|\vphantom{\sum_{i=1}^{j-1}} g(\mathbf{x}+\mathbf{z})-g(\mathbf{x}) -\right.&\left.\left(\sum_{i=1}^{j-1}T_{i}(\mathbf{z})\mathbf{w}_{i}+\lambda^{(j)}(\mathbf{x},\mathbf{z})T_{j}(\mathbf{z})\mathbf{w}_{j}\right)\right\|_{Y}\\ &\qquad \leq \sum_{i=1}^{j}\sum_{k=1}^{\infty}\left(\frac{2\varepsilon_{k}^{(i)}\left\|T_{i}\right\|_{X^{\ast}}}{1-\varepsilon_{k}^{(i)}}+\left\|T_i\right\|_{X^*}\varepsilon_{k}^{(i)}\operatorname{Lip}(\varphi_{k}^{(i)})\right)\left\|\mathbf{w}_{i}\right\|_{Y}\left\|\mathbf{z}\right\|_{X}\leq \theta\left\|\mathbf{z}\right\|_{X}, \end{align*} $$

where the last inequality holds due to (B). Therefore, in the latter case of (4.13),

(4.19) $$ \begin{align} \left\|g(\mathbf{x}+\mathbf{z})-g(\mathbf{x})\right\|_{Y}&\leq \left(\left\|\sum_{i=1}^{j-1}T_{i}(\cdot)\,\mathbf{w}_{i}+\lambda^{(j)}(\mathbf{x},\mathbf{z})T_{j}(\cdot)\,\mathbf{w}_{j}\right\|_{\operatorname{op}}+\theta\right)\left\|\mathbf{z}\right\|_{X} \nonumber \\ &\leq \left(\max\left\{\left\|\sum_{i=1}^{j-1}T_{i}(\cdot)\,\mathbf{w}_{i}\right\|_{\operatorname{op}},\left\|\sum_{i=1}^{j}T_{i}(\cdot)\,\mathbf{w}_{i}\right\|_{\operatorname{op}}\right\}+\theta\right)\left\|\mathbf{z}\right\|_{X} \nonumber\\ &\leq(\mathfrak{C}(T)+\theta)\left\|\mathbf{z}\right\|_{X} , \end{align} $$

where the penultimate inequality is due to $\lambda ^{(j)}(\mathbf {x},\mathbf {z})\in [0,1]$ and the convexity of the $\left \|\cdot \right \|_{\operatorname {op}}$ norm. Combining (4.14) and (4.19), see also Remark 4.3(ii), we conclude that the inequality

$$ \begin{align*} \left\|g(\mathbf{x}+\mathbf{z})-g(\mathbf{x})\right\|_{Y}\leq (\mathfrak{C}(T)+\theta)\left\|\mathbf{z}\right\|_{X} \end{align*} $$

holds in all cases from (4.13) covering all $\mathbf {x}\in U_0\setminus \bigcup _{j=1}^{l}\partial U_{j}$ and $\mathbf {z}\in B(\mathbf {0}_X,r(\mathbf {x}))$ . Together with (4.11), this proves that g is locally $(\mathfrak {C}(T)+\theta )$ -Lipschitz on the complement of the closed Lebesgue null set $\bigcup _{j=0}^{l}\partial {U_{j}}$ , hence $\left \|Dg(\mathbf {x})\right \|_{\operatorname {op}}\le \mathfrak {C}(T)+\theta $ for almost all $\mathbf {x}\in X$ . Since we have already established that g is also locally Lipschitz on X, we deduce (e), that $g\colon X\to Y$ is a $(\mathfrak {C}(T)+\theta )$ -Lipschitz mapping; see Corollary A.2.

Proof. The statement of the lemma is trivial if $W=\{\mathbf {0}_{Y}\}$ . Also, if the open set $U:=\{\mathbf {x}\in V\colon \xi (\mathbf {x})>0\}$ is empty, then $Dg$ coincides, almost everywhere on V, with a continuous mapping $\psi T$ by (4.20). Therefore $g\in C^1(V,Y)$ and $Dg(\mathbf {x})=\psi (\mathbf {x})T$ for all $\mathbf {x}\in V$ (see Theorem B.3), so $f:=g$ satisfies conditions (a)–(d). Hence assume, without loss of generality, that $W\ne \{\mathbf {0}_{Y}\}$ and $U\ne \emptyset $ .

