Prove that the modulus of continuity is continuous Let $f:[a,b]\to \Bbb R$ be a uniformly continuous function and define
$$ \omega(\delta) = \sup\{|f(x)-f(y)|:\ x,y\in [a,b], \ \text{ and } |x-y|<\delta\}$$
which we call the modulus of continuity.  Prove that $\omega$ is continuous on $(0,\infty)$.

My attempt: Let $\delta\in(0,\infty)$ and $\varepsilon>0$.  We want a neighborhood $N_r(\delta)$ such that $\eta\in N_r(\delta)$ implies $|\omega(\eta)-\omega(\delta)|<\varepsilon$.  I don't see any reason to pick any particular $r$ so maybe we try to play with $f$ and see what happens.
We know that if $|x-y|<\delta$ then $|f(x)-f(y)|\le \omega(\delta)$.
Perhaps it's worth looking at $\omega(\delta)\pm\varepsilon$.  Ignoring for now any concern about $\omega(\delta)-\varepsilon<0$, we could say that there correspond some two $\delta_1,\delta_2$ such that
$$ |x-y|<\delta_1 \Rightarrow |f(x)-f(y)|\le \omega(\delta_1) $$
$$ |x-y|<\delta_2 \Rightarrow |f(x)-f(y)|\le \omega(\delta_2) $$
I don't see any way to use this, though.
 A: We can at least show left continuity fairly easily. Let's let $A(\delta)$ be the set we are taking the supremum over, namely
$$A(\delta) = \{|f(x)-f(y)| : x,y\in[a,b], \text{and } |x-y|<\delta\}.$$
Given $\epsilon > 0$ the supremum properties says that there exists an element $z \in A(\delta)$ so that $\omega(\delta) - \epsilon < z$.
But $z$ is of the form $z = |f(x_0) - f(y_0)|$ for some $x_0,y_0\in[a,b]$ with $|x_0-y_0| < \delta.$
We may thus fix $r = \frac{1}{2}(\delta - |x_0-y_0|) > 0$. Let $\eta\in(0,\infty)$ be given so that $0 \leq \delta - \eta < r$.
It should be clear that $A(\eta) \subseteq A(\delta)$ (if $|x-y| < \eta$ then certainly $|x-y| < \delta$). But this means $\omega(\eta) \leq \omega(\delta)$. Notice
$$\eta > \delta - r = \frac{1}{2}(\delta + |x_0-y_0|)> \frac{1}{2}(|x_0-y_0| + |x_0-y_0|) = |x_0-y_0|.$$
So since we know $\omega(\delta) - \epsilon < |f(x_0)-f(y_0)|$, it follows that $\omega(\delta) - \epsilon$ cannot be an upper bound on $A(\eta)$. Therefore $\omega(\delta) - \epsilon < \omega(\eta)$. But now $\omega(\delta) - \epsilon < \omega(\eta) \leq \omega(\delta)$ and so of course $|\omega(\eta) - \omega(\delta)| < \epsilon$. This therefore demonstrates left continuity.
For right continuity, we would need a different stipulation on $r$ to ensure $\omega(\eta)$ stays less than $\omega(\delta) + \epsilon$ when $\eta \geq \delta$ (we'd get $\omega(\delta) \leq \omega(\eta)$ for free in this case).
A: I think right-continuity can be proved as follows. First observe than in the definition of $\omega(\delta)$, we may as well optimize over the closure
$\overline{A(\delta)}$= {$|()−()|:, \in [,]\text{ and }|−| \le  $}
of $A(\delta)$. The advantage is that this is a compact set, so the supremum is achieved.
Then let $\delta_n$ be a sequence tending to $\delta$ from above. For each $n$ there is $x_n$,$y_n$ such that $$|x_n-y_n| \le \delta_n$ and $|f(x_n)-f(y_n)| =\omega(\delta_n)$$ Up to taking a subsequence, we may assume that $x_n$ and $y_n$ converge to some $x_0$ and $y_0$
We have $|x_0-y_0| \le \delta$, so that
$$ \omega(\delta) \ge |f(x_0)-f(y_0)| = \lim_{n \to +\infty} \omega(\delta_n),$$
which is enough to prove right-continuity since $\omega$ is increasing.
