# A New Theorem?

I can’t actually take much credit for this. I made the initial conjecture that sparked this theorem, but my friend John Bytheway was the first person to codify the actual theorem and prove it. My proof of it is pretty independent of his, but I probably wouldn’t have come up with it unless I already knew the theorem was true. (The neccesary part of the theorem is stolen from him, but the sufficiency part is entirely mine).

Note: In this post I take the slightly unconventional approach that a Normal space needn’t be $$T_1$$. i.e. a Normal topological space is one in which any two disjoint closed sets have disjoint neighbourhoods, but points needn’t be closed. I could invent a new term for this, like quasinormal, but I really can’t be bothered. Feel free to replace ‘normal’ with ‘quasinormal’ wherever you see it if this makes you more comfortable. It may be readily verified that the usual proof of Urysohn’s lemma in no way depends on the closedness of points, so Urysohn’s lemma holds for this new definition of normal.

First we need some preliminary definitions.

### Definition 1:

Let $$X$$ be a topological space and $$f : X \to \mathbb{R}$$ a function, not neccesarily continuous. For $$x \in X$$ we define the oscillation of $$f$$ at $$x$$ \to be

$$\omega_f(x) = \inf\limits_{U \ni x} \mathrm{diam} f(U)$$

where $$U$$ ranges over open sets.

We define the total oscillation of $$f$$ to be

$$\delta(f) = \sup\limits_{x \in X} \omega_f (x)$$

We now prove some preliminary results about $$\delta$$

### Proposition 2:

Let $$X$$ be an arbitrary topological space and consider $$B(X)$$ the Banach space of bounded functions on $$X$$ and $$C(X)$$ the closed subspace consisting of the continuous bounded functions. Consider $$\delta : B(X) \to [0, \infty)$$

1. $$\delta(f) = 0$$ iff $$f$$ is continuous.
2. $$\delta(f + g) \leq \delta(f) + \delta(g)$$
3. $$\delta(tf) = |t| \delta(f)$$
4. $$\delta(f) \leq 2 ||f||$$
5. $$\delta(f) \leq 2 d(f, C(X))$$

These are all perfectly trivial to prove, so I’m not going to bother.

What I noticed is that in almost every case I considered the final inequality was in fact an equality. I could prove a weaker result for Compact hausdorff spaces (I showed that $$||f|| \leq \delta(f)$$, but I couldn’t do any better, so I passed this over to John to see what he could come up with. He came up with the following theorem, which is the main theorem of this post:

$$\delta(f) = 2 d(f, C(X))$$ for every $$f \in B(X)$$ iff $$X$$ is normal.

I’ll prove this in two parts. For the right to left implication I’ll in fact prove something stronger:

### Theorem 4:

Let $$X$$ be a normal topological space and let $$f \in B(X)$$. Then there is a (not usually unique) continuous function $$h$$ with $$||f – h|| = 2 \delta(f)$$.

As is my wont, the proof of this will precede by a couple clever definitions and then drop out as practically a corollary of a Big Theorem.

The big theorem in question is the following, which is due to Tong. I conjectured it, tinkered around with proving it for a bit, went to look up something about semicontinuous functions in Engelking’s general topology book and saw the theorem staring out at me from one of the exercises.

### Theorem 5:

Let $$X$$ be a normal topological space. Let $$f, g : X \to \mathbb{R}$$ be upper semicontinuous and lower semicontinuous respectively with $$f \leq g$$. There is a continuous function $$h$$ with $$f \leq h \leq g$$.

Note the direction of the inequality and which are upper and lower semicontinuous respectively. If you reverse this then the theorem becomes false.

I’m not actually going to prove this here – I’ve not yet totally sorted out the details of the proof in my mind. It’s basically a modified version of the standard proof of Urysohn’s lemma, but with additional constraints on the sets constructed. (Update: See here for a proof.)

