$f$ is continuous on $[a,b]$ and $|f(x)-f(y)|<\varepsilon$. Show that $\sup f – \inf f \leq \varepsilon$
Suppose that $\sup f – \inf f > \varepsilon$.
$|f(x)|-|f(y)| \leq |f(x)-f(y)| < \varepsilon < \sup f – \inf f \Rightarrow |f(x)|-|f(y)| < \sup f – \inf f$
It follows that $\begin{cases}|f(x)|<\sup f\\-|f(y)|<-\inf f \end{cases} \Rightarrow \begin{cases}|f(x)|<\sup f\\|f(y)|>\inf f \end{cases}$
In particular $f(x)<\sup f$ for all $x \in [a,b]$. But $f(x)$ has a maximum on $[a,b]$ since it's continuous and $\max(f)<\sup f$, hence $\sup f$ is not a least upper bound. I've derived a contradiction, so $\sup f – \inf f \leq \varepsilon$.
Not sure if this proof is correct. Is there a better one?
Best Answer
A bettter proof: $f(x)-\epsilon <f(y)$ . Take sup over $x$ to get $\sup f -\epsilon \leq f(y)$. This is true for all $y$. Take inf over $y$ to get $\sup f \leq \inf f+\epsilon$.