I remember (or misremember) the following theorem and I am looking for a reference and/or proof: (More or slightly different conditions may be needed than I have specified)
Let $X$ and $Y$ be metric spaces such that $X$ is a dense subset of $Y$.
Let $f_n : Y \rightarrow \mathbb{R}$, and $f: X \rightarrow \mathbb{R}$ be sequence of functions where $n \in \mathbb{N}$ and $ f_n \rightarrow f $ pointwise on $X$.
Let $g, g_n : Y \rightarrow \mathbb{R}$, be sequence of functions where $n \in \mathbb{N}$ and $ g_n \rightarrow g$ uniformly on $Y$.
Also $ f_n \rightarrow g$ pointwise on $X$. In other words, $g(x) = f(x)$ on $X$.
Then $f_n \rightarrow g$ on $Y$ pointwise (does it?)
Has anyone seen this or similar theorem? If yes, I am looking for the name of the theorem and/or book reference. If this is not provable and I am misremembering, let me know.
Thanks in advance.
I have a counter example. Let $X$ be ($0, 1$) and $Y$ be [$0, 1$] and the metric for both of them are just the usual one in $\mathbb{R}$. Let $f_n = x^n$ defined on [$0, 1$], f(x) = $0$ when $x \in [0, 1)$ but $1$ when $x = 1$. Let $g_n = \frac{1}{n(1 + x^2)}$ defined in $[0, 1], g = 0$ (the zero constant function) defined in [$0, 1$].
Now in $X$, $g = f, \sup|g_n - g| \to 0, f_n \to g$ pointwise and $f_n \to f$ pointwise. However, $f_n(1) = 1 \forall n \in \mathbb{N}$ but $g(1) = 0$.