For each $n \geq 2$, define $f_n \in C([0,1])$ as such, $$ f_n(x):= \begin{cases} 0, &\text{if }x =0\\ 1/n, &\text{if } x \in [1/n^2,1]\\ \text{linear}, &\text{otherwise}. \end{cases} $$
I was asked to (and was successfull) show that
- $\{f_n\} \to f,$ point-wise.
- $\{f_n\} \to f$, uniformly.
- Each $f_n$ is uniformly continuous.
For 1. I nominate the point-wise limit $f(x) = 0$, the proof was not difficult, then 2. came for free since my proof of 1. did not depend on $x$, and 3. also came for free since each $f_n$ is continuous on compact domain.
The last question was
- Is $\{f_n\}$ equicontinuous?
I tried to tackle it intuitively at first, but intuition left me on the fence. First I thought because the "linear sections" are getting arbitrarily steep I should be able to make it fail at $x=0$, but the fact that my linear sections are shrinking in length rapidly and bounded above by $1/n$ ruined that plan.
I ended up being successful in using properties 2. and 3. above to prove that equicontinuity was satisfied, using an $\epsilon / 3$ style argument. But this holds for any family of functions with properties 2. and 3. and although I was convinced, I still felt in the dark as to why this specific family was equicontinuous. Could anyone meander a proof directly from the definition of equicontinuous to shed some constructive light on me?
I suspect what's bothering you is that all the functions are Lipschitz (in particular, $f_n$ has a Lipschitz constant of $n$), there is no universal Lipschitz constant, but the functions are equicontinuous anyway. But you are right, and we can see it this way.
Given $\varepsilon>0$, for $n>1/\varepsilon$ all the values of $f_n$ are within $\varepsilon$ of each other. Therefore the "equi-modulus of continuity" $\delta(\varepsilon)$ can be taken to be $\min \{ \delta_n(\varepsilon) : n \leq 1/\varepsilon \}$ which is guaranteed to be positive because you're taking the minimum of a finite set of positive numbers. (Also note that these don't depend on $x$ because the domain is compact.) In this particular case $\delta_n(\varepsilon)$ can be taken to be $\varepsilon/n$ by the Lipschitz property, so $\delta(\varepsilon)$ can be taken to be $\varepsilon/(1/\varepsilon)=\varepsilon^2$. Try to prove this directly: if $|x-y|<\varepsilon^2$ then $|f_n(x)-f_n(y)|<\varepsilon$ for all $n,x,y$.