Selections associated to the metric projection

(Cluj-Napoca)

Let $X$$X$XX$X$ be a normed space, $M$$M$MM$M$ a subspace of $X$$X$XX$X$ and $x$$x$xx$x$ an element of $X$$X$XX$X$. The distance from $x$$x$xx$x$ to $M$$M$MM$M$ is defined by

An element $y\in M$$y\in M$y in My \in M$y\in M$ verifying the equality $\Vert x-y\Vert =d(x,M)$$\Vert x-y\Vert =d(x,M)$||x-y||=d(x,M)\|x-y\|=d(x, M)$\Vert x-y\Vert =d(x,M)$ is called an element of best approximation for $x$$x$xx$x$ by elements in $M$$M$MM$M$. The set of all elements of best approximation for $x$$x$xx$x$ is denoted by $P_M(x)$$P_M(x)$P_(M)(x)P_{M}(x)$P_M(x)$, i.e.

If $P_M(x)\ne \varnothing$$P_M(x)\ne \varnothing$P_(M)(x)!=O/P_{M}(x) \neq \varnothing$P_M(x)\ne \varnothing$ (respectively $P_M(x)$$P_M(x)$P_(M)(x)P_{M}(x)$P_M(x)$ is a singleton) for all $x\in X$$x\in X$x in Xx \in X$x\in X$, then $M$$M$MM$M$ is called a proximinal (respectively a Chebyshevian) subspace of $X$$X$XX$X$.

The set-valued application $P_M:X\to 2^M$$P_M:X\to 2^M$P_(M):X rarr2^(M)P_{M}: X \rightarrow 2^{M}$P_M:X\to 2^M$ is called the metric projection of $X$$X$XX$X$ on $M$$M$MM$M$ and a function $p:X\to M$$p:X\to M$p:X rarr Mp: X \rightarrow M$p:X\to M$ such that $p(x)\in P_M(x)$$p(x)\in P_M(x)$p(x)inP_(M)(x)p(x) \in P_{M}(x)$p(x)\in P_M(x)$, for all $x\in X$$x\in X$x in Xx \in X$x\in X$, is called a selection for the metric projection $P_M$$P_M$P_(M)P_{M}$P_M$. Observe that the existence of a selection for $P_M$$P_M$P_(M)P_{M}$P_M$ implies $P_M(x)\ne \varnothing$$P_M(x)\ne \varnothing$P_(M)(x)!=O/P_{M}(x) \neq \varnothing$P_M(x)\ne \varnothing$, for all $x\in X$$x\in X$x in Xx \in X$x\in X$, i.e. the subspace $M$$M$MM$M$ is necessarily proximinal.

The set

is called the kernel of the metric projection $P_M$$P_M$P_(M)P_{M}$P_M$.
In many situations for a given subspace $M$$M$MM$M$ of $X$$X$XX$X$ the problem of best approximation is not considered for the whole space $X$$X$XX$X$ but rather for a subset $K$$K$KK$K$ of $X$$X$XX$X$. This is the case which we consider in this paper and to this end we need some definitions and notation.

If $K$$K$KK$K$ is a subset of the normed space X and $P_M(x)\ne \varnothing$$P_M(x)\ne \varnothing$P_(M)(x)!=O/P_{M}(x) \neq \varnothing$P_M(x)\ne \varnothing$ (respectively $P_M(x)$$P_M(x)$P_(M)(x)P_{M}(x)$P_M(x)$ is a singleton) for all $x\in K$$x\in K$x in Kx \in K$x\in K$, then the subspace $M$$M$MM$M$ is called $K$$K$KK$K$-proximinal (respectively $K$$K$KK$K$-Chebyshevian). The restriction of the metric projection $P_M$$P_M$P_(M)P_{M}$P_M$ to $K$$K$KK$K$ is denoted by $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$ and its kernel by $\mathrm{Ker}{P}_{M\mid K}$$\mathrm{Ker}{P}_{M\mid K}$KerP_(M∣K)\operatorname{Ker} P_{M \mid K}$\mathrm{Ker}{P}_{M\mid K}$ :

For two nonvoid subsets $U,V$$U,V$U,VU, V$U,V$ of $X$$X$XX$X$ denote by $U+V:=\{u+v:u\in U,v\in V\}$$U+V:=\{u+v:u\in U,v\in V\}$U+V:={u+v:u in U,v in V}U+V:=\{u+v: u \in U, v \in V\}$U+V:=\{u+v:u\in U,v\in V\}$ their algebraic sum. If every $x\in U+V$$x\in U+V$x in U+Vx \in U+V$x\in U+V$ can be uniquely written in the form $x=u+v$$x=u+v$x=u+vx=u+v$x=u+v$ with $u\in U$$u\in U$u in Uu \in U$u\in U$ and $v\in V$$v\in V$v in Vv \in V$v\in V$, then $U+V$$U+V$U+VU+V$U+V$ is called the direct algebraic sum of the sets $U$$U$UU$U$ and $V$$V$VV$V$ and is denoted by $U+V$$U+V$U+VU+V$U+V$. If $K=U+V$$K=U+V$K=U+VK=U+V$K=U+V$ and the application $(u,v)\to u+v,u\in U,v\in V$$(u,v)\to u+v,u\in U,v\in V$(u,v)rarr u+v,u in U,v in V(u, v) \rightarrow u+v, u \in U, v \in V$(u,v)\to u+v,u\in U,v\in V$, is a topological homeomorphism between $U\times V$$U\times V$U xx VU \times V$U\times V$ (endowed with the product topology) and $K$$K$KK$K$ then $K$$K$KK$K$ is called the direct topological sum of the sets $U$$U$UU$U$ and $V$$V$VV$V$, denoted by $K=U\oplus V$$K=U\oplus V$K=U o+VK=U \oplus V$K=U\oplus V$.
F. Deutsch [2] proved that if $M$$M$MM$M$ is a proximinal subspace of $X$$X$XX$X$ then the metric projection $P_M$$P_M$P_(M)P_{M}$P_M$ admits a continuous and linear selection if and only if the subspace $M$$M$MM$M$ is complemented in $X$$X$XX$X$ by a closed subspace of $\mathrm{Ker}{P}_M$$\mathrm{Ker}{P}_M$KerP_(M)\operatorname{Ker} P_{M}$\mathrm{Ker}{P}_M$ ([2], Theorem 2.2).

