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It is known that a sequence of continuous functions on a metric space that converges pointwise on a dense subset need not converge pointwise on the full space. But what about if one assumes uniform continuity? Let me be more precise:

Let $X$ be a metric space and let $r_\alpha$ (for $\alpha=1,2,\ldots$) be a sequence of uniformly continuous functions $r_\alpha:X\to\mathbb{R}$. Furthermore, assume that $r:X\to\mathbb{R}$ is a uniformly continuous function such that $\lim_{\alpha\to\infty}r_\alpha(x)=r(x)$ for all $x$ in a dense subset $A\subseteq X$. Does this imply that $\lim_{\alpha\to\infty}r_\alpha(x)=r(x)$ for all $x\in X$?

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3 Answers 3

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If your sequence of functions $r_\alpha$ is uniformly equicontinuous, then this result should hold. That is, there should be one modulus of continuity for all functions in the sequence. Note that the sequence of @i707107 does not satisfy this stronger property. The proof goes along the same lines as the proof that C([0,1]) with supremum norm is a Banach (i.e. complete) space.

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That's a good point. If you have equicontinuity on compact metric space $X$, then the convergence should be for every point of $X$, and the sequence is uniformly convergent. If you don't have equicontinuity, then my example shows that the pointwise convergence does not necessarily hold for every point. There is even more extreme example. Such as $f_n(x)=\sin n!x$ on $[-\pi,\pi]$. W.Rudin's "Principles of Analysis", page 334, exercise 16 shows that the set of all points $x\in [-\pi,\pi]$ of pointwise convergence has measure zero, and it contains all rational multiples of $\pi$. –  i707107 Mar 7 '12 at 21:48
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I don't know what example that you have in mind in the first statement, but you can find such example in your question on $X=[0,1]$ as pointed out by Peter, any continuous function on $X$ is uniformly continuous. Consider $f_n(x)=0$ on $x\in [\frac{1}{n},1]$, and $(-1)^nn(x-\frac{1}{n})$ on $x\in [0,\frac{1}{n}]$. This sequence is pointwise convergent to $0$ on $(0,1]$ which is a dense set in $[0,1]$.

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Since all continuous functions on a compact metric space are uniformly continuous one can construct an easy conterexample on $X=[0,1]$.

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