Anyway, we now make some more definitions:

### Definition 6:

Let $$f \in B(X)$$. Define

$$f^*(x) = \inf\limits_{U \ni x} \sup f(U)$$

$$f_*(x) = \sup\limits_{U \ni x} \inf f(U)$$

where U ranges over open sets.

### Proposition 7:

These satisfying the following properties:

1. $$f_* \leq f \leq f^*$$
2. $$f_*$$ is lower semicontinuous. $$f^*$$ is upper semicontinuous.
3. If $$f$$ is upper semicontinuous then $$f^* = f$$. Similary if $$f$$ is lower semicontinuous then $$f_* = f$$.
4. The maps $$f \to f^*$$ and $$f \to f_*$$ are monotone with respect to the pointwise ordering.
5. If $$g, h$$ are lower, upper semicontinuous respectively and $$g \leq f \leq h$$ then $$g \leq f_* \leq f \leq f^* \leq h$$
6. $$\omega_f(x) = f^*(x) – f_*(x)$$

Again, these are all really very easy to prove (assuming you prove them in order), so I’m not going to do it. I’ll actually not use most of these, but those that I don’t use are of independent interest. i.e. they’re cool. :-)
Now, we have:

#### Proof of theorem 4:

Note that $$\delta(f) = \sup (f^*(x) – f_*(x)$$. Consequently we have that

$$f^* – \frac{1}{2}\delta(f) \leq f_* + \frac{1}{2} \delta(f)$$

Now, the left hand side is upper semicontinuous and the right hand side is lower semicontinuous. Thus we have an interpolating continuous function $$h$$.

So

$$f^* – \frac{1}{2}\delta(f) \leq h \leq f_* + \frac{1}{2} \delta(f)$$

But we have that $$f \leq f^*$$ and $$f_* \leq f$$.

So

$$f – \frac{1}{2}\delta(f) \leq h \leq f + \frac{1}{2} \delta(f)$$

i.e. $$||f – h|| \leq \frac{1}{2} \delta$$
But we know that $$||f – h|| \geq \frac{1}{2} \delta$$ by our very first proposition. Hence we have equality.

QED

### Lemma 8:

If for every $$f \in B(X)$$ we have $$d(f, C(X)) = \frac{1}{2} \delta(f)$$ then $$X$$ is normal.

#### Proof:

This proof is entirely John’s.

Let $$A, B$$ be disjoint closed sets. Define $$f : X \to \mathbb{R}$$ by $$f|_A = -1$$, $$f|_B = 1$$ and $$f(x) = 0$$ otherwise.

By considering appropriate neighbourhoods we may see that $$\omega_f(x) \leq 1$$ for every $$x$$. i.e. $$\delta(f) \leq 1$$. Consequently we have $$d(f, C(X)) \leq \frac{1}{2}$$ and may find a continuous function $$h$$ with $$||f – h|| \leq \frac{3}{4}$$.

Thus we have that $$h|_A \leq – \frac{1}{4}$$ and $$h|_B \geq \frac{1}{4}$$. Composing with some appropriate function from $$[-1, 1] \to [0, 1]$$ we get a continuous function $$g$$ which separates the two closed sets. Then $$g^{-1}([0, \frac{1}{2}))$$ and $$g^{-1}((\frac{1}{2}, 1])$$ are appropriate disjoint neighbourhoods of $$A$$ and $$B$$.

QED

This entry was posted in Numbers are hard on by .

## 2 thoughts on “A New Theorem?”

1. Etrev

I find this scratch interesting but please let me ask in relation to proposition 7. Given that f is an upper semicontinuous function on the reals, can we clasify the points where limits from the right or left does not exist?.

And by its own definition it is natural to consider the function F(x)=lim sup f(x) (from the right), is it a nice function?.

This is problem i have to deal with, and i dont know if this a standar result!!!!. Please help!.

2. david

Hi Etrev. Sorry I didn’t answer earlier, but I’ve been trying to think of something useful to add. Unfortunately I haven’t come up with anything especially pertinent, and I’m not aware of any relevant theory.

What exactly do you mean by ‘classifying’ these points?