real analysis - Alternative Uniform-Continuity theorem proof by Luroth

Can please someone elaborately give the proof of Uniform-Continuity theorem ( every continuous function on a closed bounded real interval is uniformly continuous) by Luroth ? thanks in advance

Answer

This is the proof that does not use the Heine-Borel Theorem directly.

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 $y,z \in I(x)$.

First, 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.

To prove continuity, 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. Hence,

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

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. Hence,

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

Therefore $|\delta(x+\omega)-\delta(x)| < 2\omega$ for every $\omega >0$ and $\delta(x)$ is continuous. Since $[a,b]$ is compact 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 $|f(x)-f(y)|< \epsilon$.

QED