MSC 2000. 41A65, 46A22, 46B20.
Keywords. Asymmetric normed spaces, extensions preserving asymmetric norm, best approximation.

Let $X$$X$XX$X$ be a real linear space. A function $\Vert \cdot \mid :X\to [0,\mathrm{\infty })$$\Vert \cdot \mid :X\to [0,\mathrm{\infty })$||*∣:X rarr[0,oo)\| \cdot \mid: X \rightarrow[0, \infty)$\Vert \cdot \mid :X\to [0,\mathrm{\infty })$ is called an asymmetric norm if it satisfies all the usual axioms of a norm, excepting the absolute homogeneity, which is replaced by positive homogeneity, i.e.,

The asymmetric norm is called with extended values if there exists $x\in X$$x\in X$x in Xx \in X$x\in X$ such that $\Vert x\mid =+\mathrm{\infty }$$\Vert x\mid =+\mathrm{\infty }$||x∣=+oo\| x \mid=+\infty$\Vert x\mid =+\mathrm{\infty }$. The pair ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ) is called a space with asymmetric norm (see [8], [2]).

In a space with asymmetric norm it is possible that $\Vert x\Vert \ne \Vert -x\mid$$\Vert x\Vert \ne \Vert -x\mid$||x||!=||-x∣\|x\| \neq \|-x \mid$\Vert x\Vert \ne \Vert -x\mid$ for some $x\in X$$x\in X$x in Xx \in X$x\in X$. The asymmetric norm generates a topology having as a neighborhood base the balls $B(x,r)=\{y\in X:\Vert y-x\mid <r\},x\in X,r\ge 0$$B(x,r)=\{y\in X:\Vert y-x\mid <r\},x\in X,r\ge 0$B(x,r)={y in X:||y-x∣<r},x in X,r >= 0B(x, r)=\{y \in X: \| y-x \mid<r\}, x \in X, r \geq 0$B(x,r)=\{y\in X:\Vert y-x\mid <r\},x\in X,r\ge 0$, but the topological space $(X,{\tau }_{\Vert \cdot \mid })$$\left(X,{\tau }_{\Vert \cdot \mid }\right)$(X,tau_(||*∣))\left(X, \tau_{\| \cdot \mid}\right)$(X,{\tau }_{\Vert \cdot \mid })$ is not a linear topological space, because the multiplication by scalars is not a continuous operation (see [2, p. 199]).

Let $X^{\mathrm{\#}}$$X^{\mathrm{\#}}$X^(#)X^{\#}$X^{\mathrm{\#}}$ be the algebraic dual of $X$$X$XX$X$, i.e. the space of all linear functional on $X$$X$XX$X$. We say that $f\in X^{\mathrm{\#}}$$f\in X^{\mathrm{\#}}$f inX^(#)f \in X^{\#}$f\in X^{\mathrm{\#}}$ is bounded on $X$$X$XX$X$ if

It is immediate that if $f,g\in X^{\mathrm{\#}}$$f,g\in X^{\mathrm{\#}}$f,g inX^(#)f, g \in X^{\#}$f,g\in X^{\mathrm{\#}}$ are bounded then their sum $f+g$$f+g$f+gf+g$f+g$ and the product $\lambda f$$\lambda f$lambda f\lambda f$\lambda f$ for $\lambda \ge 0$$\lambda \ge 0$lambda >= 0\lambda \geq 0$\lambda \ge 0$ are bounded too. This shows that the set of all bounded linear functionals on $X$$X$XX$X$ is a cone (see [2]), or an ac-space, according to [7].

