Is it possible to find a universal $\delta_U$, such that $|x-y|<\delta_U\Longrightarrow |f_n(x)-f_n(y)|<\epsilon, \forall n=\color{red}0,1,2\dots$

Question

Suppose $f_n(x)$ ($n=1,2,\dots$) are continuous on $[a, b]$, $f_n(x)\rightarrow f(x)$ pointwisely, and $f(x)$ is continous on $[a, b]$. Define $f_0(x)=f(x)$. We know continuous on compact set implies uniformly continuous, hence $f_n(x)$ ($n=\color{red}0, 1,2,\dots$) are uniformly continuous on $[a,b]$.

Then for any fixed $\epsilon>0$, we can find $\delta_0>0, s. t,~|x-y|<\delta_0\Longrightarrow|f(x)-f(y)|<\epsilon$. Similarly, we can find $\delta_n>0, s. t,~|x-y|<\delta_n\Longrightarrow|f_n(x)-f_n(y)|<\epsilon$.

My question is,

If we define the set $S=\{\delta_k|k=0,1,2\dots\}$, how can we say about $\inf S$? For exmaple, we can always make $\inf S=0$,

we can find $\delta'>0, s. t,~|x-y|<\delta'\Longrightarrow|f_1(x)-f_1(y)|<\epsilon$. Set $\delta_1=\min(\delta', \frac12\delta_{0})$

we can find $\delta''>0, s. t,~|x-y|<\delta''\Longrightarrow|f_2(x)-f_2(y)|<\epsilon$. Set $\delta_2=\min(\delta'', \frac12\delta_{1})$

Keep going, we can guarantee $\delta_n\le\frac12\delta_{n-1}$, hence $\inf S=0$. But is it possible to construct $\delta_n$ such that $\inf S>0$? If this is possible (or with some additional conditions), then it means we can find a universal $\delta_U$, such that whenever $|x-y|<\delta_U\Longrightarrow |f_n(x)-f_n(y)|<\epsilon, \forall n=\color{red}0,1,2\dots$

Answer 1

The condition you are talking about is called equicontinuity.

A simple characterization of when such a $\delta_U$ can be found is given by the Arzela-Ascoli theorem, which states roughly that a sequence of continuous functions $f_n$ is equicontinuous (i.e. such a $\delta_U$ exists for any $\epsilon>0$), and $sup_{n \in \mathbb{N}}sup_{x\in [a,b]}f_n(x)<\infty$, if and only if for any subsequence of $n$, there exists a further subsequence $n_j$ and a continuous function $f$, so that $f_{n_j}\to f$ uniformly, i.e. $$ \sup_{x\in [a,b]}|f_n(x)-f(x)|\to 0 $$ as $n \to \infty$.

In your case the assumption that $sup_{n \in \mathbb{N}}sup_{x\in [a,b]}f_n(x)<\infty$ is automatically satisfied if we have equicontinuity. This can be seen by noting that in particular $f_n(a)$ converges, and is thus bounded by say $M$. Thus, pick $\delta>0$ so that for all $n$, $|x-y|<\delta\implies |f_n(x)-f_n(y)|<1$. Then, given $x \in [a,a+\delta)$, it holds $|f_n(x)|<M+1$. Proceeding inductively, we may cover the interval $[a,b]$ by finitely many intersecting intervals of radius $\delta$ and conclude there exists a universal constant $R$ so that $|f_n(x)|\leq R$, for all $x\in [a,b], n \in \mathbb{N}$.

Therefore, in your case, one can find such a $\delta_U$ for any $\epsilon>0$ $\iff$ every subsequence has a further subsequence along which $f_{n_j}$ converges uniformly to $f$. However, by a standard argument this just implies that the full sequence actually converges, so indeed a necessary and sufficient condition is that $f_n$ converges uniformly to $f$.

Answer 2

I think you will need further assumptions. Consider the sequence $f_n \colon [0,1] \to [0,1]$ defined by $$ f_n(x) = \begin{cases} nx \quad \text{if } &x \leq \frac{1}{n}, \\ 2-nx \quad \text{if } &x \in [\frac{1}{n}, \frac{2}{n}], \\ 0 \quad &\text{else}. \end{cases} $$ (The graph of the function is a straight line from $(0,0)$ to $(\frac{1}{n}, 1)$ and back down to $(0,\frac{2}{n})$.)

We have that the $f_n$ are continuous on $[0,1]$ and converge (pointwise) to $f_0 \equiv 0$. But

$$ |f_n(0) - f_n\left(\frac{1}{n}\right)| = 1 $$ for all $n \geq 1$, so a uniform $\delta$ cannot exist.

I think what you want is that your sequence along with the limit forms an equicontinuous family of functions. One condition that works is uniform convergence, see e.g., here: Uniformly continuous implies equicontinuous.

Another thing that works is if the functions are Hölder (or Lipschitz) continous with the same constant(s), see e.g., here: Hölder Condition and equicontinuity.

Is that what you are looking for?