In [4], one of the authors of the present paper considered a similar problem for a closed convex cone $K$$K$KK$K$ in $X$$X$XX$X$ and a $K$$K$KK$K$-proximinal subspace $M$$M$MM$M$ of $X$$X$XX$X$, asking for $P_{M\{K}$$P_{M\{K}$P_(M{K)P_{M\{K}$P_{M\{K}$ to admit a continuous, positively homogeneous and additive selection. The following sufficient condition for the existence of such a selection was obtained:

If there exist two closed convex cones $C\subset \mathrm{Ker}{P}_M$$C\subset \mathrm{Ker}{P}_M$C sub KerP_(M)C \subset \operatorname{Ker} P_{M}$C\subset \mathrm{Ker}{P}_M$ and $U\subset M$$U\subset M$U sub MU \subset M$U\subset M$, such that $K=C\oplus U$$K=C\oplus U$K=C o+UK=C \oplus U$K=C\oplus U$, then the metric projection $P_{MJK}$$P_{MJK}$P_(MJK)P_{M J K}$P_{MJK}$ admits a continuous, positively homogeneous and additive selection ([4], Theorem A).

This condition is not necessary for the existence of such a selection. In this paper we shall give a reformulation (called Theorem $A^{\prime }$$A^{\prime }$A^(')A^{\prime}$A^{\prime }$ ) of Theorem $A$$A$AA$A$ from [4] and prove that if $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$ admits a continuous, positively homogeneous and additive selection, satisfying some suplementary conditions, then the cone $K$$K$KK$K$ admits a decompositions $K=C\oplus U$$K=C\oplus U$K=C o+UK=C \oplus U$K=C\oplus U$, with $C\subset \mathrm{Ker}{P}_{M\mid K}$$C\subset \mathrm{Ker}{P}_{M\mid K}$C sub KerP_(M∣K)C \subset \operatorname{Ker} P_{M \mid K}$C\subset \mathrm{Ker}{P}_{M\mid K}$ and $U\subset M$$U\subset M$U sub MU \subset M$U\subset M$, closed convex cones (Theorem B). Although the conditions in theorems A' and B are very close to be necessary and sufficient for the existence of a continuous, positively homogeneous and additive selection for $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$, we weren't able to find such conditions. Some situations which may occur are illustrated by some examples following Theorem B.

By a convex cone in $X$$X$XX$X$ we understand a nonvoid subset $K$$K$KK$K$ of $X$$X$XX$X$ such that:
a) $x_1+x_2\in K$$x_1+x_2\in K$x_(1)+x_(2)in Kx_{1}+x_{2} \in K$x_1+x_2\in K$, for all $x_1,x_2\in K$$x_1,x_2\in K$x_(1),x_(2)in Kx_{1}, x_{2} \in K$x_1,x_2\in K$, and
b) $\lambda \cdot x\in K$$\lambda \cdot x\in K$lambda*x in K\lambda \cdot x \in K$\lambda \cdot x\in K$, for all $x\in K$$x\in K$x in Kx \in K$x\in K$, and $\lambda \ge 0$$\lambda \ge 0$lambda >= 0\lambda \geq 0$\lambda \ge 0$.

A carefull examination of the statement and of the proof of Theorem $A$$A$AA$A$ in [4] yields the following more detailed reformulation:

Theorem A'. Let $M$$M$MM$M$ be a closed linear subspace of a normed space $X$$X$XX$X$ and $K$$K$KK$K$ a closed convex cone in $X$$X$XX$X$. If there exist two closed convex cones $C\subset \mathrm{Ker}{P}_{M/K}$$C\subset \mathrm{Ker}{P}_{M/K}$C sub KerP_(M//K)C \subset \operatorname{Ker} P_{M / K}$C\subset \mathrm{Ker}{P}_{M/K}$ and $U\subset M$$U\subset M$U sub MU \subset M$U\subset M$ such that $K=C\oplus U$$K=C\oplus U$K=C o+UK=C \oplus U$K=C\oplus U$, then the application $p:K\to M$$p:K\to M$p:K rarr Mp: K \rightarrow M$p:K\to M$,
defined by $p(x)=z$$p(x)=z$p(x)=zp(x)=z$p(x)=z$, for $x=y+z\in K,y\in C,z\in U$$x=y+z\in K,y\in C,z\in U$x=y+z in K,y in C,z in Ux=y+z \in K, y \in C, z \in U$x=y+z\in K,y\in C,z\in U$, is a continuous, positively homogeneous and additive selection of the metric projection $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$. The subspace $M$$M$MM$M$ is $K$$K$KK$K$-proximinal and $C=p^{-1}(0),U=p(K)$$C=p^{-1}(0),U=p(K)$C=p^(-1)(0),U=p(K)C=p^{-1}(0), U=p(K)$C=p^{-1}(0),U=p(K)$.