In general, it is possible that for a bounded linear functional $f$$f$ff$f$ on ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ) the linear functional $-f$$-f$-f-f$-f$ be not bounded. Such an example is given by the functional $f(x)=x(1)$$f(x)=x(1)$f(x)=x(1)f(x)=x(1)$f(x)=x(1)$ defined on the space

equipped with the asymmetric norm

Taking $x_n(t)=1-n{t}^{n-1}$$x_n(t)=1-n{t}^{n-1}$x_(n)(t)=1-nt^(n-1)x_{n}(t)=1-n t^{n-1}$x_n(t)=1-n{t}^{n-1}$ it follows $\Vert x_n\Vert =1$$‖x_n‖=1$||x_(n)||=1\left\|x_{n}\right\|=1$\Vert x_n\Vert =1$ and $(-f)\left(x_n\right)=f(-x_n)=(-x_n)(1)=n-1$$(-f)\left(x_n\right)=f\left(-x_n\right)=\left(-x_n\right)(1)=n-1$(-f)(x_(n))=f(-x_(n))=(-x_(n))(1)=n-1(-f)\left(x_{n}\right)=f\left(-x_{n}\right)= \left(-x_{n}\right)(1)=n-1$(-f)\left(x_n\right)=f(-x_n)=(-x_n)(1)=n-1$, implying

For a bounded linear functional $f$$f$ff$f$ put

It follows that the function $\Vert \cdot \mid$$\Vert \cdot \mid$||*∣\| \cdot \mid$\Vert \cdot \mid$ defined by (2) satisfies the axioms of an asymmetric norm on the cone of all bounded linear functionals on $X$$X$XX$X$ (see [2]).

Observe that

and, since $f(-x)\le \Vert f|\cdot \Vert -x|$$f(-x)\le \Vert f|\cdot \Vert -x|$f(-x) <= ||f|*||-x|f(-x) \leq\|f|\cdot \|-x|$f(-x)\le \Vert f|\cdot \Vert -x|$, one obtains the inequalities

The inequality $|f(x)|\le \Vert f|\cdot \Vert x|$$|f(x)|\le \Vert f|\cdot \Vert x|$|f(x)| <= ||f|*||x||f(x)| \leq\|f|\cdot \| x|$|f(x)|\le \Vert f|\cdot \Vert x|$ is not true in general, but if we consider the symmetric norm $\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$||x||=max{||x|,||-x|}\|x\|=\max \{\|x|, \|-x|\}$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$ on $X$$X$XX$X$, then $|f(x)|\le \Vert f\mid \Vert x\Vert ,x\in X$$|f(x)|\le \Vert f\mid \Vert x\Vert ,x\in X$|f(x)| <= ||f∣||x||,x in X|f(x)| \leq\|f \mid\| x \|, x \in X$|f(x)|\le \Vert f\mid \Vert x\Vert ,x\in X$. This shows that a bounded linear functional is always continuous with respect to the topology generated by the symmetric norm $\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$||x||=max{||x|,||-x|}\|x\|=\max \{\|x|, \|-x|\}$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$ associated to an asymmetric norm $\Vert \cdot \mid$$\Vert \cdot \mid$||*∣\| \cdot \mid$\Vert \cdot \mid$.

If both $f$$f$ff$f$ and $-f$$-f$-f-f$-f$ are bounded then the linear functional $f$$f$ff$f$ is continuous with respect to the topology generated by the asymmetric norm.

Consider on $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ the asymmetric norm $u(x)=x\vee 0=max\{x,0\}$$u(x)=x\vee 0=max\{x,0\}$u(x)=x vv0=max{x,0}u(x)=x \vee 0=\max \{x, 0\}$u(x)=x\vee 0=max\{x,0\}$ (see [7]). A functional $f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$$f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$f:(X,||*||)rarr(R,u)f:(X,\|\cdot\|) \rightarrow(\mathbb{R}, u)$f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$ is continuous in $x_0\in X$$x_0\in X$x_(0)in Xx_{0} \in X$x_0\in X$ if for every $\epsilon >0$$\epsilon >0$epsi > 0\varepsilon>0$\epsilon >0$ there exists $\delta >0$$\delta >0$delta > 0\delta>0$\delta >0$ such that for every $x\in X$$x\in X$x in Xx \in X$x\in X$ with $\Vert x-x_0\mid <\delta$$\Vert x-x_0\mid <\delta$||x-x_(0)∣<delta\| x-x_{0} \mid<\delta$\Vert x-x_0\mid <\delta$ we have $(f(x)-f\left(x_0\right))\vee 0<\epsilon$$\left(f(x)-f\left(x_0\right)\right)\vee 0<\epsilon$(f(x)-f(x_(0)))vv0 < epsi\left(f(x)-f\left(x_{0}\right)\right) \vee 0<\varepsilon$(f(x)-f\left(x_0\right))\vee 0<\epsilon$.