Let $\mathbf {w}_{1},\ldots ,\mathbf {w}_{l}\in Y$ be a basis of W, such that $\left \|\mathbf {w}_i\right \|_{Y}=1$ for all $1\le i\le l$ and $\mathbf {w}_{1}^{\ast },\ldots ,\mathbf {w}_{l}^{\ast }\in W^*$ be the corresponding biorthogonal functionals; let $g_i=\mathbf {w}_i^*\circ g$ , so that $g(\mathbf {x})=\sum _{i=1}^lg_i(\mathbf {x})\mathbf {w}_i$ for all $\mathbf {x}\in Q$ . We also fix a basis of X and use this to identify X with $\mathbb {R}^{\dim X}$ in the standard way. This identification allows us to define the Lebesgue measure on X. Accordingly all integrals on subsets of X which appear in this proof should be understood via this identification. Let also $C>0$ be a constant of equivalence between the norm $\left \|\cdot \right \|_{X}$ and the Euclidean norm $\left \|\cdot \right \|_{\text {E}}$ on $\mathbb {R}^{\dim X}=X$ so that for all $\mathbf {z}\in X$

(4.21) $$ \begin{align} \frac1C\left\|\mathbf{z}\right\|_{\text{E}}\leq\left\|\mathbf{z}\right\|_{X}\le C\left\|\mathbf{z}\right\|_{\text{E}}. \end{align} $$

Let $\rho \colon X\to [0,\infty )$ denote the standard smooth (Euclidean) mollifier in $\mathbb {R}^{\dim X}=X$ and for $\varepsilon>0$ let $\rho _{\varepsilon }(\mathbf {x}){:=} \varepsilon ^{-\dim X}\rho (\mathbf {x}/\varepsilon )$ . In what follows we consider, for each $\mathbf {x}\in U$ and $\varepsilon \in (0,\varepsilon (\mathbf {x}))$ , where $\varepsilon (\mathbf {x})>0$ is such that $\mathbf {x}+\mathbf {y}\in U$ for any $\left \|\mathbf {y}\right \|_{\text {E}}\le \varepsilon (\mathbf {x})$ , the convolution

$$ \begin{align*} g* \rho_{\varepsilon}(\mathbf{x})= \sum_{i=1}^l(g_i*\rho_{\varepsilon})(\mathbf{x})\mathbf{w}_i=\sum_{i=1}^l\left(\int_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon}g_i(\mathbf{x-y})\rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y}\right)\mathbf{w}_i, \end{align*} $$

where the integration is with respect to the Lebesgue measure. Note that $g*\rho _{\varepsilon }(\mathbf {x})\in W$ for any $\varepsilon \in (0,\varepsilon (\mathbf {x}))$ . We also let ${U}_{\varepsilon }=\left \{\mathbf {x}\in U\colon \operatorname {dist}_{\left \|\cdot \right \|_{\text {E}}}(\mathbf {x},X\setminus U)> \varepsilon \right \}$ , for $\varepsilon>0$ .

Note that for any $\varepsilon>0$ the convolution $g*\rho _{\varepsilon }$ is defined on ${U}_{\varepsilon }$ , belongs to $\operatorname {Lip}({U}_{\varepsilon },Y)\cap C^1({U}_{\varepsilon },Y)$ and approximates g well: for any $\mathbf {x}\in {U}_{\varepsilon }$

(4.22) $$ \begin{align} \left\|g* \rho_{\varepsilon}(\mathbf{x})-g(\mathbf{x})\right\|_{Y} &= \left\|\sum_{i=1}^l (g_i*\rho_{\varepsilon}(\mathbf{x})-g_i(\mathbf{x}))\mathbf{w}_i \right\|_{Y} \nonumber \\ &= \left\|\sum_{i=1}^l \left(\int_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon}\left( g_i(\mathbf{x}-\mathbf{y})-g_i(\mathbf{x})\right)\rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y}\right)\mathbf{w}_i \right\|_{Y} \nonumber\\ &\le \sum_{i=1}^{l}\int_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon}\left|g_{i}(\mathbf{x}-\mathbf{y})-g_{i}(\mathbf{x})\right|\rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y} \leq l\max_{1\le i\le l}\left\|\mathbf{w}_i^*\right\|_{X^*}C\operatorname{Lip}(g)\varepsilon. \end{align} $$

We note, for future reference, that using $D(g_i*\rho _\varepsilon )(\mathbf {x}) =(Dg_i*\rho _\varepsilon )(\mathbf {x})$ and (4.20), we have for all $\varepsilon>0$ , $\mathbf {x}\in {U}_{\varepsilon }$ and $\left \|\mathbf {v}\right \|_{X}\le 1$

(4.23) $$ \begin{align} &{D(g* \rho_{\varepsilon})(\mathbf{x})(\mathbf{v})-\psi(\mathbf{x})T(\mathbf{v})} ={ \sum_{i=1}^{l}D(g_i*\rho_\varepsilon)(\mathbf{x})(\mathbf{v})\mathbf{w}_i-\psi(\mathbf{x})T(\mathbf{v}) } \\ & ={ \sum_{i=1}^{l}(Dg_i*\rho_\varepsilon)(\mathbf{x})(\mathbf{v})\mathbf{w}_i-\psi(\mathbf{x})T(\mathbf{v}) }\notag ={ \int\limits_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon}\left( Dg(\mathbf{x}-\mathbf{y})(\mathbf{v})-\psi(\mathbf{x})T(\mathbf{v})\right)\rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y} }. \notag \end{align} $$

