Proof of Radial Retraction onto a Closed Ball in ℝⁿ and in the Real Matrices

Let $V$ be $\mathbb{R}^{n}$ or $\mathcal{M}_{p\times q}(\mathbb{R})$, with its sum, scalar multiple and difference (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv), respectively Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations (iv)), which obey the vector-space laws of the pointwise operations (Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iii)), and with its norm $\lVert\cdot\rVert$ as in the statement. Only the following three properties of the norm are used: for all $z,z'\in V$ and $t\in\mathbb{R}$,

(N1) $\lVert z\rVert\ge0$;

(N2) $\lVert tz\rVert=|t|\,\lVert z\rVert$;

(N3) $\lVert z+z'\rVert\le\lVert z\rVert+\lVert z'\rVert$.

For $V=\mathbb{R}^{n}$ these are Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v) and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi); for $V=\mathcal{M}_{p\times q}(\mathbb{R})$ they are Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (i), Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (v) and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (vi). Two consequences: since $z'-z=(-1)(z-z')$, (N2) gives

$$\lVert z'-z\rVert=\lVert z-z'\rVert;\qquad(\mathrm{N4})$$

and since $z=(z-z')+z'$ and $z'=(z'-z)+z$, (N3) and (N4) give $\lVert z\rVert\le\lVert z-z'\rVert+\lVert z'\rVert$ and $\lVert z'\rVert\le\lVert z-z'\rVert+\lVert z\rVert$, that is,

$$\bigl|\lVert z\rVert-\lVert z'\rVert\bigr|\le\lVert z-z'\rVert.\qquad(\mathrm{N5})$$

We write $\rho$ for $\pi_{R}$ if $V=\mathbb{R}^{n}$ and for $\Pi_{R}$ if $V=\mathcal{M}_{p\times q}(\mathbb{R})$. For $z\in V$, by totality of the order of $\mathbb{R}$, either $\lVert z\rVert\le R$ or $\lVert z\rVert>R$, and not both.

The map ((i)). Apply Maps Defined by Cases (i) with $V$ as both the domain and the codomain set, the property $P(z)$: $\lVert z\rVert\le R$, expressed by a predicative formula, $t_{1}(z)=z$ and $t_{2}(z)=(R/\lVert z\rVert)\,z$. For $z$ with $P(z)$, $t_{1}(z)=z\in V$. For $z$ without $P(z)$, $\lVert z\rVert>R>0$, so $\lVert z\rVert\ne0$, the quotient $R/\lVert z\rVert$ is a real number and $t_{2}(z)\in V$. The lemma gives exactly one map $\rho:V\to V$ with $\rho(z)=z$ when $\lVert z\rVert\le R$ and $\rho(z)=(R/\lVert z\rVert)\,z$ when $\lVert z\rVert>R$.

The values ((ii)). Let $z\in V$. If $\lVert z\rVert\le R$, then $\rho(z)=z$ and $\lVert\rho(z)\rVert=\lVert z\rVert\le R$. If $\lVert z\rVert>R$, then $R/\lVert z\rVert>0$, and by (N2)

$$\lVert\rho(z)\rVert=\frac{R}{\lVert z\rVert}\,\lVert z\rVert=R.$$

The Lipschitz bound ((iii)). For $z\in V$ with $\lVert z\rVert>R$ put $u_{z}=R/\lVert z\rVert$; then $0<u_{z}<1$ and $u_{z}\lVert z\rVert=R$. Let $z,z'\in V$. There are four cases.

Case 1: $\lVert z\rVert\le R$ and $\lVert z'\rVert\le R$. Then $\rho(z)-\rho(z')=z-z'$, and $\lVert z-z'\rVert\le2\lVert z-z'\rVert$ by (N1).

Case 2: $\lVert z\rVert\le R<\lVert z'\rVert$. Put $v=u_{z'}$. Then $\rho(z)-\rho(z')=z-vz'=(z-z')+(1-v)z'$, and $1-v>0$. By (N3), (N2), $v\lVert z'\rVert=R$, $R\ge\lVert z\rVert$ and (N5),

$$\lVert\rho(z)-\rho(z')\rVert\le\lVert z-z'\rVert+(1-v)\lVert z'\rVert=\lVert z-z'\rVert+\lVert z'\rVert-R\le\lVert z-z'\rVert+\bigl(\lVert z'\rVert-\lVert z\rVert\bigr)\le2\lVert z-z'\rVert.$$

Case 3: $\lVert z'\rVert\le R<\lVert z\rVert$. Case 2 with $z$ and $z'$ exchanged gives $\lVert\rho(z')-\rho(z)\rVert\le2\lVert z'-z\rVert$, and by (N4), applied to $\rho(z),\rho(z')$ and to $z,z'$, this is $\lVert\rho(z)-\rho(z')\rVert\le2\lVert z-z'\rVert$.

Case 4: $\lVert z\rVert>R$ and $\lVert z'\rVert>R$. Put $u=u_{z}$ and $v=u_{z'}$. Then $\rho(z)-\rho(z')=uz-vz'=u(z-z')+(u-v)z'$. By (N2) and $0<u<1$, $\lVert u(z-z')\rVert=u\lVert z-z'\rVert\le\lVert z-z'\rVert$. Moreover

$$(u-v)\lVert z'\rVert=\frac{R\lVert z'\rVert}{\lVert z\rVert}-R=u\bigl(\lVert z'\rVert-\lVert z\rVert\bigr),$$

so by (N2), $0<u<1$ and (N5), $\lVert(u-v)z'\rVert=|u-v|\,\lVert z'\rVert=u\,\bigl|\lVert z'\rVert-\lVert z\rVert\bigr|\le\lVert z-z'\rVert$. By (N3), $\lVert\rho(z)-\rho(z')\rVert\le2\lVert z-z'\rVert$.

The four cases exhaust all $z,z'\in V$, which proves (iii) for $\pi_{R}$ ($V=\mathbb{R}^{n}$) and for $\Pi_{R}$ ($V=\mathcal{M}_{p\times q}(\mathbb{R})$).