It is clear that a linear functional $f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$$f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$f:(X,||*||)rarr(R,u)f:(X,\|\cdot\|) \rightarrow(\mathbb{R}, u)$f:(X,\Vert \cdot \Vert )\to (\mathbb{R},u)$ is continuous if and only if there exists $M>0$$M>0$M > 0M>0$M>0$ such that $f(x)\vee 0<M\Vert x\mid$$f(x)\vee 0<M\Vert x\mid$f(x)vv0 < M||x∣f(x) \vee 0<M \| x \mid$f(x)\vee 0<M\Vert x\mid$.

According to [7], the set

is called the asymmetric dual of the space with asymmetric norm ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ).
If $\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$||x||=max{||x|,||-x|}\|x\|=\max \{\|x|, \|-x|\}$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$ is the norm generated by $\Vert \cdot \mid$$\Vert \cdot \mid$||*∣\| \cdot \mid$\Vert \cdot \mid$ on $X$$X$XX$X$ and $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ is equipped with the usual absolute-value norm $|\cdot |$$|\cdot |$|*||\cdot|$|\cdot |$, then the set

is the (symmetric) dual of ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ). In other words, $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$ is the usual topological algebraic dual of the normed space $(X,\Vert \cdot \Vert )$$(X,\Vert \cdot \Vert )$(X,||*||)(X,\|\cdot\|)$(X,\Vert \cdot \Vert )$, where $\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$||x||=max{||x|,||-x|}\|x\|=\max \{\|x|, \|-x|\}$\Vert x\Vert =max\{\Vert x|,\Vert -x|\}$.

Observe that $X^{\ast }$$X^{\ast }$X^(**)X^{*}$X^{\ast }$ is a cone in the linear space $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$.
Equip the linear space $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$ with the extended asymmetric norm

whose restriction to the asymmetric dual $X^{\ast }$$X^{\ast }$X^(**)X^{*}$X^{\ast }$ is

It is important to remark the fact that a linear functional $f$$f$ff$f$ belongs to $X^{\ast }$$X^{\ast }$X^(**)X^{*}$X^{\ast }$ if and only if it is an upper semicontinuous linear functional on ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ), and that $X^{\ast }=\{f\in X^{\ast s}:\Vert {f|}^{\ast s}<+\mathrm{\infty }\}$$X^{\ast }=\left\{f\in X^{\ast s}:\Vert {\left.f\right|}^{\ast s}<+\mathrm{\infty }\right\}$X^(**)={f inX^(**s):||f|^(**s) < +oo}X^{*}=\left\{f \in X^{* s}: \|\left. f\right|^{* s}<+\infty\right\}$X^{\ast }=\{f\in X^{\ast s}:\Vert {f|}^{\ast s}<+\mathrm{\infty }\}$ (see [7]).

Let $(Y,\Vert \cdot \Vert )$$(Y,\Vert \cdot \Vert )$(Y,||*||)(Y,\|\cdot\|)$(Y,\Vert \cdot \Vert )$ be a subspace of the space with asymmetric norm ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ), and let $Y^{\ast }$$Y^{\ast }$Y^(**)Y^{*}$Y^{\ast }$ and $Y^{\ast s}$$Y^{\ast s}$Y^(**s)Y^{* s}$Y^{\ast s}$ be the dual cone and the (symmetric) dual of $Y$$Y$YY$Y$, respectively.