We now use (4.20) to estimate the norm of (4.23) from above as

(4.24) $$ \begin{align} &\left\|D(g* \rho_{\varepsilon})(\mathbf{x})(\mathbf{v})-\psi(\mathbf{x})T(\mathbf{v})\right\|_{Y}\notag\\ &\le\int\limits_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon} \left( \left\|Dg(\mathbf{x}-\mathbf{y})-\psi(\mathbf{x}-\mathbf{y})T\right\|_{\operatorname{op}} + \left|\psi(\mathbf{x}-\mathbf{y})-\psi(\mathbf{x})\right| \left\|T\right\|_{\operatorname{op}} \right) \rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y} \notag \\ &\leq \xi(\mathbf{x})+ \int\limits_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon} \left(\left| \xi(\mathbf{x}-\mathbf{y})-\xi(\mathbf{x}) \right| +\left|\psi(\mathbf{x}-\mathbf{y})-\psi(\mathbf{x})\right|\left\|T\right\|_{\operatorname{op}}\right) \rho_{\varepsilon}(\mathbf{y})\,d\mathbf{y} \notag \\ &\le \xi(\mathbf{x})+ \sup_{\left\|\mathbf{y}\right\|_{\text{E}}\le\varepsilon} \left(\left| \xi(\mathbf{x}-\mathbf{y})-\xi(\mathbf{x}) \right| +\left|\psi(\mathbf{x}-\mathbf{y})-\psi(\mathbf{x})\right|\left\|T\right\|_{\operatorname{op}}\right). \end{align} $$

Moreover, for any compact set $\emptyset \ne K\subseteq U$ , we may use that $\xi $ and $\psi $ are continuous on U and $\xi>0$ on U to choose $\delta _{K}>0$ sufficiently small so that for all $\varepsilon \in (0,\delta _{K})$ we have $K\subseteq {U}_{\varepsilon }$ and

$$ \begin{align*} \sup_{\mathbf{x}\in K}\sup_{\left\|\mathbf{y}\right\|_{E}\le\varepsilon}\left(\left| \xi(\mathbf{x}-\mathbf{y})-\xi(\mathbf{x}) \right| +\left|\psi(\mathbf{x}-\mathbf{y})-\psi(\mathbf{x})\right|\left\|T\right\|_{\operatorname{op}}\right)\leq \frac{\theta}{2}\min_{\mathbf{z}\in K}\xi(\mathbf{z}). \end{align*} $$

Combining this with (4.24) we get

(4.25) $$ \begin{align} \left\|D(g* \rho_{\varepsilon})(\mathbf{x})(\mathbf{v})-\psi(\mathbf{x})T(\mathbf{v})\right\|_{Y}\leq \xi(\mathbf{x})\left(1+\frac{\theta}{2}\right) \end{align} $$

for any compact $K\subseteq U$ , $\mathbf {x}\in K$ and $\varepsilon \in (0,\delta _{K})$ .

Let $(\varphi _{k})_{k\in \mathbb {N}}$ be a smooth, locally finite partition of unity on $U=\{\mathbf {x}\in V\colon \xi (\mathbf {x})>0\}$ and for each $k\in \mathbb {N}$ set

(4.26) $$ \begin{align} \theta_{k}&:=\frac{\theta\min_{\mathbf{z}\in \operatorname{supp}\varphi_{k}}\xi(\mathbf{z})}{2^{k}\left(1+\operatorname{Lip}(\varphi_{k})\right)},\quad \varepsilon_{k}:=\frac{1}{2}\min\left\{\delta_{\operatorname{supp} \varphi_{k}}, \frac{\theta_{k}}{l w C\left(\operatorname{Lip}(g)+1\right)}\right\},\quad A_{k}:=U_{\varepsilon_{k}},\nonumber\\ h_{k}(\mathbf{x})&=\begin{cases} g* \rho_{\varepsilon_{k}}(\mathbf{x}) & \text{if }\mathbf{x}\in A_{k},\\ \mathbf{0}_{Y} & \text{if }\mathbf{x}\in U\setminus A_{k} \end{cases}\quad \text{so that }h_{k}\in\operatorname{SLA}(g,A_{k},Y,\theta_k), \end{align} $$

where C is fixed in (4.21), $w=\max _{1\le i\le l}\left \|\mathbf {w}_i^*\right \|_{X^*}$ and $\operatorname {SLA}(g,A_{k},Y,\theta _{k})$ is defined as the class of smooth Lipschitz mappings $A_{k}\to Y$ which approximate g uniformly within error $\theta _{k}$ ; a precise description of this class is given in Definition 5.2, and $h_k\in \operatorname {SLA}(g,A_{k},Y,\theta _k)$ follows from (4.22) and the choice of $\varepsilon _k$ above. The desired mapping f may now be defined by $\sum _{k\in \mathbb {N}}\varphi _{k}h_{k}$ in U and set equal to g in $Q\setminus U$ ; then a ‘smooth approximation result’ for mappings of this form will establish the remaining properties of f. We postpone this result, Lemma 5.3, until Section 5. We now describe precisely how to apply Lemma 5.3. Let $h:=g$ , $P(\mathbf {x}):=\psi (\mathbf {x})T\in \mathcal {L}(X,Y)$ and $\eta (\mathbf {x}):=\xi (\mathbf {x})\left (1+\frac {\theta }{2}\right )$ for each $\mathbf {x}\in U$ . The conditions of Lemma 5.3, including the additional condition of (ii), are now satisfied, by the definition of U and (4.25), and we may let $f{:=} \tilde {h}$ be given by the conclusion of Lemma 5.3. Then conditions (a), (b) of the present lemma and $f\in C^1(U,Y)$ , which is a part of (c), are satisfied. Since all values of g and of $h_k$ are in W, we get $f\colon Q\to W$ . Moreover, Lemma 5.3(ii), (4.26) and (4.25) give, for any $\mathbf {x}\in U$ ,

