Finding the set of best approximation

Question

Given $X$=$l^1$ and its dual space $X^*=l^\infty$. Now take $f=(1, 1/2, 2/3, 3/4,...) \in X^*$. Then clearly $\|f\|_\infty = 1$. I have found that $H=\ker f$ is a proximinal hyperplane in $X$.

Note: A subspace $Y$ of a normed linear space $X$ is called proximinal if for all $x \in X$, $P_Y(x) \neq \emptyset$, where $P_Y(x)= \{ y \in Y : \|x-y\| = d(x, Y) \}$. Here $y\in P_Y(x)$ will be called a best approximation; the distance function is defined as $d(x, Y)= \inf \|x-y_0\|$, $\forall y_0 \in Y$. I need to find the set of best approximations $B_Y(x)$ of $x$ to $B_Y$, where $B_Y$ is the closed unit ball in $Y$ defined as $B_Y=\{y_0 \in Y: \|y_0\| \leq 1\}$. I am stuck at the confusion that whether $B_Y$ exists or not and if it exists, what would $P_{B_Y}(x)$ be? Kindly help me. Thank you in advance.

Answer 1

Similar to $P_Y(x)$, there is no such ready formula for evaluating $P_{B_Y}(x)$, when $Y=\ker (f)$, and so is for $d(x,B_Y)$. In some cases, for instance when $d(x,Y)=d(x,B_Y)$, it is easier to understand $P_{B_Y}(x)$. In fact in few cases one can write $P_{B_Y}(x)=B_Y\cap P_Y(x)$, when $d(x,Y)=d(x,B_Y)$.

Now the point $f=(1,1/2,2/3,\ldots)\in\ell_\infty$ is a smooth point of $\ell_\infty$ (in fact Frechet smooth) and $f (e_1)=1$. Hence $J_X (f)\neq\emptyset$ and a singleton. It now follows that $P_Y(e_1)=\{0\}$ and clearly $P_{B_Y}(e_1)=\{0\}$ as $d(e_1,Y)=d(e_1,B_Y)$. Here $Y=\ker (1,1/2,2/3,\ldots)$ and $J_X(f)=\{x\in S_X: f(x)=1\}$.

Now $Y$ is a strongly proximinal subspace of $\ell_1$ as $(1,1/2,2/3,\ldots)$ is an SSD point of $\ell_\infty$. This follows that $d(e_i,B_Y)>d(e_i,Y)$, whenever $i>1$. So in these cases one needs to evaluate separately $d(e_i,B_Y)$ and figure out whether $P_{B_Y}(e_i)\neq\emptyset$. See (Proposition 2.10, Indumathi, Lalithambigai, J. Convex Analysis, {vol 18}(2), 353--366, (2011)).

For a point $x\in\ell_1$, suppose $\inf\{\|y\|:y\in P_Y(x)\}\leq 1$. Then $d(x,Y)=d(x,B_Y)$ if and only if $P_{B_Y}(x)=P_Y(x)\cap B_Y$.