# Radii in function spaces

Let $$V$$ be a metric space and $$A, W \subseteq V$$. Define

$r(A, W) = \inf \{ R : \exists w \in W, A \subseteq \overline{B}(w, R) \}$

Where $$\overline{B}(w, R)$$ is the closed ball of radius $$R$$ centered on $$w$$.

i.e. we’re looking at the radius of $$A$$ when the center is restricted to lie in $$W$$.

When $$W = V$$ we will simply write $$r(A)$$

A $$W$$-center for $$A$$ is a point $$w \in W$$ such that $$A \subseteq \overline{B}(w, r(A, W))$$. Note that it is not guaranteed that such a point exists (example: Let $$V = W = [-1, 1] \setminus \{0\}$$. $$r(V) = 1$$ but V has no center).

The case we’re interested in for this post is where we have some topological space $$X$$ and set $$V = l^\infty(X)$$, the space of bounded functions $$X \to \mathbb{R}$$, and $$W = C(X)$$, the subspace consisting of all continuous functions.

Why are we interested in this? Good question. I have absolutely no practical reason for studying this question and am not aware of any applications. There were three reasons I found it interesting:

1. It can be regarded as the question “Given a bunch of possibly discontinuous functions, how well can you simultaneously approximate them by a continuous function” (this is the problem I started with which lead to the whole paper, although with only one discontinuous function)
2. Studying radius is a fairly natural geometric problem and $$C(X)$$ is a natural example of a Banach space
3. There’s a lot of surprisingly neat theory involved in the question

In this post we’ll establish three results:

1. A very nice characterization of $$r(\mathcal{F}, C(X))$$
2. If $$X$$ is normal then every bounded $$\mathcal{F} \subseteq l^\infty(X)$$ has a $$C(X)$$-center
3. A theorem about how the C(X) radius of $$\mathcal{F} \subseteq l^\infty(X)$$ can be determined from “sufficiently large” subsets (see later)

### A characterization of $$r(\mathcal{F}, C(X))$$ and the existence of centers

Let $$\mathcal{F} \subseteq l^\infty(X)$$ be bounded. We define the following functions:
$$\mathcal{F}^*(x) = \inf\limits_{U \ni x} \sup \bigcup\limits_{f \in \mathcal{F}} f(U)$$ $$\mathcal{F}_*(x) = \sup\limits_{U \ni x} \inf \bigcup\limits_{f \in \mathcal{F}} f(U)$$

where U is restricted to ranging over open sets.

Clearly for $$f \in \mathcal{F}$$ we have $$\mathcal{F}_*(x) \leq f(x) \leq \mathcal{F}^*(x)$$. The idea is that these functions provide better behaved tight bounds on $$\mathcal{F}$$. Specifically:

Theorem: $$\mathcal{F}^*(x)$$ is upper semicontinous. $$\mathcal{F}_*(x)$$ is lower semicontinuous.
Proof:

We’ll just show that $$\mathcal{F}^*(x)$$ is upper semicontinous. The other case will follow similarly.

So we must show that the set $$A = \{ x : \mathcal{F}^*(x) < t \}$$ is open. Let $$x \in A$$. Then there exists $$U \ni x$$ such that $$\forall f \in F, y \in U, f(y) < t$$ (by definition of $$\mathcal{F}^*$$). But then for any $$y \in U$$ we must have $$\mathcal{F}^*(y) < t$$. Therefore $$A$$ contains an open neighbourhood of x. x was arbitrary, hence A must be open. QED Believe it or not, this is almost all we need to construct centers: Theorem: $$\frac{1}{2} \sup\limits_x (\mathcal{F}^*(x) – \mathcal{F}_*(x)) \leq r(\mathcal{F}, C(X))$$

If $$X$$ is normal then this inequality is an equality and $$\mathcal{F}$$ has a C(X) center.

Proof:

It may seem that this theorem is two unconnected results, but in fact our proof of the latter will simply fall out of the proof of the former.

Let $$R = r(\mathcal{F}, C(X))$$ and $$S = \frac{1}{2} (\sup\limits_x \mathcal{F}^*(x) – \mathcal{F}_*(x))$$.

We will first show $$S \leq R$$.