(4.27) $$ \begin{align} \left\|Df(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}}=\left\|D\tilde{h}(\mathbf{x})-P(\mathbf{x})\right\|_{\operatorname{op}} \leq \eta(\mathbf{x})+\sum_{k\in\mathbb{N}}\operatorname{Lip}(\varphi_{k})\mathbb{1}_{\operatorname{supp}\varphi_{k}}(\mathbf{x})\theta_{k}\leq \xi(\mathbf{x})\left(1+\theta\right), \end{align} $$

which proves the inequality of (d) for all $\mathbf {x}\in U$ . This implies, in particular, that for all $\mathbf {x}\in U$

$$ \begin{align*} \left\|Df(\mathbf{x})\right\|_{\operatorname{op}} \le \psi(\mathbf{x})\left\|T\right\|_{\operatorname{op}}+\xi(\mathbf{x}) \left(1+\theta\right)\le L:=\Psi\left\|T\right\|_{\operatorname{op}}+\Xi\left(1+\theta\right), \end{align*} $$

where $\Psi ,\Xi>0$ are upper bounds of bounded functions $\psi ,\xi $ respectively. Hence we conclude that f is locally L-Lipschitz on U, and so using (a), $g\in \operatorname {Lip}(Q,W)$ and Lemma A.3, we obtain $f\in \operatorname {Lip}(Q,W)$ . We also note from (4.27) that the mapping $\Phi \colon V\to \mathcal {L}(X,Y)$ defined by

$$ \begin{align*} \Phi(\mathbf{x})=\begin{cases} Df(\mathbf{x}) & \mathbf{x}\in U,\\ \psi(\mathbf{x})T & \mathbf{x}\in V\setminus U \end{cases} \end{align*} $$

is continuous on V, as $U=\{\mathbf {x}\in V\colon \xi (\mathbf {x})>0\}$ and $\xi $ is continuous on V. We may now apply Theorem B.1(ii) to Lipschitz mappings $f,g\colon V\subseteq X\to W$ between finite-dimensional spaces to conclude that $Df$ and $Dg$ exist and are equal almost everywhere on $\{\mathbf {x}\in V\colon f(\mathbf {x})=g(\mathbf {x})\}\supseteq V\setminus U$ . This, together with (4.20) and the definitions of $\Phi $ and U implies $Df(\mathbf {x})=\Phi (\mathbf {x})$ for almost every $\mathbf {x}\in V$ . Therefore, by Theorem B.3, $f\in C^1(V,Y)$ and $Df(\mathbf {x})=\Phi (\mathbf {x})$ for all $\mathbf {x}\in V$ . This verifies (c) and, together with (4.27), implies (d).

Proof. The statement of the lemma is clear for $T=0$ , so assume, without loss of generality, that $T\ne 0$ is such that $\left \|T\right \|_{\operatorname {op}}\le 1$ . We may do so as if $\left \|T\right \|_{\operatorname {op}}>1$ and the conclusion of this lemma holds for operators of norm less than or equal to $1$ , we let $f_1,\psi $ and H correspond to E, $\eta _1=\eta /\left \|T\right \|_{\operatorname {op}}$ , $\varphi $ and $T_1=T/\left \|T\right \|_{\operatorname {op}}$ and define $f=\left \|T\right \|_{\operatorname {op}}f_1$ .

Let $C=\mathfrak {C}(T)+5$ , where $\mathfrak {C}(T)$ is the constant given by Definition 4.2,

(4.28) $$ \begin{align} \theta:=\min\left\{1,\eta/5\right\} , \qquad k\in\mathbb{N}\cap\Big[\frac{C}{\theta},\frac{2C}{\theta}\Big], \end{align} $$

$G_{1}:=H_{1}:=B_{X}(E,\eta )$ and note that $E\subseteq H_{1}$ . For each $i=2,\ldots ,k-1$ , whenever the open set $H_{i-1}$ containing compact $E\cap \left \{\mathbf {x}\in X\colon \varphi (\mathbf {x})\geq \frac {i}{k}\right \}$ has been defined, let

(4.29) $$ \begin{align} G_{i}:=H_{i-1}\cap \left\{\mathbf{x}\in X\colon\varphi(\mathbf{x})>\frac{i-1}{k}\right\}. \end{align} $$

