By the set of linear functions I mean the functions of the form $$f(x)=\alpha x,$$ for $\alpha\in\mathbb{R}$. We clearly have for any $x,y\in[0,1]$ $$|f(x)-f(y)|=|\alpha||x-y|.$$ So for any $\epsilon>0$, $\delta=\frac{\epsilon}{|\alpha|}$ gives uniform continuity. However this $\delta$ is different depending on $f$, so it cannot give equicontinuity. I am having trouble figuring out how to show that there would be no $\delta$ that would work.
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the very concept of equicontinuity is that the functions have the "same" growth rate in a small neighborhood, where it is obvious that these don't :) – Alan Apr 28 '22 at 14:54
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Here is an argument.
Suppose that your set of functions is equicontinuous on $[0,1]$. Then, given $\epsilon >0,$ there exists a $\delta >0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$. But if we pick $x,y$ such that $|x-y| = \delta/2$, then
$$ |f(x)-f(y)| = |a||x-y| = |a|\delta/2. $$
If choose $a$ large enough, say $a=4\epsilon/ \delta$, then $|f(x)-f(y)| = 2 \epsilon > \epsilon$, a contradiction.
(Note that our assumption that there exists $x,y$ such that $|x-y| = \delta/2$ is harmless.)
Steven
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