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Let X be a metric space and {$f_n$} be a sequence in $C(X)$ that converges uniformly on $X$ to $f \in C(X)$. Show that {$f_n$} is equicontinuous.

So I know uniform convergence means given $\epsilon > 0 \exists N \in \mathbb{N}$ such that $\rho (f_n(x),f(x))< \epsilon$ $\forall x \in X$ whenever $n > N$ but I don't know where to go from here

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    $\begingroup$ $X$ is compact? $\endgroup$ – user284331 Jan 25 '18 at 22:53
  • $\begingroup$ @user284331 thats not given. are you saying that follows from the info? $\endgroup$ – Vinny Chase Jan 25 '18 at 22:55
  • $\begingroup$ If $X$ is compact, then the assertion follows. For non-compact $X$, I don't know how to do. $\endgroup$ – user284331 Jan 25 '18 at 22:55
  • $\begingroup$ It doesn't. The assertion holds when $X$ is any topological space, but having a metric structure on $X$ makes people think of uniform equicontinuity, which would generally not hold without compactness. $\endgroup$ – Daniel Fischer Jan 25 '18 at 22:57
  • $\begingroup$ Ah... I take the definition of equi here to be uniformly equi, thanks for reminding. $\endgroup$ – user284331 Jan 25 '18 at 22:59
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Take $x\in X$ and $\varepsilon>0$; you want to prove that, for some $\delta>0$, you have$$(\forall n\in\mathbb{N})(\forall y\in X):d(x,y)<\delta\implies\bigl|f_n(y)-f(_nx)\bigr|<\varepsilon.\tag1$$Take $N\in\mathbb N$ such that$$(\forall n\in\mathbb{N})(\forall y\in X):n\geqslant N\implies\bigl|f(y)-f_n(y)\bigr|<\frac\varepsilon3.$$Take $\delta'>0$ such that$$(\forall y\in X):d(x,y)<\delta'\implies\bigl|f(y)-f(x)\bigr|<\frac\varepsilon3.$$Then, if $y\in X$ and $n\geqslant N$, you'll have$$d(x,y)<\delta'\implies\bigl|f_n(y)-f_n(x)\bigr|\leqslant\bigl|f_n(y)-f(y)\bigr|+\bigl|f(y)-f(x)\bigr|+\bigl|f(x)-f_n(x)\bigr|<\varepsilon.$$So, $(1)$ holds, except perhaps when $n\in\{1,2,\ldots,N-1\}$. Take, take $\delta_1$ that woarks for $f_1$, $\delta_2$ that woarks for $f_2$, … and $\delta_{N-1}$ that works for $f_{N-1}$. Now, define$$\delta=\min\{\delta_1,\delta_2,\ldots,\delta_{N-1},\delta'\}$$and you're done.

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