Next apply Lemma 4.4 to X, Y, the compact, purely $1$ -unrectifiable subset $E\cap \left \{\mathbf {x}\in X\colon \varphi (\mathbf {x})\geq i/k\right \}$ of the open set $U=G_{i}$ , $\theta $ and T to obtain a mapping $g_{i}\in \operatorname {Lip}_{C}(X,T(X))$ , as $C\geq \mathfrak {C}(T)+\theta $ , and an open set $H_{i}\subseteq X$ such that

1. (a) $\operatorname {supp} g_{i}\subseteq G_{i}$ and $Dg_i(\mathbf {x})=0$ for any $\mathbf {x}\notin G_i$ .

2. (b) $\sup _{\mathbf {x}\in X} \left \|g_{i}(\mathbf {x})\right \|_{Y}\leq \theta $ .

3. (c) $E\cap \left \{\mathbf {x}\in X\colon \varphi (\mathbf {x})\geq \frac {i}{k}\right \}\subseteq H_{i}\subseteq \overline {H_{i}}\subseteq G_{i}$ .

4. (d) $\left \|Dg_{i}(\mathbf {x})-T\right \|_{\operatorname {op}}\leq \theta $ for Lebesgue almost all $\mathbf {x}\in H_{i}$ .

Note that (c) implies that $E\cap \left \{\mathbf {x}\in X\colon \varphi (\mathbf {x})\geq \frac {i+1}{k}\right \}\subseteq H_i$ and hence we construct inductively mappings $g_{2},\ldots ,g_{k-1}\in \operatorname {Lip}_C(X,T(X))$ and nested sequences of open sets $H_{1}\supseteq \ldots \supseteq H_{k-1}$ and $G_{1}\supseteq \dots \supseteq G_{k-1}$ such that properties (a)–(d) hold for every $2\le i\le k-1$ . Consider $g:=\frac {1}{k}\sum _{i=2}^{k-1}g_{i}\in \operatorname {Lip}_C(X,T(X))$ . Properties (a) and (b) of $g_{2},\ldots ,g_{k-1}$ , and the nested property of $G_i$ lead to

(4.30) $$ \begin{align} \sup_{\mathbf{x}\in X}\left\|g(\mathbf{x})\right\|_{Y}\leq \theta, \end{align} $$

and

(4.31) $$ \begin{align} \operatorname{supp} g\subseteq G_{2}\subseteq \operatorname{supp}\varphi. \end{align} $$

Define $j\colon G_{1}\to \left \{1,\ldots ,k-1\right \}$ by

(4.32) $$ \begin{align} j(\mathbf{x})=\max\left\{j\in\left\{1,\ldots,k-1\right\}\colon \mathbf{x}\in G_{j}\right\} \end{align} $$

and note that for all $\mathbf {x}\in G_1$ we have $\varphi (\mathbf {x})\ge \frac {j(\mathbf {x})-1}{k}$ due to the non-negativity of $\varphi $ when $j(\mathbf {x})=1$ and to the definitions (4.29) and (4.32) when $j(\mathbf {x})>1$ . Observe also that whenever all $g_i$ are differentiable at $\mathbf {x}\in G_{1}$ ,

(4.33) $$ \begin{align} Dg(\mathbf{x})=\frac{1}{k}\sum_{i=2}^{j(\mathbf{x})}Dg_{i}(\mathbf{x}), \end{align} $$

because of (a), and the definitions of g and $j(\mathbf {x})$ . Let $\psi \colon X\to [0,1]$ be given by

(4.34) $$ \begin{align} \psi(\mathbf{x})= \begin{cases} \min\left\{\frac{j(\mathbf{x})+2}{k},\varphi(\mathbf{x})\right\},&\text{if }\mathbf{x}\in G_{1};\\ 0,&\text{otherwise}. \end{cases} \end{align} $$

Then the last inequality of (v) is satisfied. Note also for future reference that for every $\mathbf {x}\in G_{1}$ we have

(4.35) $$ \begin{align} \frac{j(\mathbf{x})-1}{k} \le\psi(\mathbf{x}) \le\frac{j(\mathbf{x})+2}{k}, \end{align} $$

using $\varphi (\mathbf {x})\ge \frac {j(\mathbf {x})-1}{k}$ . Define an open set $H\subseteq X$ by

$$ \begin{align*} H:= \bigcup_{j=1}^{k-1} S_j, \text{ where } S_j=\left\{\mathbf{x}\in H_{j}\colon \varphi(\mathbf{x})<\frac{j+2}{k}\right\}. \end{align*} $$

The second inclusion of (i), $H\subseteq B_X(E,\eta )=H_{1}$ , is now clear, due to the nested property of $H_i$ . We prove the first inclusion: Let $\mathbf {x}\in E$ . Then there is an $m\in \left \{0,\ldots ,k-1\right \}$ such that $\frac {m}{k}\leq \varphi (\mathbf {x})\leq \frac {m+1}{k}$ . If $m\leq 1$ , then since $\mathbf {x}\in E\subseteq H_1$ and $\varphi (\mathbf {x})\le \frac 2k<\frac 3k$ , we have $\mathbf {x}\in S_1$ ; for $2\le m\le k-1$ we have $\mathbf {x}\in E\cap \left \{\mathbf {x}\in X\colon \varphi (\mathbf {x})\geq m/k\right \}\subseteq H_{m}$ by (c) and $\varphi (\mathbf {x})\leq \frac {m+1}{k}<\frac {m+2}{k}$ so $\mathbf {x}\in S_m$ . This proves the first inclusion of (i).