The following theorem shows that, in some cases, the existence of a continuous, positively homogeneous and additive selection for $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$ implies the decomposability of K in the form $K=C\oplus U$$K=C\oplus U$K=C o+UK=C \oplus U$K=C\oplus U$, with $C$$C$CC$C$ and $U$$U$UU$U$ closed convex cones.

THEOREM B. Let $X$$X$XX$X$ be a normed space, $M$$M$MM$M$ a closed subspace of $X$$X$XX$X$ and $K$$K$KK$K$ a closed convex cone in $X$$X$XX$X$. Suppose that the metric projection $P_{M\mid K}$$P_{M\mid K}$P_(M∣K)P_{M \mid K}$P_{M\mid K}$ admits a continuous, positively homogeneous and additive selection $p$$p$pp$p$ such that:
a) $p(K)$$p(K)$p(K)p(K)$p(K)$ is closed and contained in $K$$K$KK$K$, and
b) $x-p(x)\in K$$x-p(x)\in K$x-p(x)in Kx-p(x) \in K$x-p(x)\in K$, for all $x\in K$$x\in K$x in Kx \in K$x\in K$.

Then $p^{-1}(0)$$p^{-1}(0)$p^(-1)(0)p^{-1}(0)$p^{-1}(0)$ and $p(K)$$p(K)$p(K)p(K)$p(K)$ are closed convex cones contained in $\mathrm{Ker}{P}_{M\mid K}$$\mathrm{Ker}{P}_{M\mid K}$KerP_(M∣K)\operatorname{Ker} P_{M \mid K}$\mathrm{Ker}{P}_{M\mid K}$ and $M$$M$MM$M$ respectively, and $K=p^{-1}(0)\oplus p(K)$$K=p^{-1}(0)\oplus p(K)$K=p^(-1)(0)o+p(K)K=p^{-1}(0) \oplus p(K)$K=p^{-1}(0)\oplus p(K)$.

If $p(K)$$p(K)$p(K)p(K)$p(K)$ is a closed subspace of $K$$K$KK$K$ or $M\subset K$$M\subset K$M sub KM \subset K$M\subset K$ then the conditions a) and b) are automatically fulfilled.

Proof. By the additivity, positive homogeneity of $p$$p$pp$p$ and the fact that $K$$K$KK$K$ is a convex cone, it follows immediately that $p(K)$$p(K)$p(K)p(K)$p(K)$ is a convex cone contained in $M$$M$MM$M$. By hypothesis $a$$a$aa$a$ ) it is also closed.

By the continuity of $p$$p$pp$p$ the set $p^{-1}(0)\subset \mathrm{Ker}{P}_{M\mid K}$$p^{-1}(0)\subset \mathrm{Ker}{P}_{M\mid K}$p^(-1)(0)sub KerP_(M∣K)p^{-1}(0) \subset \operatorname{Ker} P_{M \mid K}$p^{-1}(0)\subset \mathrm{Ker}{P}_{M\mid K}$ is closed. If $y\in p^{-1}(0)$$y\in p^{-1}(0)$y inp^(-1)(0)y \in p^{-1}(0)$y\in p^{-1}(0)$ and $\lambda \ge 0$$\lambda \ge 0$lambda >= 0\lambda \geq 0$\lambda \ge 0$ then $p(\lambda \cdot y)=\lambda \cdot p(y)=0$$p(\lambda \cdot y)=\lambda \cdot p(y)=0$p(lambda*y)=lambda*p(y)=0p(\lambda \cdot y)=\lambda \cdot p(y)=0$p(\lambda \cdot y)=\lambda \cdot p(y)=0$, showing that $\lambda \cdot y\in p^{-1}(0)$$\lambda \cdot y\in p^{-1}(0)$lambda*y inp^(-1)(0)\lambda \cdot y \in p^{-1}(0)$\lambda \cdot y\in p^{-1}(0)$. Similarly, $y_1,y_2\in p^{-1}(0)$$y_1,y_2\in p^{-1}(0)$y_(1),y_(2)inp^(-1)(0)y_{1}, y_{2} \in p^{-1}(0)$y_1,y_2\in p^{-1}(0)$ and the additivity of $p$$p$pp$p$ imply $p(y_1+y_2)=p\left(y_1\right)+p\left(y_2\right)=0$$p\left(y_1+y_2\right)=p\left(y_1\right)+p\left(y_2\right)=0$p(y_(1)+y_(2))=p(y_(1))+p(y_(2))=0p\left(y_{1}+y_{2}\right)=p\left(y_{1}\right)+p\left(y_{2}\right)=0$p(y_1+y_2)=p\left(y_1\right)+p\left(y_2\right)=0$, showing that $p^{-1}(0)$$p^{-1}(0)$p^(-1)(0)p^{-1}(0)$p^{-1}(0)$ is a closed convex cone contained in $\mathrm{Ker}{P}_{M\mid K}$$\mathrm{Ker}{P}_{M\mid K}$KerP_(M∣K)\operatorname{Ker} P_{M \mid K}$\mathrm{Ker}{P}_{M\mid K}$.

