I am using the following fact as part of a lemma (it is not a HW problem), and I am wondering if someone could check my proof.
Proposition: Let $I\subseteq\mathbb{R}$ be a nonempty interval and $f:I\to\mathbb{R}$ be bounded. Show that $\underset{s,t\in I}\sup |f(s)-f(t)|=\underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t)$.
Here is my proof:
First, fixing $s,t\in I$, we have \begin{multline*} |f(s)-f(t)|=\max\{f(s)-f(t),f(t)-f(s)\} \\ \leq \max\{\underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t), \underset{t\in I}\sup f(t) - \underset{s\in I}\inf f(s)\}\end{multline*} but then we are taking the max of the same two numbers, which yields $$|f(s)-f(t)|\leq \underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t)$$ and hence, $\underset{s,t\in I}\sup|f(s)-f(t)|\leq \underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t)$. On the other hand, given $\epsilon>0$, we may find $t_{\epsilon}$ and $s_{\epsilon}$ such that \begin{multline*} \underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t) < f(s_{\epsilon})+\epsilon - (f(t_{\epsilon})-\epsilon)=f(s_{\epsilon})-f(t_{\epsilon})+2\epsilon \\ \leq |f(s_{\epsilon})-f(t_{\epsilon})|+2\epsilon \leq \underset{s,t\in I}\sup|f(s)-f(t)|+2\epsilon\end{multline*} and taking $\epsilon\to 0^{+}$, we have that $\underset{s\in I}\sup f(s) - \underset{t\in I}\inf f(t) \leq \underset{s,t\in I}\sup|f(s)-f(t)|$, and we are done.
This certainly seems like one of those ``follow your nose" proofs, but I would very much appreciate someone taking the time to check over my work!
It's fine. The only thing that I would do in another way is this: I would have avoided limits. Since$$(\forall\varepsilon>0):\sup_{s\in I}f(s)-\inf_{t\in I}f(t)\leqslant\sup_{s,t\in I}\bigl|f(s)-f(t)\bigr|+2\varepsilon,\tag1$$then of course that$$\sup_{s\in I}f(s)-\inf_{t\in I}f(t)\leqslant\sup_{s,t\in I}\bigl|f(s)-f(t)\bigr|;$$otherwise, you could just take some $\varepsilon>0$ smaller that half of $$\sup_{s\in I}f(s)-\inf_{t\in I}f(t)-\sup_{s,t\in I}\bigl|f(s)-f(t)\bigr|$$ and get a contradiction with $(1)$.