Counterexample of the existence of length minimizer in Banach space

Question

Consider the Banach space $c_0$ of all sequences that converges to $0$ with supremum norm. Let $$ E=\{x=(x_n)\in c_0:\sum_{n=1}^\infty 2^{-n}x_n=0\}\\ r=(2,0,0,0,...)$$ How can I show that for all $x\in E$, $d(x,r) \neq d(E,r)=\inf\{d(x,r)|x\in E\}$? i.e. Length minimizer of $r$ respect to $E$ does not exist.

Answer 1

It's usually important to first figure out the distance $d(r, E)$. Knowing this number often helps formulate a proof. In this case, we have (for reasons that we're about to see) $d(r, E) = 1$.

Consider a sequence $(x_n) \in c_0$ such that $\|x_n - r\| \le 1$. That is, $$\max\left\lbrace |2 - x_1|, \sup_{n \ge 2} |x_n| \right\rbrace \le 1.$$ We wish to show that $(x_n) \notin E$. If this were the case, then $$\sum_{n=1}^\infty 2^{-n}x_n = 0.$$ Since $|2 - x_1| \le 1$, we have $x_1 \ge 1$. We also have $x_n \ge -1$ for $n \ge 2$. So, $$0 = \frac{x_1}{2} + \sum_{n=2}^\infty 2^{-n} x_n \ge \frac{1}{2} - \sum_{n=2}^\infty 2^{-n} = 0$$ Hence, the inequality is an equality. If $x_1 > 1$ or if $x_n > -1$ for some $n \ge 2$, then the inequality would have been strict, so $x_1 = 1$, and $x_n = -1$ for all $n \ge 2$. This is not possible, as $x_n \to 0$. Therefore, $x_n \notin E$.

On the other hand, for all $n, k$, let $$x_n^k = \begin{cases}1 - 2^{-k} & \text{if } n = 1 \\ -1 & \text{if } 1 < n \le k + 1 \\ 0 & \text{if } n \ge k + 1 \end{cases}.$$ Note that, for all $k$, we have $(x_n^k)_{n=1}^\infty \in E$, and $\|x_n^k - r\| = 2^{-k} \to 0$.