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Let $M=\{(x_n)_{n\in\mathbb{N}}\in c_0: \sum_{n=1}^{\infty}\frac{x_n}{2^n}=0\}$ be closed subspace of a Banach space $c_0=\{(x_n)\in l^{\infty}: \lim_{n\rightarrow\infty}x_n=0\}$. Let $x\not\in M$. If there is a $z\in M$ such that $d(x,M)=d(x,z)$ ¿What can you say about $z$? According to what I read at some point there is no said z, but I have no idea how to test it, any idea how to start?

I think that if $M$ were compact there would be no $z$ that satisfies that, but in this situation nothing occurs to me. If $M$ were finite dimension then if said z would exist, but this is not the case.

Here is a fairly similar question but with a fairly general answer

Zaragosa
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  • Your first question is too vague. Your statement about $M$ being compact does not make much sense. Perhaps you should just stick to the specific case you are interested in. – Kavi Rama Murthy Nov 01 '21 at 23:28
  • The condition $d(x, M) = d(x, z)$, where $z \in M$ means that $z$ is considered the metric projection of $x$ onto $M$ (this is true whether or not $M$ is a subspace). If $M$ is a compact subset (only the trivial subspace is compact), then such a projection always exists (though there may be multiple projections if the space is not strictly convex, or if $M$ is not convex). This is as much information as I can provide, as I don't understand how $M$ or the projection problem relates to proving $c_0$ is a closed subspace of $\ell^\infty$. – Theo Bendit Nov 01 '21 at 23:31
  • @KaviRamaMurthy You are very right, my question was very vague, I already edited it. – Zaragosa Nov 01 '21 at 23:33

1 Answers1

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$M$ is a hyperplane; it is the kernel of the linear functional $f : (x_n) \mapsto \sum_{n=1}^\infty \frac{x_n}{2^n}$, which is bounded, as $$\left|\sum_{n=1}^\infty \frac{x_n}{2^n}\right| \le \sum_{n=1}^\infty \frac{|x_n|}{2^n} \le \sum_{n=1}^\infty \frac{\|x\|}{2^n} = \|x\|. \tag{$\star$}$$ This implies that it is a subspace, and is closed (which I know you already know).

Generally speaking, the metric projection onto the kernel of a bounded linear functional exists if and only if the functional achieves a maximum on the unit sphere. I would like to show this is not true for our particular functional.

First, let us show that $(\star)$ never achieves equality on the unit sphere. If $\|x\| = 1$ is a counterexample, then we would have $$\sum_{n=1}^\infty \frac{|x_n|}{2^n} = \sum_{n=1}^\infty \frac{\|x\|}{2^n}$$ which occurs when $x_n$ is constantly equal to $\|x\|$, which is impossible, since $\|x\| = 1$, and $x_n \to 0$. That said, we can get arbitrarily close by considering $$x^m = \sum_{i=1}^m e^i = (\underbrace{1, 1, \ldots, 1}_{m \text{ times}}, 0, 0, \ldots) \in S_X,$$ on which we have $$f(x^m) = \sum_{n=1}^\infty\frac{x_n}{2^n} = \sum_{n=1}^m\frac{1}{2^n} = 1 - \frac{1}{2^m} \to 1 = \|x^m\|.$$ Thus, $\|f\| = 1$, but $f$ never achieves its supremum on the unit sphere.

Now, let's generalise the problem. Suppose $f \in X^*$ fails to achieve a maximum on $S_X$ and $M = \ker f$. Further, suppose $x \notin M$ and $y \in P_M(x)$ (i.e. the set of nearest points in $M$ to $x$).

Because $f$ is continuous, $M$ is closed, so $d(x, M) > 0$. By dividing through by $d(x, M)$, assume without loss of generality that $d(x, M) = 1$ (implicit here: $d(\alpha x, M) = \alpha d(x, M)$, which is true because $M$ is a subspace). Note that $\|x - y\| = 1$. I claim that $|f|$ achieves a maximum on the sphere at $x - y$, against assumption. Specifically, $|f(x - y)| = |f(x)| = 1$.

Indeed, given $u \in S_X$, let $z=x-\frac{f(x)}{f(u)}u$. Then $$f(z)=f(x)-f(x)\frac{f(u)}{f(u)}=0\implies z\in M.$$ We then have $$1 = d(z, M) \le \|x-z\| = \left|\frac{f(x)}{f(u)}\right|\|u\| = \frac{|f(x)|}{|f(u)|}\implies |f(x-y)|=|f(x)|\ge|f(u)|.$$ This implies that $|f|$ achieves a maximum at $x-y$, so $f$ achieves a maximum at $x-y$ or $y-x$. Either way, this is a contradiction; no nearest point to $x$ can exist in $M$.

Theo Bendit
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