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.



Finding the set of best approximation
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 $P\_{B\_Y}(x)=P\_Y(x)\cap B\_Y$.