Now we prove that $K=p^{-1}(0)+p(K)$$K=p^{-1}(0)+p(K)$K=p^(-1)(0)+p(K)K=p^{-1}(0)+p(K)$K=p^{-1}(0)+p(K)$. If $x\in K$$x\in K$x in Kx \in K$x\in K$ then by Condition $b$$b$bb$b$ ), $y:=x-p(x)\in K$$y:=x-p(x)\in K$y:=x-p(x)in Ky:=x-p(x) \in K$y:=x-p(x)\in K$. By Condition a), $p(x)\in K$$p(x)\in K$p(x)in Kp(x) \in K$p(x)\in K$ implying $x=y+p(x)$$x=y+p(x)$x=y+p(x)x=y+p(x)$x=y+p(x)$ with $y,p(x)\in K$$y,p(x)\in K$y,p(x)in Ky, p(x) \in K$y,p(x)\in K$ : Using the additivity of the function $p$$p$pp$p$ and the fact that $p(p(x))=p(x)$$p(p(x))=p(x)$p(p(x))=p(x)p(p(x))=p(x)$p(p(x))=p(x)$ (in fact $p(m)=m$$p(m)=m$p(m)=mp(m)=m$p(m)=m$ for all $m\in M$$m\in M$m in Mm \in M$m\in M$ ) we obtain $p(x)=p(y)+p(p(x))=p(y)+p(x)$$p(x)=p(y)+p(p(x))=p(y)+p(x)$p(x)=p(y)+p(p(x))=p(y)+p(x)p(x)=p(y)+p(p(x))=p(y)+p(x)$p(x)=p(y)+p(p(x))=p(y)+p(x)$. It follows $p(y)=0$$p(y)=0$p(y)=0p(y)=0$p(y)=0$, i.e. $y\in p^{-1}(0)$$y\in p^{-1}(0)$y inp^(-1)(0)y \in p^{-1}(0)$y\in p^{-1}(0)$ and $K\subset p^{-1}(0)+p(K)$$K\subset p^{-1}(0)+p(K)$K subp^(-1)(0)+p(K)K \subset p^{-1}(0)+p(K)$K\subset p^{-1}(0)+p(K)$. Since $p^{-1}(0)$$p^{-1}(0)$p^(-1)(0)p^{-1}(0)$p^{-1}(0)$ and $p(K)$$p(K)$p(K)p(K)$p(K)$ are contained in $K$$K$KK$K$ and $K$$K$KK$K$ is a convex cone, it follows that $p^{-1}(0)+p(K)\subset K$$p^{-1}(0)+p(K)\subset K$p^(-1)(0)+p(K)sub Kp^{-1}(0)+p(K) \subset K$p^{-1}(0)+p(K)\subset K$ and $K=p^{-1}(0)+p(K)$$K=p^{-1}(0)+p(K)$K=p^(-1)(0)+p(K)K=p^{-1}(0)+p(K)$K=p^{-1}(0)+p(K)$.

To show that this is a direct algebraic sum, suppose that an element $x\in K$$x\in K$x in Kx \in K$x\in K$ admits two representations: $x=y+p(x)$$x=y+p(x)$x=y+p(x)x=y+p(x)$x=y+p(x)$ and $x=y^{\prime }+z^{\prime }$$x=y^{\prime }+z^{\prime }$x=y^(')+z^(')x=y^{\prime}+z^{\prime}$x=y^{\prime }+z^{\prime }$, with $y,y^{\prime }\in p^{-1}(0)$$y,y^{\prime }\in p^{-1}(0)$y,y^(')inp^(-1)(0)y, y^{\prime} \in p^{-1}(0)$y,y^{\prime }\in p^{-1}(0)$ and $z^{\prime }\in p(K)\subset M$$z^{\prime }\in p(K)\subset M$z^(')in p(K)sub Mz^{\prime} \in p(K) \subset M$z^{\prime }\in p(K)\subset M$. It follows $p\left(z^{\prime }\right)=z^{\prime }$$p\left(z^{\prime }\right)=z^{\prime }$p(z^('))=z^(')p\left(z^{\prime}\right)=z^{\prime}$p\left(z^{\prime }\right)=z^{\prime }$ and, by the additivity of $p,p(x)=p\left(y^{\prime }\right)+p\left(z^{\prime }\right)=0+z^{\prime }=z^{\prime }$$p,p(x)=p\left(y^{\prime }\right)+p\left(z^{\prime }\right)=0+z^{\prime }=z^{\prime }$p,p(x)=p(y^('))+p(z^('))=0+z^(')=z^(')p, p(x)=p\left(y^{\prime}\right)+p\left(z^{\prime}\right)=0+z^{\prime}=z^{\prime}$p,p(x)=p\left(y^{\prime }\right)+p\left(z^{\prime }\right)=0+z^{\prime }=z^{\prime }$, implying $y^{\prime }=x-p(x)=y$$y^{\prime }=x-p(x)=y$y^(')=x-p(x)=yy^{\prime}=x-p(x)=y$y^{\prime }=x-p(x)=y$ and $z^{\prime }=p(x)$$z^{\prime }=p(x)$z^(')=p(x)z^{\prime}=p(x)$z^{\prime }=p(x)$.