Let $$\epsilon > 0$$. Let $$g \in C(X)$$ be such that $$\mathcal{F} \subseteq \overline{B}(g, R + \epsilon)$$.

Fix $$x \in X$$ and find $$U \ni x$$ open such that for $$y \in U, |f(x) – f(y)| < \epsilon$$. Then for $$y \in U, f \in \mathcal{F}$$ we have $$|f(y) - g(x)| \leq |f(y) - g(y)| + |g(y) - g(x)| \leq R + 2\epsilon$$. This we must have $$\mathcal{F}^*(x) \leq g(x) + R + 2 \epsilon$$ and $$\mathcal{F}_*(x) \geq g(x) - R - 2 \epsilon$$. Hence we have $$\mathcal{F}^*(x) - \mathcal{F}_*(x) \leq 2R + 4\epsilon$$. But $$\epsilon$$ was arbitrary, so we must have $$\mathcal{F}^*(x) - \mathcal{F}_*(x) \leq 2R$$ and hence $$S \leq R$$. Now assume $$X$$ is normal. From the definition, we have $$\mathcal{F}^*(x) - S \leq \mathcal{F}_*(x) + S$$ The left hand side is upper semicontinuous, the right hand side is lower semi continuous, therefore by the Katětov–Tong insertion theorem there exists a continuous function $$g$$ with $$\mathcal{F}^*(x) – S \leq g \leq \mathcal{F}_*(x) + S$$

Now let $$f \in \mathcal{F}$$

We have $$f(x) – S \leq \mathcal{F}^*(x) – S \leq g \leq \mathcal{F}_*(x) + S \leq f(x) + R$$

Or, rearranging, $$g(x) – S \leq f(x) \leq g(x) + S$$

Hence $$||f – g|| \leq R$$, and $$\mathcal{F} \subseteq \overline{B}(g, S)$$

Thus $$R \leq S$$ and hence $$R = S$$

But additionally we have shown that actually $$\mathcal{F} \subseteq \overline{B}(g, R)$$, and thus $$g$$ is a C(X)-center for $$\mathcal{F}$$.

QED

### Finding radii from radii of subsets

The diameter of a set is very nicely behaved in terms of the diameter of its subsets: It is the supremum of the diameter of all two-element subsets simply by matter of its basic definition. The radius is not so well behaved.

Consider the Banach space $$c_0$$, the space of all sequences converging to 0. Consider the set $$E = \{ e_n \}$$ where $$(e_n)_k$$ is 1 if $$k \leq n$$ and 0 elsewhere. Every finite subset of this has radius half (it is contained in the ball of radius half around $$\frac{1}{2} e_N$$ for some sufficiently large N), but the overall set has radius $$1$$ because any center for it with smaller radius than that would not converge to 0.

It’s fairly easy to turn this into an example of a $$C(X)$$ with similar behaviour.

Let $$X = \{ \pm \frac{1}{n} : n \in \mathbb{N} \} \cup \{ 0 \} \}$$ and let $$\mathcal{F} = \{f_n\}$$ where $$f_n(x) = \mathrm{sign}(x)$$ if \$|x| \geq \frac{1}{n}\), else $$f_n(x) = 0$$.

Then once again any finite subset is contained within $$\overline{B}(\frac{1}{2} f_N, \frac{1}{2})$$ for some sufficiently large N, but $$\mathcal{F}^*(0) = 1$$ and $$\mathcal{F}_*(0) = -1$$ so $$r(\mathcal{F}) = 1$$.

(The idea of this example is that this space looks like two copies of the space of convergent sequences glued together by requiring that they converge to the same thing. We then force that value to be 0 by considering sequences where the desired center converges to opposite things).

So radius with centers in $$C(X)$$ isn’t determined by finite subsets in general (incidentally, it is for centers in $$l^\infty(X)$$). How large do the subsets need to be in order to determine it?

There turns out to be a nice theorem in this regard.

First we’ll need a definition:

Let $$X$$ be a topological space. The character of a point $$x \in X$$ written $$\chi(x)$$ is the smallest cardinality of a neighbourhood base of $$x$$. The character of the space X, written $$\chi(X)$$, is $$\sup\limits_{x \in X} \chi(x)$$.