The following Hahn-Banach type theorem holds:
Theorem 1. Let ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ) be a real space with asymmetric norm and ( $Y,\Vert \cdot \Vert$$Y,\Vert \cdot \Vert$Y,||*||Y,\|\cdot\|$Y,\Vert \cdot \Vert$ ) a subspace of it. Then for every $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ there exists $F\in X^{\ast }$$F\in X^{\ast }$F inX^(**)F \in X^{*}$F\in X^{\ast }$ such that

Proof. For $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ let $p:X\to \mathbb{R}$$p:X\to \mathbb{R}$p:X rarrRp: X \rightarrow \mathbb{R}$p:X\to \mathbb{R}$ be defined by $p(x)=\Vert {f|}^{\ast }\cdot \Vert x\mid ,x\in X$$p(x)=‖{\left.f\right|}^{\ast }\cdot ‖x\mid ,x\in X$p(x)=||f|^(**)*||x∣,x in Xp(x)=\left\|\left.f\right|^{*} \cdot\right\| x \mid, x \in X$p(x)=\Vert {f|}^{\ast }\cdot \Vert x\mid ,x\in X$.
The functional $p$$p$pp$p$ is convex, positively homogeneous and $f(y)\le \Vert {f|}^{\ast }\cdot \Vert y\mid$$f(y)\le ‖{\left.f\right|}^{\ast }\cdot ‖y\mid$f(y) <= ||f|^(**)*||y∣f(y) \leq\left\|\left.f\right|^{*} \cdot\right\| y \mid$f(y)\le \Vert {f|}^{\ast }\cdot \Vert y\mid$, for every $y\in Y$$y\in Y$y in Yy \in Y$y\in Y$. By the Hahn-Banach extension theorem ([8, p. 484]) there exists a linear functional $F:X\to \mathbb{R}$$F:X\to \mathbb{R}$F:X rarrRF: X \rightarrow \mathbb{R}$F:X\to \mathbb{R}$ such that ${F|}_Y=f$${\left.F\right|}_Y=f$F|_(Y)=f\left.F\right|_{Y}=f${F|}_Y=f$ and $F(x)\le \Vert {f|}^{\ast }\Vert x\mid$$F(x)\le ‖{\left.f\right|}^{\ast }‖x\mid$F(x) <= ||f|^(**)||x∣F(x) \leq\left\|\left.f\right|^{*}\right\| x \mid$F(x)\le \Vert {f|}^{\ast }\Vert x\mid$, for every $x\in X$$x\in X$x in Xx \in X$x\in X$. It follows

On the other hand

showing that $\Vert F{\Vert }^{\ast }=\Vert {f|}^{\ast }$$\Vert F{\Vert }^{\ast }=\Vert {\left.f\right|}^{\ast }$||F||^(**)=||f|^(**)\|F\|^{*}=\|\left. f\right|^{*}$\Vert F{\Vert }^{\ast }=\Vert {f|}^{\ast }$.

For $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ denote by

the set of all extensions that preserve the asymmetric norm.
By Theorem 1, the set $\mathcal{E}(f)$$\mathcal{E}(f)$E(f)\mathcal{E}(f)$\mathcal{E}(f)$ is always nonempty.
The problem of finding necessary or/and sufficient conditions in order that every $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ have a unique norm preserving extension is closely related to a best approximation problem in the space $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$ equipped with the asymmetric norm $\Vert {F|}^{\ast s}=sup\{F(x):\Vert x\mid \le 1\}$$\Vert {\left.F\right|}^{\ast s}=sup\{F(x):\Vert x\mid \le 1\}$||F|^(**s)=s u p{F(x):||x∣≤1}\|\left. F\right|^{* s}=\sup \{F(x): \| x \mid \leq 1\}$\Vert {F|}^{\ast s}=sup\{F(x):\Vert x\mid \le 1\}$.

Concerning the following notions, in the case of usual spaces, one can consult Singer's book [16].