It remains to show that the comespondence $(y,z)\to y+z,y\in p^{-1}(0),z\in p(K)$$(y,z)\to y+z,y\in p^{-1}(0),z\in p(K)$(y,z)rarr y+z,y inp^(-1)(0),z in p(K)(y, z) \rightarrow y+z, y \in p^{-1}(0), z \in p(K)$(y,z)\to y+z,y\in p^{-1}(0),z\in p(K)$, is a homeomophism between $p^{-1}(0)\times p(K)$$p^{-1}(0)\times p(K)$p^(-1)(0)xx p(K)p^{-1}(0) \times p(K)$p^{-1}(0)\times p(K)$, equipped with the product topology,
and $K$$K$KK$K$. To this end consider a sequence $(y_n,z_n)\in p^{-1}(0)\times p(K),n\in N$$\left(y_n,z_n\right)\in p^{-1}(0)\times p(K),n\in N$(y_(n),z_(n))inp^(-1)(0)xx p(K),n in N\left(y_{n}, z_{n}\right) \in p^{-1}(0) \times p(K), n \in N$(y_n,z_n)\in p^{-1}(0)\times p(K),n\in N$, converging to $(y,z)\in p^{-1}(0)\times p(K)$$(y,z)\in p^{-1}(0)\times p(K)$(y,z)inp^(-1)(0)xx p(K)(y, z) \in p^{-1}(0) \times p(K)$(y,z)\in p^{-1}(0)\times p(K)$. It follows $y_n\to y$$y_n\to y$y_(n)rarr yy_{n} \rightarrow y$y_n\to y$ and $z_n\to z$$z_n\to z$z_(n)rarr zz_{n} \rightarrow z$z_n\to z$, implying $(y_n,z_n)\to y+z$$\left(y_n,z_n\right)\to y+z$(y_(n),z_(n))rarr y+z\left(y_{n}, z_{n}\right) \rightarrow y+z$(y_n,z_n)\to y+z$, which proves the continuity of the application $(y,z)\to y+z$$(y,z)\to y+z$(y,z)rarr y+z(y, z) \rightarrow y+z$(y,z)\to y+z$.

To prove the continuity of the inverse application $x\mapsto (y,z)$$x\mapsto (y,z)$x|->(y,z)x \mapsto(y, z)$x\mapsto (y,z)$, where $x=y+z,y\in p^{-1}(0),z\in p(K)$$x=y+z,y\in p^{-1}(0),z\in p(K)$x=y+z,y inp^(-1)(0),z in p(K)x=y+z, y \in p^{-1}(0), z \in p(K)$x=y+z,y\in p^{-1}(0),z\in p(K)$, take again a sequence $x_n=y_n+z_n\in K,y_n\in p^{-1}(0),z_n\in p(K)$$x_n=y_n+z_n\in K,y_n\in p^{-1}(0),z_n\in p(K)$x_(n)=y_(n)+z_(n)in K,y_(n)inp^(-1)(0),z_(n)in p(K)x_{n}=y_{n}+z_{n} \in K, y_{n} \in p^{-1}(0), z_{n} \in p(K)$x_n=y_n+z_n\in K,y_n\in p^{-1}(0),z_n\in p(K)$, converging to $x=y+z\in K$$x=y+z\in K$x=y+z in Kx=y+z \in K$x=y+z\in K$, where $y\in p^{-1}(0)$$y\in p^{-1}(0)$y inp^(-1)(0)y \in p^{-1}(0)$y\in p^{-1}(0)$ and $z\in p(K)$$z\in p(K)$z in p(K)z \in p(K)$z\in p(K)$. It follows $z_n=p\left(x_n\right),n\in N,z=p(x)$$z_n=p\left(x_n\right),n\in N,z=p(x)$z_(n)=p(x_(n)),n in N,z=p(x)z_{n}=p\left(x_{n}\right), n \in N, z=p(x)$z_n=p\left(x_n\right),n\in N,z=p(x)$, and, by the continuity of the application $p,z_n=p\left(x_n\right)\to p\left(x_n\right)=z$$p,z_n=p\left(x_n\right)\to p\left(x_n\right)=z$p,z_(n)=p(x_(n))rarr p(x_(n))=zp, z_{n}=p\left(x_{n}\right) \rightarrow p\left(x_{n}\right)=z$p,z_n=p\left(x_n\right)\to p\left(x_n\right)=z$. But then $y_n=x_n-z_n\to x-z=y$$y_n=x_n-z_n\to x-z=y$y_(n)=x_(n)-z_(n)rarr x-z=yy_{n}=x_{n}-z_{n} \rightarrow x-z=y$y_n=x_n-z_n\to x-z=y$, proving that the sequence ${\left((y_n,z_n)\right)}_{n\in N}$${\left(\left(y_n,z_n\right)\right)}_{n\in N}$((y_(n),z_(n)))_(n in N)\left(\left(y_{n}, z_{n}\right)\right)_{n \in N}${\left((y_n,z_n)\right)}_{n\in N}$ converges to $(y,z)$$(y,z)$(y,z)(y, z)$(y,z)$ with respect to the product topology of $p^{-1}(0)\times p(K)$$p^{-1}(0)\times p(K)$p^(-1)(0)xx p(K)p^{-1}(0) \times p(K)$p^{-1}(0)\times p(K)$. This shows that the application $x\mapsto (y,z)$$x\mapsto (y,z)$x|->(y,z)x \mapsto(y, z)$x\mapsto (y,z)$, $x=y+z\in p^{-1}(0)+p(K)$$x=y+z\in p^{-1}(0)+p(K)$x=y+z inp^(-1)(0)+p(K)x=y+z \in p^{-1}(0)+p(K)$x=y+z\in p^{-1}(0)+p(K)$, is continuous too and, consequently, the application $(y,z)\mapsto y+z$$(y,z)\mapsto y+z$(y,z)|->y+z(y, z) \mapsto y+z$(y,z)\mapsto y+z$ is a homeomorphism between $p^{-1}(0)\times p(K)$$p^{-1}(0)\times p(K)$p^(-1)(0)xx p(K)p^{-1}(0) \times p(K)$p^{-1}(0)\times p(K)$ and $K$$K$KK$K$.

