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.


  • 1
    $\begingroup$ 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. $\endgroup$ – Joppy Oct 1 '18 at 14:24
  • $\begingroup$ 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 $\endgroup$ – user359302 Oct 1 '18 at 15:08
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Sarthak123 Oct 3 '18 at 7:33
  • $\begingroup$ @Sarthak123: You're welcome. $\endgroup$ – RRL Oct 3 '18 at 22:52

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