Lemma: Let $$x$$ be a point with $$\chi(x) \leq \kappa$$ and let $$\mathcal{F} \subseteq l^\infty(X)$$. There exists $$\mathcal{G} \subseteq \mathcal{F}$$ with $$|\mathcal{G}| \leq \kappa$$, $$\mathcal{G}^*(x) = \mathcal{F}^*(x)$$ and $$\mathcal{G}_*(x) = \mathcal{F}_*(x)$$.

Proof:

We’ll construct $$\mathcal{G}_n$$ of size $$\leq \kappa$$ with $$\mathcal{G}^*(x) \geq \mathcal{F}^*(x) – \frac{1}{n}$$ and $$\mathcal{G}_*(x) \leq \mathcal{F}_*(x) + \frac{1}{n}$$. Then $$\mathcal{G} = \bigcup\limits_n \mathcal{G}_n$$ will produce the desired set.

In fact it will suffice to construct a set $$\mathcal{H}$$ with $$\mathcal{H}^*(x) \geq \mathcal{F}^*(x) – \frac{1}{n}$$. We can then repeat an identical construction to get the other condition and take the union of the two sets.

So, let $$\{U_\alpha : \alpha < \kappa \}$$ be a neighbourhood base for $$x$$. By definition of $$\mathcal{F}^*(x)$$ for each $$\alpha$$ we can find $$f_\alpha \in \mathcal{F}$$ with $$\sup f_\alpha(U_\alpha) \geq \mathcal{F}^*(x) - \frac{1}{n}$$. Let $$\mathcal{H} = \{ f_\alpha \}$$. Then if $$U \ni x$$ is open it contains some $$U_\alpha$$ and thus $$\sup f_\alpha(U) \geq \mathcal{F}^*(x) - \frac{1}{n}$$. Hence $$\mathcal{H}^*(x) \geq \mathcal{F}^*(x) - \frac{1}{n}$$ as desired. QED This lemma contains the bulk of the work we'll need to prove the following theorem: Theorem: Let $$X$$ be a topological space with $$\chi(X) \leq \kappa$$. Let $$\mathcal{F} \subseteq l^\infty(X)$$. There exists $$\mathcal{G} \subseteq \mathcal{F}$$ with $$|\mathcal{G}| \leq \kappa$$ and $$r(\mathcal{G}, C(X)) = r(\mathcal{F}, C(X))$$.

Proof:
Let $$R = r(\mathcal{G}, C(X))$$.

We’ll use our previous characterization of the radius. So we can find $$x_n$$ with $$\mathcal{F}^*(x_n) – \mathcal{F}_*(x_n) \geq R – \frac{1}{n}$$. As per the lemma we can find a subset $$\mathcal{G}_n$$ with $$|\mathcal{G}_n| \leq \kappa$$, $$\mathcal{G}_n^*(x_n) = \mathcal{F}^*(x_n)$$ and $$\mathcal{G}_{n*}(x_n) = \mathcal{F}_*(x_n)$$. Then $$r(\mathcal{G}_n) \geq R – \frac{1}{n}$$ hence if $$\mathcal{G} = \bigcup \mathcal{G}_n$$ then $$r(\mathcal{G}) = r(\mathcal{F})$$ as desired.
QED.

In particular if $$X$$ is first countable then the radius is determined by countable subsets.

It’s worth noting that this is not an equivalence. Easy counter examples happen because there are first countable spaces with non first countable compactifications.

It is however tight in that for a regular uncountable cardinal $$\kappa$$ if we take the order topology on two copies $$\{ a_\alpha : \alpha \leq \kappa \}$$ and $$\{ b_\alpha : \alpha \leq \kappa \}$$ with the quotient that $$a_\kappa = b_\kappa$$. We can then construct an example exactly like our modification of the $$c_0$$ example (this relies on the fact that every function in $$C(\kappa)$$ is eventually constant).

### References

• The basic result about centers existing isn’t at all new, but I can’t track down the exact reference right now (various forms of it were proved at various times)
• The inspiration for the theorem about characters comes from a mathoverflow answer by Sergei Ivanov. I have no idea if it’s new or not. I expect not though. It’d be surprising if no one else had proven it or something like it before.
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