We now verify (vi) and the remaining inequalities of (v). Note first that from $H_1=G_1$ , (4.32) and (c) we have, for any $1\leq j\leq k-1$ , that $\mathbf {x}\in H_{j}$ implies $j(\mathbf {x})\geq j$ . Fix any $\mathbf {x}\in H\subseteq G_{1}$ and $j\in \left \{1,\ldots ,k-1\right \}$ such that $\mathbf {x}\in S_{j}$ ; then $\varphi (\mathbf {x})< \frac {j+2}{k}\le \frac {j(\mathbf {x})+2}{k}$ thus, by (4.34), $\psi (\mathbf {x})=\varphi (\mathbf {x})$ proving (vi), which together with (4.34) gives the first two inequalities of (v).

Let now $\mathbf {x}\in G_{2}$ be any point such that all $g_i$ are differentiable at $\mathbf {x}$ and for any $1\le i\le k-1$ such that $\mathbf {x}\in H_i$ we also have the inequality (d). We remark that almost all points of $G_{2}$ are such. Thus, whenever $1\leq i\leq j(\mathbf {x})-1$ , we have, from (4.32) and (4.29), that $\mathbf {x}\in G_{j(\mathbf {x})}\subseteq H_{j(\mathbf {x})-1}\subseteq H_i$ , hence the inequality (d) applies. Therefore, we now verify that for almost all $\mathbf {x}\in G_2$ ,

(4.36) $$ \begin{align} \begin{aligned} &\left\|Dg(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}} =\left\|\tfrac{1}{k}\sum_{i=2}^{j(\mathbf{x})}Dg_{i}(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}} \\ &\le \tfrac{1}{k}\sum_{i={2}}^{j(\mathbf{x})-1}\left\|Dg_{i}(\mathbf{x})-T\right\|_{\operatorname{op}}+ \tfrac1k\left\|Dg_{j(\mathbf{x})}(\mathbf{x})\right\|_{\operatorname{op}}+\left|\tfrac{j(\mathbf{x})-2}{k}-\psi(\mathbf{x})\right|\left\|T\right\|_{\operatorname{op}} \\ &\leq \theta+\tfrac{C}{k}+\tfrac{4}{k} \le\tfrac{4C}{k} \le 4C\min\left\{\varphi(\mathbf{x}),\tfrac1k\right\}. \end{aligned} \end{align} $$

The first equality follows from (4.33); the first inequality of the last line is guaranteed by (d), $g_{j(\mathbf {x})}\in \operatorname {Lip}_C(X,T(X))$ and (4.35); finally, we use (4.28), $\mathbf {x}\in G_2$ and (4.29) for the remaining inequalities.

Using (a) of the present proof, the nested property of $G_i$ and (4.33), we conclude that $Dg(\mathbf {x})=0$ for every $\mathbf {x}\in X\setminus G_{2}$ . From (4.29) and the already verified (v) we also have that $0\le \psi (\mathbf {x})\le \varphi (\mathbf {x})\mathbb {1}_{H_1}(\mathbf {x})\le \frac 1k$ for all $\mathbf {x}\in X\setminus G_{2}$ . Hence, for every $\mathbf {x}\in X\setminus G_{2}$ we have $\left \|Dg(\mathbf {x})-\psi (\mathbf {x})T\right \|_{\operatorname {op}}=\psi (\mathbf {x})\left \|T\right \|_{\operatorname {op}}\le \psi (\mathbf {x})\le \min \left \{\varphi (\mathbf {x}),\frac 1k\right \}$ . This, together with (4.36) and $C>1$ , establishes

(4.37) $$ \begin{align} \left\|Dg(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}}\leq 4C\min\Big\{\varphi(\mathbf{x}),\frac1k\Big\} \quad\text{for almost all } \mathbf{x}\in X. \end{align} $$

Let $f\colon X\to T(X)\subseteq Y$ be the Lipschitz mapping given by the conclusion of Lemma 4.5 applied to X, Y, $Q=X$ , $V=H$ , T, $W=T(X)$ , $g\in \operatorname {Lip}(X,T(X))$ , continuous bounded functions $\psi |_H$ , $\xi (\mathbf {x})=4C\min \Big \{\varphi (\mathbf {x}),\frac 1k\Big \}$ for $\mathbf {x}\in H$ and $\frac {\theta }{4C}$ , where the validity of (4.20) is guaranteed by (4.37). Note that from Lemma 4.5(c) we have $f\in C^1(H,Y)$ , so (ii) is satisfied.

