How to formalize my intution of this theorem on continuous functions?

Theorem : If a function $$f$$ is continuous on a closed and bounded interval $$[a, b]$$ then $$f$$ must be uniformly continuous in $$[a, b]$$

My Idea : I get the intuition that for a continuous function on a closed interval, range of $$\delta$$ for a given $$\epsilon$$ must be finite and in particular if we scan the function with 'window' of $$\epsilon$$ and obtain all deltas than $$\inf \{\delta\}$$ must be greater than zero.

Formalizing the idea

Let $$\epsilon \in \mathbb {R} : \epsilon > 0$$

Given, $$f$$ is continuous in $$[a,b]$$, therefore $$\exists \delta (\alpha, \epsilon) > 0$$, s.t.,

$$|x-\alpha| < \delta \implies |f(x) - f(\alpha)| < \epsilon$$

Now, denote $$\delta (\epsilon) = \inf ~\{ \delta (\alpha, \epsilon) ~\forall ~\alpha \in [a,b] \}$$

Now, the challenge remains to prove that this $$\delta (\epsilon) > 0$$ and we are (almost) done. But this is where I get stuck. I have an intuition that if delta goes zero than function must go asymptotic at that point but I am unable to formalize it.

I know there are other standrad proofs of this theorem, but I need to complete it this way because it is closer to what I imagine in my head.

Thanks

• The idea of making a $\delta(\alpha, \varepsilon)$ for each point $\alpha \in [a, b]$ is on the right track, but doesn't immediately give you a way to apply the compactness of the interval. If you can prove that $\delta(\alpha, \varepsilon)$ (or maybe half that) actually works in an open interval around $\alpha$, then you can use the compactness to argue that all of $[a, b]$ is covered by finitely many of these regions, and then take the minimum of the (finitely many) $\delta$ that were used. – Joppy Oct 1 '18 at 14:24
• I find it actually way easier to understand in terms of covers: by Heine-Borel-Lebesgue, closed intervals are compact; continuous images of compacts are themselves compact; now, given any epsilon, since the function is continuous, you have deltas for each point forming an open cover; by compactness, you can extract a finite subcover and then pick the least delta – user359302 Oct 1 '18 at 15:08
• Based on what I understand, I tried to re-attempt the proof by somehow trying to show what @Joppy had to say, but I was unsuccessful. As far as alkchf suggestion is concerned, I have not yet been defined compactness formally in my course so I am hesitant to proceed. A one level higher hint would be appreciated. – Sarthak123 Oct 2 '18 at 4:04

You can prove the uniform continuity theorem without invoking Heine-Borel. The only property related to compactness needed is that a continuous function on a closed, bounded interval attains its minimum at a point in the interval -- and this can be proved without any reference to open covers.

The reasoning is based on finding the infimum of a "maximal" delta for each $$x \in [a,b]$$ and so follows your intuition.

Suppose $$f:[a,b] \rightarrow \mathbb{R}$$ is continuous. Given $$\epsilon > 0$$, for each $$x \in [a,b]$$ let

$$\delta(x) = \sup \{ \delta>0: |f(y)-f(z)| < \epsilon \,\,\forall \,y,z \in [x-\delta/2,x+\delta/2]\}$$

Here $$\delta(x)$$ is the length of the largest interval $$I(x)$$ centered at $$x$$ such that $$|f(y)-f(z)| < \epsilon$$ when $$y,z \in I(x)$$.

We can dispense with the case $$\delta(x) = \infty$$ for some $$x$$ where any $$\delta$$ works for uniform continuity. If $$\delta(x) < \infty$$ for every $$x$$, we can show that $$\delta(x)$$ is continuous.

Consider the intervals centered at points, $$x$$ and $$x + \omega$$ where $$\omega >0$$. If $$x+\omega - \delta(x+\omega)/2 \leq x -\delta(x)/2$$ then $$I(x) \subset I(x+\omega)$$. In this case, the interval $$I(x)$$ could be made larger, which contradicts the construction of $$I(x)$$ as the maximal interval, and it follows that

$$x+\omega - \delta(x+\omega)/2 > x -\delta(x)/2 \implies \delta(x+\omega)-\delta(x) < 2\omega$$

Similarly, if $$x+\delta(x)/2 \geq x +\omega +\delta(x+\omega)/2$$ then $$I(x+\omega) \subset I(x)$$. Again, in this case, the interval $$I(x+\omega)$$ could be made larger, which contradicts the construction of $$I(x+\omega)$$ as the maximal interval , and it follows that

$$x+\delta(x)/2 < x +\omega +\delta(x+\omega)/2 \implies \delta(x+\omega)-\delta(x) > -2\omega.$$

Therefore $$|\delta(x+\omega)-\delta(x)| < 2\omega$$ for every $$\omega >0$$ and $$\delta(x)$$ is continuous.

Since $$[a,b]$$ is closed and bounded, a minimum value $$\delta(c)$$ is attained at some point $$c$$ in the interval.

Hence, for any $$y,z \in [a,b]$$, if $$|y-z|< \delta(c)$$ then $$y,z \in [\xi-\delta(\xi)/2,\xi+\delta(\xi)/2]$$ where $$\xi = (y+z)/2$$ and $$|f(x)-f(y)|< \epsilon$$. This proves that $$f$$ is uniformly continuous.

• Amazing! That's much like the picture I had in my mind put into words. Your insight of claiming $\delta(x)$ to be continuous to obtain it's minimum is wonderful. Thanks a lot. – Sarthak123 Oct 3 '18 at 7:33
• @Sarthak123: You're welcome. – RRL Oct 3 '18 at 22:52