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 $x\in X$?
Best Answer
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.