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Let $\{f_n\}$ be a sequence of continuous functions $f_n\colon X \to R$ where $X \subset R$. Prove that if every point $x_0 \in X$ has an interval $ I_{x_0}=B\left({x_0 ,\varepsilon _{x_0}}\right)\cap X$ for some $\varepsilon _{x_0 } >0$, such that $\{f_n\}$ converges uniformly in $I_{x_0 } \cap X$, then $\{f_n\}$ it´s also equicontinuous.

Is this result true for a non-countable set of continuous functions? Is the reciprocal true? Is this result true for other kind of topological, or metric spaces? Please help me with this, to start, it's the only that i could not do.

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Definition of Equicontinuity: We say that the set of continuous functions $ E = \left\{ {f\,:X \to R\,\,\,\,\,f\,continuous} \right\} $ is equicontinuous, if for every $ x_0 \in X $ and $ \forall \varepsilon > 0\,\,\,\exists \,\,\delta > 0\,\,:\,\forall \,x \in \,B\left( {x_0 ,\delta } \right) $ such that $ \left| {f\left( x \right) - f\left( {x_0 } \right)} \right| < \varepsilon $ –  August Nov 12 '11 at 16:05
    
@DavideGiraudo: Why don't you write your comment as a solution? (Also, it seems to me that your argument could be generalized to more general spaces.) –  Jesse Madnick Nov 12 '11 at 17:46
    
@JesseMadnick Yes, it's seems it works in any metric space (at least, we didn't use neither separability nor local compactness of $\mathbb R$). –  Davide Giraudo Nov 12 '11 at 17:50
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1 Answer 1

We have the generalization:

Let $(Y,d)$ a metric space and $X$ a non-empsty subset of $Y$ and $\{f_n\}$ a sequence of continuous functions from $X$ to $\mathbb R$. If for all $x_0\in X$, we can find a ball $B:=B(x_0,\varepsilon_{x_0})$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $B\cap X$, then the family $\{f_n\}$ is equicontinuous.

We fix $x_0\in X$ and $\varepsilon>0$. Let $\delta_1$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $X\cap B(x_0,\delta_1)$. Let $N$ an integer such that $\sup_{x\in B(x_0,\delta_1)}|f(x)-f_n(x)|\leq\frac{\varepsilon}3$. For each $i\in\{1,\ldots,N\}$ we can find a $\eta_i$ such that if $d(x,x_0)\leq \eta_i$ then $|f_i(x)-f_i(x_0)|\leq \frac{\varepsilon}3$. Put $\delta_2:=\min\{\eta_i,1\leq i\leq N\}$. Finally, since $f$ is continuous at $x_0$ as a uniform limit of such functions, let $\delta_3$ such that if $d(x,x_0)\leq \delta_3$ then $|f(x)-f(x_0)|\leq\frac{\varepsilon}3$. Now put $\delta:=\min(\delta_1,\delta_2,\delta_3)$. Let $x\in B(x_0,\delta)$. Let $n\in\mathbb N$. If $n\leq N$ we are done since $\eta_n\leq\delta$, and if $n>N$ we have \begin{align}|f_n(x)-f_n(x_0)|&\leq |f_n(x)-f(x)|+|f(x)-f(x_0)|+|f(x_0)-f_n(x_0)|\\&\leq 2\sup_{y\in B(x_0,\delta)}|f_n(y)-f(y)|+|f(x)-f(x_0)|\\ &\leq 3\frac{\varepsilon}3, \end{align} hence the family $\{f\}\cup\{f_n,n\geq 1\}$ is equicontinuous at each point of $X$.

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