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Let $X$ and $Y$ be two metric spaces, and $F$ a family of functions from $X$ to $Y$. The family $F$ is equicontinuous at a point $x_0 ∈ X$ if for every $ε > 0$, there exists a $δ > 0$ such that $d(f(x_0), f(x)) < ε$ for all $f ∈ F$ and all $x$ such that $d(x_0, x) < δ$.

Is it equivalent to $$\lim_{\delta \to 0} \sup_{f \in F, x \in \{x \in X: d(x_0, x) < δ\}} d(f(x_0), f(x)) = 0?$$ Or some other correct form?

Thanks and regards!

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up vote 2 down vote accepted

Yes, it's equivalent. If we have the equicontinuity at $x_0$ of $F$ then for a fixed $\varepsilon>0$ we can find $\delta_0$ such that if $d_X(x,x_0)<\delta_0$ and $f\in F$ then $d_Y(f(x_0),f(x))<\frac{\varepsilon}2$, so we have $\sup_{f\in F,x\in B(x_0,\delta_0)}d_Y(f(x_0),f(x))\leq \frac{\varepsilon}2<\varepsilon$. Since the map $\delta\mapsto \sup_{f\in F,x\in B(x_0,\delta)}d_Y(f(x_0),f(x))$ is increasing we have the wanted result. Conversely, if $\lim_{\delta\to 0}\sup_{f\in F,x\in B(x_0,\delta)}d_Y(f(x_0),f(x))=0$ then for a fixed $\varepsilon>0$ we can find $\delta$ such that $\sup_{f\in F,x\in B(x_0,\delta)}d_Y(f(x_0),f(x))<\varepsilon$ which implies that if $f\in F$ and $d_X(x,x_0)<\delta$ then $d_Y(f(x_0),f(x))<\varepsilon$.

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Thanks! I wonder if the proof can be without "the map $\delta\mapsto \sup_{f\in F,x\in B(x_0,\delta)}d_Y(f(x_0),f(x))$ is increasing"? – Tim Feb 23 '12 at 17:28

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