One-point extension of a nonexpansive map in finite dimensions

If $n=0$, then $E=\{0\}$. If $D$ is empty choose $q=0$; if it is nonempty then $D=\{0\}$, $y=p=T(p)=0$, and $q=0$ satisfies the inequality. Hence assume $n\ge 1$. If $D=\mathrm{\varnothing }$, any $q$ works. Otherwise consider any finite nonempty list $p_1,\dots ,p_m\in D$ and set $h_i(q)=\mathrm{\parallel }q-T{p}_i{\mathrm{\parallel }}_2^2-\mathrm{\parallel }y-p_i{\mathrm{\parallel }}_2^2,H(q)={\max }_{1\le i\le m}{h}_i(q)\mathrm{.}$ By [F1], $\mathrm{\mid }\parallel u{\parallel }_2-\parallel v{\parallel }_2\mathrm{\mid }\le \parallel u-v{\parallel }_2=d_2(u,v)$, so each $h_i$ and their finite maximum $H$ are continuous. Also, when $\parallel q{\parallel }_2\ge \parallel T{p}_i{\parallel }_2$, $h_i(q)\ge (\parallel q{\parallel }_2-\parallel T{p}_i{\parallel }_2{)}^2-\parallel y-p_i{\parallel }_2^2,$ which tends to $+\mathrm{\infty }$ as $\parallel q{\parallel }_2\to \mathrm{\infty }$; thus $H$ is coercive. The nonempty sublevel $\{H\le H(0)\}$ is closed and bounded, hence compact by [F2]; [F3] gives a global minimizer $q_0$ of $H$.

Let $I=\{i:h_i(q_0)=H(q_0)\}$ be the nonempty active set. Its gradients' convex hull is the image of the closed bounded simplex $\{\alpha \in {\mathbb{R}}^m:{\alpha }_i\ge 0,{\sum }_i{\alpha }_i=1\}$ under the continuous affine map $\alpha \mapsto {\sum }_{i\in I}{\alpha }_{i}2(q_0-T{p}_i)$ (with zero weights outside $I$). The simplex is compact by [F2]; pulling an open cover back to it shows its continuous image is compact, and [F2] then makes that image closed. If $0$ were not in the convex hull, project $0$ onto it using [F4]. The projection vector $v\mathrm{\ne }0$ obeys $⟨v,2(q_0-T{p}_i)⟩\ge \mathrm{\parallel }v{\mathrm{\parallel }}_2^2$ for each $i\in I$, by the variational inequality. For sufficiently small $ϵ>0$, every active $h_i(q_0-ϵv)$ decreases strictly; the finitely many inactive terms have a strict gap below $H(q_0)$ and remain below it by continuity. This contradicts minimality. Therefore there are weights ${\alpha }_i\ge 0$, supported on $I$, with ${\sum }_i{\alpha }_i=1$ and $q_0={\sum }_i{\alpha }_{i}T{p}_i$.

Put $\bar{p}={\sum }_i{\alpha }_i{p}_i$. Expanding squares and using $q_0={\sum }_i{\alpha }_{i}T{p}_i$ gives $H(q_0)={\sum }_i{\alpha }_i{h}_i(q_0)=\frac{1}{2}{\sum }_{i,j}{\alpha }_i{\alpha }_j\mathrm{\parallel }T{p}_i-T{p}_j{\mathrm{\parallel }}_2^2-{\sum }_i{\alpha }_i\mathrm{\parallel }y-p_i{\mathrm{\parallel }}_2^2\mathrm{.}$ The nonexpansive hypothesis bounds the double sum by $\frac{1}{2}{\sum }_{i,j}{\alpha }_i{\alpha }_j\mathrm{\parallel }{p}_i-p_j{\mathrm{\parallel }}_2^2={\sum }_i{\alpha }_i\mathrm{\parallel }{p}_i-\bar{p}{\mathrm{\parallel }}_2^2$. The variance identity also gives ${\sum }_i{\alpha }_i\mathrm{\parallel }y-p_i{\mathrm{\parallel }}_2^2=\mathrm{\parallel }y-\bar{p}{\mathrm{\parallel }}_2^2+{\sum }_i{\alpha }_i\mathrm{\parallel }{p}_i-\bar{p}{\mathrm{\parallel }}_2^2$. Hence $H(q_0)\le -\mathrm{\parallel }y-\bar{p}{\mathrm{\parallel }}_2^2\le 0$. Thus every finite family of the closed balls $\bar{B}(Tp,\mathrm{\parallel }y-p{\mathrm{\parallel }}_2)$ has nonempty intersection.

Choose one $p_0\in D$. The ball $C_0=\bar{B}(T{p}_0,\mathrm{\parallel }y-p_0{\mathrm{\parallel }}_2)$ is compact by [F2]. The sets $C_0\cap \bar{B}(Tp,\mathrm{\parallel }y-p{\mathrm{\parallel }}_2)$ are closed in $C_0$, and every finite intersection is nonempty by step 3.1 applied with $p_0$ included. [F5] therefore supplies a point $q$ in their total intersection, which satisfies every required inequality. The proof used only finite-dimensional compactness and finite convex combinations; no choice principle is invoked.