Alternative proof of continuous on a closed bounded interval implies uniformly continuous.

Question

Consider a continuous function $f\colon [a,b] \to \mathbb{R}$, then for any given $\varepsilon>0$ define a function $\delta_{\varepsilon}\colon[a,b] \to \mathbb{R}$, where $\delta_{\varepsilon}(x)=\min(b-a,\sup\{\delta>0 :y \in [a,b] \text{ and }|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon\})$ essentially taking the largest value of $\delta$ in the $\varepsilon-\delta$ definition that works for a given $x$ and fixed $\varepsilon$. Is it true that this function is continuous?
If so then $\delta_{\varepsilon}(x)$ attains its infimum on $[a,b]$ and by continuity of $f$ we must have that $\delta_{\varepsilon}(x)>0$ and thus we can take $\delta=\inf_{x \in [a,b]}(\delta_{\varepsilon}(x))>0$ in the definition of uniform continuity. I tried to prove that $\delta_{\varepsilon}$ is continuous, but I am not getting anywhere.

Answer 1

$\delta_\varepsilon(x)$ isn't even defined for all $x \in [a,b]$ for all continuous functions.

Let $\varepsilon > 0$, $[a,b] \subseteq \Bbb{R}$, and $f(x)$ be constant. Then $$ \{\delta > 0 :y \in [a,b], |x-y| < \delta \implies |f(x) - f(y)| < \varepsilon\} = (0,\infty) \text{.} $$ The consequent of the implication is always true, so there is no constraint on $\delta$. This set's supremum does not exist, so the maximum in the definition of $\delta_\varepsilon(x)$ does not exist and $\delta_\varepsilon(x)$ is undefined for all $x \in [a,b]$.

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Update: The item below may have been addressed in edits to the Question.

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When $\delta_\varepsilon(x)$ is defined for all $x \in [a,b]$, $\delta_\varepsilon(x)$ is trivially continuous because it is constant.

The maximum of $|x-y|$ is $b-a$, so $\sup \{\delta > 0 \dots\}$ is less than or equal to $b-a$. Therefore, $$ \delta_\varepsilon(x)= \max\{b-a,\sup \{\delta > 0\dots\}\} = b-a \text{.} $$

Answer 2

I think (a slightly modified version of) your delta is lower-semicontinuous. Using that you can show that it has a positive minimum on the compact interval. Then you are done!