If $p(K)$$p(K)$p(K)p(K)$p(K)$ is a closed subspace of $K$$K$KK$K$ then Condition a) holds and, for $x\in K,p(x)$$x\in K,p(x)$x in K,p(x)x \in K, p(x)$x\in K,p(x)$ and $-p(x)$$-p(x)$-p(x)-p(x)$-p(x)$ are in $p(K)\subset K$$p(K)\subset K$p(K)sub Kp(K) \subset K$p(K)\subset K$ so that $x-p(x)\in K$$x-p(x)\in K$x-p(x)in Kx-p(x) \in K$x-p(x)\in K$, showing that Condition $b$$b$bb$b$ ) holds too. If $M\subset K$$M\subset K$M sub KM \subset K$M\subset K$ then $M=p(M)\subset p(K)$$M=p(M)\subset p(K)$M=p(M)sub p(K)M=p(M) \subset p(K)$M=p(M)\subset p(K)$ and, since $p(K)\subset M$$p(K)\subset M$p(K)sub Mp(K) \subset M$p(K)\subset M$, it follows that $p(K)=M$$p(K)=M$p(K)=Mp(K)=M$p(K)=M$ is a closed subspace of $K$$K$KK$K$. Theorem B is completely proved.

Remark. Conditions $a$$a$aa$a$ ) and $b$$b$bb$b$ ) are fulfilled by the selection $p$$p$pp$p$ given in Theorem A'. Indeed, $K=p^{-1}(0)\oplus p(K)$$K=p^{-1}(0)\oplus p(K)$K=p^(-1)(0)o+p(K)K=p^{-1}(0) \oplus p(K)$K=p^{-1}(0)\oplus p(K)$ implies that $p(K)$$p(K)$p(K)p(K)$p(K)$ is a closed convex cone contained in K . Since every $x\in K$$x\in K$x in Kx \in K$x\in K$ can be written in the form $x=y+z$$x=y+z$x=y+zx=y+z$x=y+z$ with $p(y)=0$$p(y)=0$p(y)=0p(y)=0$p(y)=0$ and $z=p(x)$$z=p(x)$z=p(x)z=p(x)$z=p(x)$, it follows that $x-p(x)=x-z=y\in K$$x-p(x)=x-z=y\in K$x-p(x)=x-z=y in Kx-p(x)=x-z=y \in K$x-p(x)=x-z=y\in K$ for all $x\in K$$x\in K$x in Kx \in K$x\in K$.

In the following examples, there always exists a continuous, positively homogeneous and additive selection of the metric projections but, the equality $K=p^{-1}(0)\oplus p(K)$$K=p^{-1}(0)\oplus p(K)$K=p^(-1)(0)o+p(K)K=p^{-1}(0) \oplus p(K)$K=p^{-1}(0)\oplus p(K)$ is not true in all these cases.

