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Let $\mathrm{Min}(f) = \min_{x \in [a, b]}f(x), f \in \mathcal{C}([a, b])$. I already know a proof for showing that a similar function, $\mathrm{Max}(f) = \max_{x \in [a, b]}f(x)$, is continuous. For reference: Why is absolute difference of max of functions less than max of absolute difference of functions. (However I do not know how the tricks used in that proof apply here, as I am not convinced that they were justified. For example, $\max(\left|f(x) - g(x)\right| + g(x)) \leq \max\left|f(x) - g(x)\right| + \max g(x)$ seems to come out of nowhere.)

I was wondering that the $\epsilon - \delta$ could be used in proving that $\mathrm{Min}$ is continuous. Namely, if we could make an argument why $||f - g||_\infty = \sup_{x \in [a, b]}\left|f(x) - g(x)\right| \geq \left|\mathrm{Min}(f) - \mathrm{Min}(g)\right|$, then choosing $\delta = \epsilon$ would finish the proof. But I am not entirely convinced of the inequality, as I don't know how to justify it rigorously.

Epsilon Away
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    $\min_{x\in[a,b]}f(x)=\max_{x\in[a,b]}(-f(x))$. Can you finish it from here? – Alex R. Feb 23 '21 at 18:31
  • @AlexR. I think I do, thanks! If you don't mind, could you comment on the $\max(\left|f(x) - g(x)\right| + g(x)) \leq \max\left|f(x) - g(x)\right| + \max g(x)$? To me that is the only "unsettling" part of the proof in the reference. Specifically, I don't see how the maximum of the sum could be at most the sum of the maximums. – Epsilon Away Feb 23 '21 at 18:54

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