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I've proven the following "theorem":

Let $I \subset \mathbb{R}$ be an interval, $(f_n: I \rightarrow \mathbb{R})_{n \in \mathbb{N}}$ be a family of continuous functions converging pointwise to a continuous function $f: I \rightarrow \mathbb{R}$ on $I$. Then: $(f_n)_{n \in \mathbb{N}}$ is equicontinuous on I.

Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $f_n$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:

Proof: Let $\epsilon > 0$. Observe first: \begin{equation} | f_n(x) - f_n(y) | \leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)| \end{equation} Now there is by pointwise convergence of $(f_n)_{n \in \mathbb{N}}$ a $N \in \mathbb{N}$ such that for all $n \geq N$ we have $|f_n(x) - f(x)|<\frac{\epsilon}{3}$ and $|f_n(y) - f(y)| < \frac{\epsilon}{3}$. Further there is a $\delta > 0$ such that $|f(x) - f(y)| < \frac{\epsilon}{3}$ for $|x-y| < \delta$ by continuity of $f$. Hence we have shown, that there is a $N \in \mathbb{N}$ and a $\delta > 0$ such that for all $n \geq N$ \begin{equation} |f_n(x) - f_n(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{equation} holds. Now let $n < N$. Then, by continuity of $f_n$ there is a $\delta_n$ such that $(x-y) < \delta_n$ implies $|f_n(x) - f_n(y)| < \epsilon$. Setting \begin{equation} \tilde{\delta} = \min_{n < N} \delta_n \end{equation} (which exists and is greater than $0$) we obtain, that for all $n < N$ the following holds: \begin{equation} |x - y| < \tilde{\delta} \Rightarrow |f_n(x) - f_n(y) | < \epsilon \end{equation} Setting now $\hat{\delta} = \min \{\delta, \tilde{\delta} \}$ we have, that for all $n \in \mathbb{N}$ the following holds: \begin{equation} |x-y| < \hat{\delta} \Rightarrow |f_n(x)- f_n(y) | < \epsilon \end{equation} Hence we have shown, that for all $\epsilon > 0$ there is a $\hat{\delta} > 0$ such that forall $n \in \mathbb{N}$ we have, that $|x-y| < \delta$ implies $|f_n(x) - f_n(y)| < \epsilon$.

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  • $\begingroup$ Can your interval be open? $\endgroup$ – kimchi lover Dec 15 '18 at 16:24
  • $\begingroup$ Yes, i stated no assumptions regarding the interval. $\endgroup$ – warpfel Dec 15 '18 at 16:27
  • $\begingroup$ The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly. $\endgroup$ – kimchi lover Dec 15 '18 at 16:28
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Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!

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