Example 1. Take $X=R^2$$X=R^2$X=R^(2)X=R^{2}$X=R^2$ with the Euclidean norm and $M\{(x_1,0):x_1\in R\}$$M\left\{\left(x_1,0\right):x_1\in R\right\}$M{(x_(1),0):x_(1)in R}M\left\{\left(x_{1}, 0\right): x_{1} \in R\right\}$M\{(x_1,0):x_1\in R\}$. Then $P_M\left((x_1,x_2)\right)=\left\{(x_1,0)\right\}$$P_M\left(\left(x_1,x_2\right)\right)=\left\{\left(x_1,0\right)\right\}$P_(M)((x_(1),x_(2)))={(x_(1),0)}P_{M}\left(\left(x_{1}, x_{2}\right)\right)=\left\{\left(x_{1}, 0\right)\right\}$P_M\left((x_1,x_2)\right)=\left\{(x_1,0)\right\}$, for all $(x_1,x_2)\in R^2$$\left(x_1,x_2\right)\in R^2$(x_(1),x_(2))inR^(2)\left(x_{1}, x_{2}\right) \in R^{2}$(x_1,x_2)\in R^2$, i.e. $M$$M$MM$M$ is a Chebyshevian subspace of $X$$X$XX$X$ and the only selection of the metric projection is $p\left((x_1,x_2)\right)=(x_1,0)$$p\left(\left(x_1,x_2\right)\right)=\left(x_1,0\right)$p((x_(1),x_(2)))=(x_(1),0)p\left(\left(x_{1}, x_{2}\right)\right)=\left(x_{1}, 0\right)$p\left((x_1,x_2)\right)=(x_1,0)$, for $(x_1,x_2)\in R^2$$\left(x_1,x_2\right)\in R^2$(x_(1),x_(2))inR^(2)\left(x_{1}, x_{2}\right) \in R^{2}$(x_1,x_2)\in R^2$. Let $R_{+}^2:=\{(x_1,x_2)\in R^2:x_1\ge 0,x_2\ge 0\}$$R_{+}^2:=\left\{\left(x_1,x_2\right)\in R^2:x_1\ge 0,x_2\ge 0\right\}$R_(+)^(2):={(x_(1),x_(2))inR^(2):x_(1) >= 0,x_(2) >= 0}R_{+}^{2}:=\left\{\left(x_{1}, x_{2}\right) \in R^{2}: x_{1} \geq 0, x_{2} \geq 0\right\}$R_{+}^2:=\{(x_1,x_2)\in R^2:x_1\ge 0,x_2\ge 0\}$.
a) Take $K=\{(x_1,x_2):x_1=x_2,x_1\ge 0\}$$K=\left\{\left(x_1,x_2\right):x_1=x_2,x_1\ge 0\right\}$K={(x_(1),x_(2)):x_(1)=x_(2),x_(1) >= 0}K=\left\{\left(x_{1}, x_{2}\right): x_{1}=x_{2}, x_{1} \geq 0\right\}$K=\{(x_1,x_2):x_1=x_2,x_1\ge 0\}$. In this case $\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$$\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$KerP_(M∣K)={(0,0)}\operatorname{Ker} P_{M \mid K}=\{(0,0)\}$\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$ so that the only closed convex cone contained in $\mathrm{Ker}{P}_{M\mid K}$$\mathrm{Ker}{P}_{M\mid K}$KerP_(M∣K)\operatorname{Ker} P_{M \mid K}$\mathrm{Ker}{P}_{M\mid K}$ is $C=\{(0,0)\}$$C=\{(0,0)\}$C={(0,0)}C=\{(0,0)\}$C=\{(0,0)\}$. The subspace $M$$M$MM$M$ contains two nontrivial closed cones $U_{+}=\{(x_1,0):x_1\ge 0\}$$U_{+}=\left\{\left(x_1,0\right):x_1\ge 0\right\}$U_(+)={(x_(1),0):x_(1) >= 0}U_{+}=\left\{\left(x_{1}, 0\right): x_{1} \geq 0\right\}$U_{+}=\{(x_1,0):x_1\ge 0\}$ and $U_{-}=\{(x_1,0):x_1\le 0\}p(K)=U_{+}$$U_{-}=\left\{\left(x_1,0\right):x_1\le 0\right\}p(K)=U_{+}$U_(-)={(x_(1),0):x_(1) <= 0}quad p(K)=U_(+)U_{-}=\left\{\left(x_{1}, 0\right): x_{1} \leq 0\right\} \quad p(K)=U_{+}$U_{-}=\{(x_1,0):x_1\le 0\}p(K)=U_{+}$and $K\ne C\oplus U_{+}=U_{+}$$K\ne C\oplus U_{+}=U_{+}$K!=C o+U_(+)=U_(+)K \neq C \oplus U_{+}=U_{+}$K\ne C\oplus U_{+}=U_{+}$.
