Uniform convergence over a moving interval

Question

Let $f,f_n : [0,8] \to \mathbb{R}$ two continuisly differentiable functions defined over $[0,8]$ for all $n \geq 1$.

If $f_n$ uniformly converges to $f$ over $[4,5]$ when $n\to +\infty$.

and $f_n$ does not converge to $f$ over $[0,4) \cup (5,8]$.

Can we prove that $f_n$ uniformly converge to $f$ over 'the moving interval' : $[4-\frac{1}{n},5+\frac{1}{n}]$ ? when $n\to +\infty$.

Answer 1

The definition of uniform continuity is:

If $U$ is a set, $f : U \to \Bbb R$ is a function and for each $n \in \Bbb N, f_n : U \to \Bbb R$ is a function, and $S \subset U$, then we say the sequence $(f_n)_{n \in \Bbb N}$ converges uniformly to $f$ on $S$ if $$\forall \epsilon > 0, \exists N \in \Bbb N\text{ such that }\forall n \ge N,\forall x \in S, |f_n(x) - f(x)| < \epsilon$$

If we wanted to allow the set to vary with $n$, we need to replace $S$ with a sequence of sets $(S_n)_{n \in \Bbb N}$:

If $U$ is a set, $f : U \to \Bbb R$ is a function and for each $n \in \Bbb N, f_n : U \to \Bbb R$ is a function and $S_n \subset U$, then we say the sequence $(f_n)_{n \in \Bbb N}$ converges uniformly to $f$ on $(S_n)_{n\in\Bbb N}$ if $$\forall \epsilon > 0, \exists N \in \Bbb N\text{ such that }\forall n \ge N,\forall x \in S_n, |f_n(x) - f(x)| < \epsilon$$

So yes, the condition is definable (though I think I'd look for a better name). Is it useful for anything? Not that I can see, but maybe something escapes me.

Are the conditions in your problem enough to show that $f_n \to f$ uniformly on $\left(\left[4 -\frac 1n, 5 + \frac 1n\right]\right)_{n\in\Bbb N}$?

No. We know little about the behavior of each $f_n$ outside of $[4,5]$. It is entirely possible that the $f_n$ start growing fast outside that interval, with the rate of growth increasing with $n$, similar to how $y=x^n$ behaves near $1$. If it rises quicker with $n$ than the $\frac 1n$ extensions shink, it can blow past $\epsilon$, with larger $n$ being worse, not better.

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[Added]

A better example of rapid rising with $n$ than $y=x^n$ for this purpose would be $$f_n(x) = \begin{cases}0&x \le 0\\e^{-n/x}&x > 0\end{cases}$$ which (rather famously) is infinitely differentiable everywhere, including at $0$. It is also trivially uniformly convergent as $n\to\infty$ on $(-\infty, 0]$, but is not convergent at all for $x > 0$. And it should be clear that if $S_n = \left(-\infty, \frac 1n\right]$, then $f_n$ is not uniformly convergent on $(S_n)_{n \in \Bbb N}$.