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Related to this question I asked (I didn't want to keep changing it out of respect for the great answers). Below is a shored up attempt taking into account the comments.

Theorem: If $f(x)$ is a real-valued function that is continuous on $[a,b]$ then it is uniformly continuous on $[a,b]$

Direct proof of Theorem

Since $f(x)$ is continuous on $[a,b]$ then for any $\epsilon > 0$ and $y,x \in [a,b],\; \exists \delta_x(\epsilon)>0 $ such that $|y-x|<\delta_x(\epsilon) \implies |f(y)-f(x)|< \epsilon$.

Define the following function of $x,\epsilon$: $$\delta^*_x(\epsilon) := \sup\{d:y \in [a,b] \text{ and } |x-y|<d \implies |f(x)-f(y)|<\epsilon\}$$

Let $\delta(\epsilon) := \inf_{x \in [a,b]} \delta^*_x(\epsilon)$. Then for each $\epsilon > 0$ and $x,y \in [a,b]$, we have $|x-y|<\delta(\epsilon) \leq \delta_x^*(\epsilon) \implies |f(x)-f(y)|< \epsilon\;\square$

This verion attempts to address the issues with my earlier attempt, where I relied (incorrectly) on continuity of $\delta_x(\epsilon)$ and ignored the fact that without taking the largest $\delta_x(\epsilon)$ for each $x$ then I'd get $\delta(\epsilon)=0$. This version (I think) guarantees that $\delta_x^*(\epsilon)>0$ for all $x\in [a,b],\epsilon>0$

EDIT: The above doesn't work because we have now shown why $\inf \sup \delta_x(\epsilon)>0$. I think I'd need to show that if there is a Cauchy sequence of suprema of $\delta_x(\epsilon)$ in $[a,b]$ that converges to zero then either $f(x)$ is not continuous or $[a,b]$ is not compact. Both are false -- but then this is just a proof by contradiction again.

Thanks everyone for the great comments.

Annika
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    again, why is $\delta(\epsilon)$ positive? This proof is completely flawed because nowhere did you explicitly invoke the fact that you’re working on $[a,b]$ (a compact set). – peek-a-boo Oct 11 '22 at 22:14
  • @peek-a-boo my thinking is that by continuity each $x \in [a,b]$ and $\epsilon>0$ has a valid $\delta >0$. The inf of these suprema will also be $>0$. – Annika Oct 11 '22 at 22:16
  • wrong. inf of sup need not be positive. Again, see my comment about not using compactness. If your argument were correct, then you’re not using any properties specific to $[a,b]$, so it would follow that every continuous function on any metric space is uniformly continuous, but this is absurd (e.g $f(x)=x^2$ is not uniformly continuous on $\Bbb{R}$). – peek-a-boo Oct 11 '22 at 22:17
  • @peek-a-boo -- I see so I need to invoke compactness? – Annika Oct 11 '22 at 22:18
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    maybe this will help https://math.stackexchange.com/questions/3540234/continuity-over-a-compact-implies-uniform-continuity – jcneek Oct 11 '22 at 22:19
  • @peek-a-boo ah, I get it now... we could easily have a decreasing sequence of suprema, so I'd need to use compactness to show that we cannot have the infimum of these suprema be 0 over the compact set (othwerwise it would be open). – Annika Oct 11 '22 at 22:20
  • @User5678 please refer to this post. https://math.stackexchange.com/questions/110573/continuous-mapping-on-a-compact-metric-space-is-uniformly-continuous#:~:text=The%20answer%20is%20yes%2C%20if,%2Cf(x)) . – nicoyanovsky Oct 11 '22 at 22:21
  • @nicoyanovsky I appreciate the links to other proofs -- what I was trying to see is why my attempt is wrong. – Annika Oct 11 '22 at 22:22
  • The key thing in this proof is that, thanks to compactness of $[a, b]$ you can take some $\delta$ such that whenever $|x-y| < \delta$ then both $x$ and $y$ lie in some $(x_i-\delta_{xi}, x_i+\delta_{xi})$, for some $x_i$ so that it will imply uniform continuity. – nicoyanovsky Oct 11 '22 at 22:30

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