Let $(Y,\Vert \cdot \Vert )$$(Y,\Vert \cdot \Vert )$(Y,||*||)(Y,\|\cdot\|)$(Y,\Vert \cdot \Vert )$ be subspace of $(X,\Vert \cdot \Vert )$$(X,\Vert \cdot \Vert )$(X,||*||)(X,\|\cdot\|)$(X,\Vert \cdot \Vert )$ and let

the annihilator of $Y$$Y$YY$Y$ in the space ( $X^{\ast s},\Vert {\cdot |}^{\ast s}$$X^{\ast s},\Vert {\left.\cdot \right|}^{\ast s}$X^(**s),||*|^(**s)X^{* s}, \|\left.\cdot\right|^{* s}$X^{\ast s},\Vert {\cdot |}^{\ast s}$ ).
For $F\in X^{\ast s}$$F\in X^{\ast s}$F inX^(**s)F \in X^{* s}$F\in X^{\ast s}$, an element $G_0\in Y^{\perp }$$G_0\in Y^{\perp }$G_(0)inY^(_|_)G_{0} \in Y^{\perp}$G_0\in Y^{\perp }$ is called a best approximation element for $F$$F$FF$F$ in $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ if

The set of all best approximation elements for $F$$F$FF$F$ in $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ is denoted by $P_{Y^{\perp }}(F)$$P_{Y^{\perp }}(F)$P_(Y^(_|_))(F)P_{Y^{\perp}}(F)$P_{Y^{\perp }}(F)$. If $P_{Y^{\perp }}(F)\ne \mathrm{\varnothing }$$P_{Y^{\perp }}(F)\ne \mathrm{\varnothing }$P_(Y^(_|_))(F)!=O/P_{Y^{\perp}}(F) \neq \emptyset$P_{Y^{\perp }}(F)\ne \mathrm{\varnothing }$ for every $F\in X^{\ast s}$$F\in X^{\ast s}$F inX^(**s)F \in X^{* s}$F\in X^{\ast s}$ one says that $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ is proximinal, and if $\mathrm{card}{P}_{Y^{\perp }}(F)=1$$\mathrm{card}{P}_{Y^{\perp }}(F)=1$cardP_(Y^(_|_))(F)=1\operatorname{card} P_{Y^{\perp}}(F)=1$\mathrm{card}{P}_{Y^{\perp }}(F)=1$ for every $F\in X^{\ast s}$$F\in X^{\ast s}$F inX^(**s)F \in X^{* s}$F\in X^{\ast s}$, then one says that $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ is Chebyshevian.

Theorem 2. Let ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ) be a space with asymmetric norm and ( $Y,\Vert \cdot \Vert$$Y,\Vert \cdot \Vert$Y,||*||Y,\|\cdot\|$Y,\Vert \cdot \Vert$ ) a subspace of it. Then
a) The annihilator $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ of $Y$$Y$YY$Y$ is a proximinal subspace of $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$ and, for every $F\in X^{\ast }$$F\in X^{\ast }$F inX^(**)F \in X^{*}$F\in X^{\ast }$ we have

b) An element $G\in Y^{\perp }$$G\in Y^{\perp }$G inY^(_|_)G \in Y^{\perp}$G\in Y^{\perp }$ is in $P_{Y^{\perp }}(F)$$P_{Y^{\perp }}(F)$P_(Y^(_|_))(F)P_{Y^{\perp}}(F)$P_{Y^{\perp }}(F)$ if and only if $G=F-H$$G=F-H$G=F-HG=F-H$G=F-H$ for some $H\in \mathcal{E}\left({F|}_Y\right)$$H\in \mathcal{E}\left({\left.F\right|}_Y\right)$H inE(F|_(Y))H \in \mathcal{E}\left(\left.F\right|_{Y}\right)$H\in \mathcal{E}\left({F|}_Y\right)$, i.e.,

c) The subspace $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ is Chebyshevian in $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$ if and only if every functional $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ has a unique norm preserving extension in $X^{\ast s}$$X^{\ast s}$X^(**s)X^{* s}$X^{\ast s}$.