Combining Lemma 4.5(b) with (4.30) and (4.28), we deduce for every $\mathbf {x}\in Q=X$

$$ \begin{align*} \left\|f(\mathbf{x})\right\|_{Y} \leq \left\|f(\mathbf{x}) - g(\mathbf{x})\right\|_{Y} +\left\|g(\mathbf{x})\right\|_{Y} \le 2\theta \leq \eta. \end{align*} $$

Moreover, we deduce from Lemma 4.5(a) that $f(\mathbf {x})>0$ implies either $g(\mathbf {x})=f(\mathbf {x})>0$ or $\xi (\mathbf {x})>0$ . In both cases we conclude that $\mathbf {x}\in \operatorname {supp}(\varphi )$ , in the former case due to (4.31) and in the latter due to the definition of $\xi $ . This establishes (iii) of the present lemma. Finally, we verify the inequality of (iv) for almost every $\mathbf {x}\in X$ . First, for every $\mathbf {x}\in H$ , we may apply Lemma 4.5(d), $\xi (\mathbf {x})\le 4C/k$ and (4.28) to obtain

$$ \begin{align*} \left\|Df(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}} \le \xi(\mathbf{x})\left(1+\frac{\theta}{4C}\right)\le\xi(\mathbf{x})+\theta\le\frac{4C}k+\theta\le5\theta\le\eta. \end{align*} $$

Further, by Lemma 4.5(a) the set $X\setminus H$ is contained in the set where f and g, Lipschitz mappings $X\to T(X)$ between finite-dimensional spaces, coincide. Since $Df=Dg$ almost everywhere in the latter set (see Theorem B.1), we have $Df=Dg$ almost everywhere in $X\setminus H$ . Hence, by (4.37) and (4.28), we have for almost every $\mathbf {x}\in X\setminus H$

$$ \begin{align*} \left\|Df(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}} =\left\|Dg(\mathbf{x})-\psi(\mathbf{x})T\right\|_{\operatorname{op}} \le \frac{4C}{k} \le 4\theta\le\eta.\\[-38pt] \end{align*} $$

Proof. Because g is a $C^{1}$ smooth mapping and E is compact, we have that g is uniformly Fréchet differentiable on E; see Lemma C.3. In other words, we may choose $\delta \in (0,\eta )$ small enough so that for all $\mathbf {x}\in E$ and all $\mathbf {y}\in X$ with $\left \|\mathbf {y}\right \|_{X}\leq \delta $ one has $B(\mathbf {x},\delta )\subseteq Q$ and

$$ \begin{align*} \left\|g(\mathbf{x}+\mathbf{y})-g(\mathbf{x})-Dg(\mathbf{x})(\mathbf{y})\right\|_{Y}\leq \tfrac{\eta}{2}\left\|\mathbf{y}\right\|_{X}. \end{align*} $$

The conclusion of the lemma follows immediately.

Proof. Suppose $j\geq 1$ and an open set $H_{j-1}\supseteq E$ and a Lipschitz mapping $f_{j-1}\in \operatorname {Lip}(Q,Y)\cap C^1({H}_{j-1}, Y)$ have already been defined so that for $i=j-1$

(4.39) $$ \begin{align} &(f_{i}-f_{0})(\mathbf{x}) =\sum_{k=1}^i g_{k}(\mathbf{x}) \quad \text{for all }\mathbf{x}\in Q,\\ &\text{where } \sum_{k=1}^i g_{k}\colon X\to\operatorname{Span}\left(\bigcup_{k=1}^{i}T_{k}(X)\right) \text{\ is a Lipschitz mapping}. \notag \end{align} $$

Here we interpret an empty sum as zero, an empty union as the empty set and the linear span of the empty set as $\left \{0\right \}$ . Thus the conditions are met for $j=1$ . Then we choose

(4.40) $$ \begin{align} 0<\eta_{j}\leq \min\{2^{-j}\eta,\theta_{j}\} \end{align} $$

small enough so that $\overline {B}_{X}(E,\eta _{j})\subseteq H_{j-1}$ and apply Lemma 4.6 to X, Y, E, $\eta _j$ , $\varphi _j$ , and $T_{j}$ to get a mapping $g_{j}\colon X\to T_{j}(X)$ , a function $\psi _{j}\colon X\to [0,1]$ and an open set $H_{j}\subseteq X$ with the following properties:

1. (a) $E\subseteq H_{j}\subseteq B_{X}(E,\eta _{j})\subseteq \overline {B}_{X}(E,\eta _{j})\subseteq H_{j-1}$ .

2. (b) $g_{j}\in \operatorname {Lip}(X, Y)\cap C^1({H}_{j}, Y)$ .

3. (c) $\sup _{\mathbf {x}\in X}\left \|g_{j}(\mathbf {x})\right \|_{ Y}\leq \eta _{j}$ and $\operatorname {supp} g_{j}\subseteq \operatorname {supp} \varphi _{j}$ .

