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Let $M=\{\{ x_n \} \in c_0 : \sum_{n = 1}^\infty \frac{x_n}{2^n} = 0 \}$. Prove that $\forall x \in c_0 - M$, there is no closest point to $x$ in $M$.

I want to solve the problem by proving that for every $y_0 \in M$, there exists $y_1 \in M$ where $\Vert x - y_1 \Vert < \Vert x - y_0 \Vert$, where $\|\cdot\|$ is the supremum norm on $c_0$. But I don't know how to construct $y_1$.

This looks like this question, but the construction seems not suitable here.

Arctic Char
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namasikanam
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1 Answers1

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Let $y = \{y_n\}$ be an element in $M$. Then

$$ \|x-y\| = \sup_{n\in \mathbb N}|x_n - y_n|.$$

Let $K \subset \mathbb N$ so that $n\in K$ if and only if $|x_n - y_n| = \|x-y\|$. If $x\notin M$, we have $\|x-y\| >0$. Since $x, y\in c_0$, there is $N\ge n_0$ so that $$|x_n|, |y_n|< \frac 14\|x - y\| \text{ for all }n \ge N.$$ This implies that $n < N$ for all $n\in K$ and thus $K$ is finite. Let $$A = \sup_{n\notin K} |x_n-y_n| < \|x-y\|$$ and let $\epsilon < \|x-y\| -A$. Let $y^\epsilon\in c_0$ be given by

$$ y^\epsilon_n = \begin{cases} y_n + \epsilon & \text{ if }n\in K, x_n > y_n, \\y_n - \epsilon & \text{ if }n\in K, x_n < y_n, \\ y_n & \text{ otherwise}. \end{cases}$$

Then $$|x_n - y^\epsilon_n|=\begin{cases} \|x-y\|-\epsilon, & \text{ if } n\in K,\\ |x_n - y_n| & \text{ otherwise}.\end{cases}.$$

In particular, $\|x-y^\epsilon\| \le \|x-y\|-\epsilon$.

Using that $y\in M$, $$ \sum_{n=1}^\infty \frac{y_n^\epsilon}{2^n} = L\epsilon,$$ where $$L = \sum_{n\in K} \frac{\pm 1}{2^n}$$ and the sign $\pm 1$ depends on whether $x_n<y_n$ or $x_n >y_n$ for each $n\in K$. Note that $L$ is never zero (indeed, if $L = 0$ we are done, but it isn't). Now choose $\epsilon$ small enough so that $$ 2^N L\epsilon < \frac 14 \|x-y\|$$ and let $y^1\in c_0$, $$ y_n^1 = \begin{cases} y_n - 2^N L\epsilon & \text{ if } n= N, \\ y_n^\epsilon & \text{ otherwise}. \end{cases}.$$

Then $$|x_N - y^1_N| \le |x_N| + | y_N| + 2^N L\epsilon < \frac 34 \|x-y\|$$ and $|x_n - y^1_n| = |x_n - y^\epsilon_n|$ otherwise. Thus $$ \|x-y^1\| \le \min\left\{ \frac 34 \| x-y\| , \|x-y\|-\epsilon\right\} < \|x-y\|$$ and $y^1\in M$ since $$ \sum_{n=1}^\infty \frac{y^1_n }{2^n} = \sum_{n=1}^\infty \frac{y^\epsilon_n }{2^n} - 2^N L\epsilon = 0.$$

Arctic Char
  • 16,007