b) Let $K=\{(x_1,x_2)\in R^2:x_2\ge x_1,x_1\ge 0\}$$K=\left\{\left(x_1,x_2\right)\in R^2:x_2\ge x_1,x_1\ge 0\right\}$K={(x_(1),x_(2))inR^(2):x_(2) >= x_(1),x_(1) >= 0}K=\left\{\left(x_{1}, x_{2}\right) \in R^{2}: x_{2} \geq x_{1}, x_{1} \geq 0\right\}$K=\{(x_1,x_2)\in R^2:x_2\ge x_1,x_1\ge 0\}$. In this case Ker $P_{M\mid K}=\{(0,x_2):x_2\ge 0\}$$P_{M\mid K}=\left\{\left(0,x_2\right):x_2\ge 0\right\}$P_(M∣K)={(0,x_(2)):x_(2) >= 0}P_{M \mid K}=\left\{\left(0, x_{2}\right): x_{2} \geq 0\right\}$P_{M\mid K}=\{(0,x_2):x_2\ge 0\}$ and the only nontrivial closed convex cone contained in $\mathrm{Ker}{P}_{M\mid K}$$\mathrm{Ker}{P}_{M\mid K}$KerP_(M∣K)\operatorname{Ker} P_{M \mid K}$\mathrm{Ker}{P}_{M\mid K}$ is $C=\mathrm{Ker}{P}_{M\mid K}$$C=\mathrm{Ker}{P}_{M\mid K}$C=KerP_(M∣K)C=\operatorname{Ker} P_{M \mid K}$C=\mathrm{Ker}{P}_{M\mid K}$. Again $p(K)=U_{+}$$p(K)=U_{+}$p(K)=U_(+)p(K)=U_{+}$p(K)=U_{+}$but $K\ne C\oplus p(K)=R_{+}^2$$K\ne C\oplus p(K)=R_{+}^2$K!=C o+p(K)=R_(+)^(2)K \neq C \oplus p(K)=R_{+}^{2}$K\ne C\oplus p(K)=R_{+}^2$.
c) Let $K=\{(x_1,x_2):x_2\le x_1,x_1\ge 0\}$$K=\left\{\left(x_1,x_2\right):x_2\le x_1,x_1\ge 0\right\}$K={(x_(1),x_(2)):x_(2) <= x_(1),x_(1) >= 0}K=\left\{\left(x_{1}, x_{2}\right): x_{2} \leq x_{1}, x_{1} \geq 0\right\}$K=\{(x_1,x_2):x_2\le x_1,x_1\ge 0\}$. In this case $\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$$\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$KerP_(M∣K)={(0,0)}\operatorname{Ker} P_{M \mid K}=\{(0,0)\}$\mathrm{Ker}{P}_{M\mid K}=\{(0,0)\}$ implying $C=\{(0,0)\}$$C=\{(0,0)\}$C={(0,0)}C=\{(0,0)\}$C=\{(0,0)\}$. We have $p(K)=U_{+}\subset K\cap M$$p(K)=U_{+}\subset K\cap M$p(K)=U_(+)sub K nn Mp(K)=U_{+} \subset K \cap M$p(K)=U_{+}\subset K\cap M$ but $K\ne C\oplus p(K)=R_{+}^2$$K\ne C\oplus p(K)=R_{+}^2$K!=C o+p(K)=R_(+)^(2)K \neq C \oplus p(K)=R_{+}^{2}$K\ne C\oplus p(K)=R_{+}^2$.
d) $K=\{(x_1,x_2):x_1\ge 0,x_2\ge 0\}$$K=\left\{\left(x_1,x_2\right):x_1\ge 0,x_2\ge 0\right\}$K={(x_(1),x_(2)):x_(1) >= 0,x_(2) >= 0}K=\left\{\left(x_{1}, x_{2}\right): x_{1} \geq 0, x_{2} \geq 0\right\}$K=\{(x_1,x_2):x_1\ge 0,x_2\ge 0\}$. In this case $\mathrm{Ker}{P}_{M\mid K}=\{(0,x_2):x_2\ge 0\}$$\mathrm{Ker}{P}_{M\mid K}=\left\{\left(0,x_2\right):x_2\ge 0\right\}$KerP_(M∣K)={(0,x_(2)):x_(2) >= 0}\operatorname{Ker} P_{M \mid K}=\left\{\left(0, x_{2}\right): x_{2} \geq 0\right\}$\mathrm{Ker}{P}_{M\mid K}=\{(0,x_2):x_2\ge 0\}$, $C=p^{-1}(0)=\mathrm{Ker}{P}_{M\mid K},p(K)=U_{+}$$C=p^{-1}(0)=\mathrm{Ker}{P}_{M\mid K},p(K)=U_{+}$C=p^(-1)(0)=KerP_(M∣K),p(K)=U_(+)C=p^{-1}(0)=\operatorname{Ker} P_{M \mid K}, p(K)=U_{+}$C=p^{-1}(0)=\mathrm{Ker}{P}_{M\mid K},p(K)=U_{+}$and $K=p^{-1}(0)\oplus p(K)$$K=p^{-1}(0)\oplus p(K)$K=p^(-1)(0)o+p(K)K=p^{-1}(0) \oplus p(K)$K=p^{-1}(0)\oplus p(K)$.
e) $K=\{(x_1,x_2)\in R^2:x_2\ge 0\}$$K=\left\{\left(x_1,x_2\right)\in R^2:x_2\ge 0\right\}$K={(x_(1),x_(2))inR^(2):x_(2) >= 0}K=\left\{\left(x_{1}, x_{2}\right) \in R^{2}: x_{2} \geq 0\right\}$K=\{(x_1,x_2)\in R^2:x_2\ge 0\}$. In this case $p(K)=M\subset K,\mathrm{Ker}{P}_{K\mid M}==\{(0,x_2):x_2\ge 0\}$$p(K)=M\subset K,\mathrm{Ker}{P}_{K\mid M}==\left\{\left(0,x_2\right):x_2\ge 0\right\}$p(K)=M sub K,KerP_(K∣M)=={(0,x_(2)):x_(2) >= 0}p(K)=M \subset K, \operatorname{Ker} P_{K \mid M}= =\left\{\left(0, x_{2}\right): x_{2} \geq 0\right\}$p(K)=M\subset K,\mathrm{Ker}{P}_{K\mid M}==\{(0,x_2):x_2\ge 0\}$ and $K=C\oplus p(K)$$K=C\oplus p(K)$K=C o+p(K)K=C \oplus p(K)$K=C\oplus p(K)$, where $C=\mathrm{Ker}{P}_{M\mid K}$$C=\mathrm{Ker}{P}_{M\mid K}$C=KerP_(M∣K)C=\operatorname{Ker} P_{M \mid K}$C=\mathrm{Ker}{P}_{M\mid K}$.