I hope it is going well for you. I am experiencing some serious trouble with the following problem and am unsure of where to start. I have included some ideas that I have had down below.
Problem
Let the function $f: [a,b] \to \mathbb{R}$ be a continuous. Let $\epsilon > 0$. We want to prove that for some $n \in \mathbb{N}$ there exists points $a = x_0 < x_1 < \dotsm < x_n = b$ such that $\omega_f ([x_i, x_{i+1}]) < \epsilon$ for all $0 \leq i \leq n-1$.
(Note that $\omega_f ([x_i, x_{i+1}]) = \underset{x\in [x_i, x_{i+1}]}{\text{sup }}f(x) – \underset{x\in [x_i, x_{i+1}]}{\text{inf }}f(x) = \underset{x,y\in [x_i, x_{i+1}]}{\text{sup}}|f(x) – f(y)|$)
Thoughts
I was thinking about using the $\epsilon-\delta$ definition of continuity, i.e. given $f: A \subseteq \mathbb{R} \to \mathbb{R}$ and $x_0 \in A$, $f$ is continuous at $x_0$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that whenever $|x-x_0| < \delta$ for some $x\in A$ it follows that $|f(x) – f(x_0)| < \epsilon$, to show that each $|f(x) – f(y)| < \epsilon$ for all $x,y\in [x_i, x_{i+1}]$ for all $0 \leq i \leq n-1$ but am stuck, generally speaking. Perhaps unfirom continuity could come into play. I also read somewhere that $f$ is only continuous at $c$ (in my case this would be $x_0$) if $\omega_f(c)=0$.
Best Answer
Your thought to use uniform continuity is correct. In particular, since $[a, b]$ is compact, $f$ is uniformly continuous, so we can find a $\delta$ so that if $|x - y| \leq \delta$, then $|f(x) - f(y)| < \varepsilon$. Then, if we let $x_0 = a, x_1 = a + \delta, x_2 = a + 2\delta, ..., x_{n - 1} = a + (n - 1)\delta, x_n = b$ with $a + n\delta \geq b$, then we have that $|x_n - x_{n + 1}| = \delta$ . So, if $x, y \in [x_{i}, x_{i + 1}]$ then $|x - y| < \delta$, so by our uniform continuity, $|f(x) - f(y)| < \varepsilon$. See if you can show that this implies that $$\left|\sup_{x \in [x_i, x_{i + 1}]}f(x) - \inf_{x \in[x_{i}, x_{i + 1}]}f(x)\right| < \varepsilon$$
Hint: Extreme value theorem makes this very easy. If your course has not proven the EVT, then directly think about the definition of sup and inf to realize that you can find a sequence $x_n : f(x_n) \to \sup_{x \in [x_i, x_{i + 1}]}f(x)$, and a sequence $y_n : f(y_n) \to \inf_{x \in[x_{i}, x_{i + 1}]}f(x)$ (why?). Then, notice that $|f(x_n) - f(y_n)| < \varepsilon$ for any $n$. What happens when you take the limit?