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Find a Banach space $X$, a closed linear subspace $Y ⊂ X$, and a vector $x_0 ∈ X$ such that the closest point of $Y$ to $x_0$ is not unique.

Coiuld you please give me some idea for this question? Thank you so much.

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  • $\begingroup$ Use $\mathbb R^2$ with the $\infty$-norm (max-norm) $\endgroup$ – daw Dec 21 '17 at 9:57
  • $\begingroup$ Could you please explain more? Thanks a lot for your help $\endgroup$ – Ross Dec 21 '17 at 10:04
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Consider $X=\mathbb R^2$ with $\|x\|_\infty=\max(|x_1|,|x_2|)$. Set $Y=\{x: x_2=0\}$. Take $x_0 = (0,1)$. Now compute the set of closest points to $x_0$ on $Y$: it is an interval.

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Take $\ell^\infty$, the Banach space of all bounded real sequences. $c_0$, the space of all sequences converging to $0$ is a closed subspace of $\ell^\infty$. Consider $e = (1, 1, 1, \ldots) \in \ell^\infty$.

The distance from $c_0$ to $e$ is equal to $1$. Indeed, we have $d(e, c_0) \ge \|e - (0, 0, \ldots)\| = 1$.

Conversely, let $x = (x_n)_{n=1}^\infty \in c_0$ be arbitrary. For all $n \in \mathbb{N}$ we have: $$\|e - x\|_\infty \ge |1 - x_n| \ge 1 - |x_n| \xrightarrow{n\to\infty} 1$$

Therefore, $d(e, c_0) = \displaystyle\inf_{x \in c_0} \|e - x\|_\infty = 1$.

However, the point in $c_0$ closest to $e$ is not unique. For example, consider $$e_n = (\underbrace{0, \ldots, 0}_{n-1}, 1, 0, \ldots) \in c_0$$ for any $n \in \mathbb{N}$.

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