4. (d) $\left \|Dg_{j}(\mathbf {x})-\psi _j(\mathbf {x})T_{j}\right \|_{\operatorname {op}}\leq \eta _{j}$ for Lebesgue almost every $\mathbf {x}\in X$ .

5. (e) $0\leq \psi _{j}(\mathbf {x})\leq \mathbb {1}_{B_{X}(E,\eta _{j})}\varphi _{j}(\mathbf {x})$ for all $\mathbf {x}\in X$ and $\psi _{j}(\mathbf {x})=\varphi _{j}(\mathbf {x})$ for all $\mathbf {x}\in H_{j}$ .

Let $f_{j}:=f_{j-1}+g_{j}$ ; note that $f_{j}$ is defined on Q and that (4.39) extends to $i=j$ . Now we have that (i) and (ii) are implied by (a), (b) and (c), and (iii) follows from (e) and (a). Property (iv) of $f_{j}$ follows from (4.39), the finite-dimensionality of $X\supseteq Q$ and of all $T_{k}(X)$ in (4.39), Rademacher’s theorem and $f_{0}\in C^1(H_{0},Y)$ . Finally, we check (v). Let $j\in \mathbb {N}$ and $\mathbf {x}$ be any point from $H_{0}$ lying in the intersection of the full measure sets corresponding to the mappings $g_{1},\ldots ,g_{j}$ given by (d). Let $L\in \mathcal {L}(X, Y)$ be arbitrary. Then we have

$$ \begin{align*} \left\|Df_{j}(\mathbf{x})-L\right\|_{\operatorname{op}}&=\left\|Df_{0}(\mathbf{x})+\sum_{k=1}^{j}Dg_{k}(\mathbf{x})-L\right\|_{\operatorname{op}}\\ &\le\left\|Df_{0}(\mathbf{x})+\sum_{k=1}^{j}\psi_{k}(\mathbf{x})T_{k}-L\right\|_{\operatorname{op}}+\sum_{k=1}^{j}\left\|Dg_{k}(\mathbf{x})-\psi_{k}(\mathbf{x})T_{k}\right\|_{\operatorname{op}}\\ &\leq \left\|Df_{0}(\mathbf{x})+\sum_{k=1}^{j}\psi_{k}(\mathbf{x})T_{k}-L\right\|_{\operatorname{op}}+\eta, \end{align*} $$

where, to get the last inequality, we applied (d) and (4.40).

Proof. Choose $\rho>0$ so that $\overline {B}_{X}(E,\rho )\subseteq H$ . Next, exploit the uniform continuity of the partial derivatives of f on the compact set $\overline {B}_{X}(E,\rho )$ to find $\tau \in (0,\rho )$ such that

$$ \begin{align*} \mathbf{x},\mathbf{y}\in \overline{B}_{X}(E,\rho),\, \left\|\mathbf{y}-\mathbf{x}\right\|_{X}\leq 2\tau\qquad \Rightarrow\qquad \left\|Df(\mathbf{y})-Df(\mathbf{x})\right\|_{\operatorname{op}}\leq \zeta, \end{align*} $$

where

(4.41) $$ \begin{align} \zeta:=\min\left\{\frac{1-\max\left\{\operatorname{Lip}(f),\left\|T\right\|_{\operatorname{op}}\right\}}{3},\frac{\theta}{4}\right\}. \end{align} $$

Let $(\gamma _{k})_{k\in \mathbb {N}}$ be a locally finite, smooth partition of unity subordinated to the family $\left \{B_{X}(\mathbf {x},\tau )\colon \mathbf {x}\in E\right \}$ and for each $k\in \mathbb {N}$ choose $\mathbf {x}_{k}\in E$ such that $\operatorname {supp} \gamma _{k}\subseteq B_{X}(\mathbf {x}_{k},\tau )$ . Apply Lemma 4.8 to $X,\ Y,\ E,\ H_0=H,\ Q,\ f_{0}=f,\ \eta =\zeta $ , $T_{2k}=T$ , $T_{2k-1}=-Df(\mathbf {x}_{k})$ , $\varphi _{2k-1}=\varphi _{2k}=\gamma _{k}$ and $\theta _k=2^{-k}\theta $ for each $k\in \mathbb {N}$ to obtain sequences of open sets $(H_{j})_{j\in \mathbb {N}}$ , Lipschitz mappings $(f_{j}\colon Q\to Y)_{j\in \mathbb {N}}$ and functions $(\psi _{j}\colon X\to [0,1])_{j\in \mathbb {N}}$ . By Lemma 4.8(ii) and by the choice of $\theta _j$ , the sequence of Lipschitz mappings $(f_{j})_{j\in \mathbb {N}}$ converges, in $(C(Q),\left \|\cdot \right \|_{\infty })$ , to a continuous mapping $g\colon Q\to Y$ satisfying (ii) of the present lemma. For each $\mathbf {x}\in H$ and for each $j\ge 1$ we may write