Proof. a) Let $F\in X^{\ast }$$F\in X^{\ast }$F inX^(**)F \in X^{*}$F\in X^{\ast }$. Then ${F|}_Y\in Y^{\ast }$${\left.F\right|}_Y\in Y^{\ast }$F|_(Y)inY^(**)\left.F\right|_{Y} \in Y^{*}${F|}_Y\in Y^{\ast }$ and, by Theorem 1, $\mathcal{E}\left({F|}_Y\right)\ne \mathrm{\varnothing }$$\mathcal{E}\left({\left.F\right|}_Y\right)\ne \mathrm{\varnothing }$E(F|_(Y))!=O/\mathcal{E}\left(\left.F\right|_{Y}\right) \neq \emptyset$\mathcal{E}\left({F|}_Y\right)\ne \mathrm{\varnothing }$. If $H\in \mathcal{E}\left({F|}_Y\right)$$H\in \mathcal{E}\left({\left.F\right|}_Y\right)$H inE(F|_(Y))H \in \mathcal{E}\left(\left.F\right|_{Y}\right)$H\in \mathcal{E}\left({F|}_Y\right)$ then

so that $F-H\in Y^{\perp }$$F-H\in Y^{\perp }$F-H inY^(_|_)F-H \in Y^{\perp}$F-H\in Y^{\perp }$. We have

and, for any $G\in Y^{\perp }$$G\in Y^{\perp }$G inY^(_|_)G \in Y^{\perp}$G\in Y^{\perp }$,

Taking the infimum with respect to $G\in Y^{\perp }$$G\in Y^{\perp }$G inY^(_|_)G \in Y^{\perp}$G\in Y^{\perp }$ we get

It follows that $d(F,Y^{\perp })=\Vert {{F|}_Y|}^{\ast }$$d\left(F,Y^{\perp }\right)=\Vert {\left.{\left.F\right|}_Y\right|}^{\ast }$d(F,Y^(_|_))=||F|_(Y)|^(**)d\left(F, Y^{\perp}\right)=\|\left.\left. F\right|_{Y}\right|^{*}$d(F,Y^{\perp })=\Vert {{F|}_Y|}^{\ast }$.
b) If $G\in P_{Y^{\perp }}(F)$$G\in P_{Y^{\perp }}(F)$G inP_(Y^(_|_))(F)G \in P_{Y^{\perp}}(F)$G\in P_{Y^{\perp }}(F)$ then

and

which shows that $F-G\in \mathcal{E}\left({F|}_Y\right)$$F-G\in \mathcal{E}\left({\left.F\right|}_Y\right)$F-G inE(F|_(Y))F-G \in \mathcal{E}\left(\left.F\right|_{Y}\right)$F-G\in \mathcal{E}\left({F|}_Y\right)$. The conclusion holds with $H=F-G$$H=F-G$H=F-GH=F-G$H=F-G$.
c) Follows from b).

Remark. $1^{\circ }$$1^{\circ }$1^(@)1^{\circ}$1^{\circ }$ Let $X^{\ast }$$X^{\ast }$X^(**)X^{*}$X^{\ast }$ be the usual topological algebraic dual of the normed space ( $X,\Vert \cdot \Vert$$X,\Vert \cdot \Vert$X,||*||X,\|\cdot\|$X,\Vert \cdot \Vert$ ), and $Y^{\ast }$$Y^{\ast }$Y^(**)Y^{*}$Y^{\ast }$ the topological algebraic dual of ( $Y,\Vert \cdot \Vert$$Y,\Vert \cdot \Vert$Y,||*||Y,\|\cdot\|$Y,\Vert \cdot \Vert$ ), where $Y$$Y$YY$Y$ is a subspace of $X$$X$XX$X$. R. R. Phelps [14] showed that $Y^{\perp }$$Y^{\perp }$Y^(_|_)Y^{\perp}$Y^{\perp }$ (the annihilator of $Y$$Y$YY$Y$ in $X^{\ast })$$\left.X^{\ast }\right)${:X^(**))\left.X^{*}\right)$X^{\ast })$ is Chebyshevian if and only if every $f\in Y^{\ast }$$f\in Y^{\ast }$f inY^(**)f \in Y^{*}$f\in Y^{\ast }$ has a unique norm-preserving extension $F\in X^{\ast }$$F\in X^{\ast }$F inX^(**)F \in X^{*}$F\in X^{\ast }$.
$2^{\circ }$$2^{\circ }$2^(@)2^{\circ}$2^{\circ }$ Some Phelps type duality results and applications can be found in [3], and in the bibliography quoted there.

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Received by the editors: February 3, 2003.