In $\ell^{\infty}$, there is exactly one sequence in $c_0$ that has the minimum distance to $(1,1,1,...)$.

I believe this statement is actually false. However, I am having trouble formulating two such sequences with minimal distance to disprove the statement. Could anyone provide a counter-example? Or is this statement actually true?

  • $\begingroup$ $c_0 = \left\{ \displaystyle\lim_{n\to\infty} a_n = 0 \space\ | \space\ a_n \space\ \text{any sequence} \right\}$ $\endgroup$ – Dragonite May 1 '17 at 12:13
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    $\begingroup$ First, what is $\operatorname{dist}(\mathbb{1}, c_0)$? Second, what condition does $x \in c_0$ have to fulfil to realise that distance? $\endgroup$ – Daniel Fischer May 1 '17 at 12:20

As you suspected, the statement IS false: if $\lim_{n\to\infty}a_n=0$, then $|a_n-1|\stackrel{n\to\infty}{\to} 1$. Therefore, for each $\epsilon>0$ there is an $N\in\mathbb{N}$ with $|a_N-1|\geq 1-\epsilon$ so that (let $d$ denote the metric on $\ell^\infty$ induced by the familiar supremum-norm - I tacitly assume that you have this norm in mind) $$d((a_n)_{n\in\mathbb{N}},(1)_{n\in\mathbb{N}})=\sup_{n\in\mathbb{N}}|a_n-1|\geq 1-\epsilon.$$ We conclude $d((a_n)_{n\in\mathbb{N}},(1)_{n\in\mathbb{N}})\geq 1$, hence $dist(c_0,(1)_{n\in\mathbb{N}})\geq 1$. Taking $(a_n)_{n\in\mathbb{N}}=(0)_{n\in\mathbb{N}}$, we see that actually $dist(c_0,(1)_{n\in\mathbb{N}})=1$. Now, we can think of any number of sequences $(a_n)_{n\in\mathbb{N}}\in c_0$ with $d((a_n)_{n\in\mathbb{N}},(1)_{n\in\mathbb{N}})=1$, take for example the sequence of zeros with finitely many zeros replaced by ones.

NOTE: for closed convex subsets $C$ of Hilbert spaces and a vector $x$ in the complement of $C$, it is indeed true that there is a unique vector in $C$ which minimises the distance between $x$ and $C$. So your question furnishes an example that Banach spaces are not sufficient (as I recall, the parallelogram rule of the inner product plays a vital role in the proof).


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