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Let $C$ be the space of all continuous functions $f: \Omega \to \mathbb{R}$ from $\Omega \subseteq \mathbb{R}^n$ to $\mathbb{R}$ and $(C,d)$ be a metric space on $C$ with the metric $d:C \times C \to \mathbb{R}$.

Suppose that $\{f_j \in C\}_{j=1}^\infty$ be a sequence that converges to some function $f_*$ pointwisely (which is not necessarily in $C$) and at the same time converges to $g_* \in C$ in metric $d$, i.e., \begin{equation} \lim_{j \to \infty} d(f_j,g_*) = 0. \end{equation} Then, can we conclude that both types of convergence have the same limit point, i.e., $f_* = g_*$?

Here, the equivalence $f = g$ always means that they are equal pointwisely so that $f_1(x) = f_2(x)$ for all $x \in \Omega$ (I assume the metric property "$d(f,g) = 0$ $\Longleftrightarrow$ $f = g$" is also defined up to this pointwise equivalence).

I know that if $\Omega$ is compact and $d$ is the uniform metric: \begin{equation} d(f,g) := \sup_{x \in \Omega} |f(x) - g(x)|, \end{equation} then the convergence $f_j \to g_*$ in metric $d$ actually means the uniform convergence, which implies the pointwise convergence $f_j \to g_*$, in which case $f_* = g_*$. I want to know whether this is true for a general metric $d$ on $C$ (and a possibly non-compact set $\Omega \subseteq \mathbb{R}^n$).

If not, can you given a counter-example with a specific metric? Or, Can we say that $f_* = g_*$ at least in a.e. sense?

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  • $\begingroup$ I am not sure if I understand you correctly, but in general it is not true that $f_*=g_*$. That's the reason why we have many types of convergence for functions in the first place. $\endgroup$
    – BigbearZzz
    Oct 3, 2016 at 21:00
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    $\begingroup$ You definitely cannot just have an arbitrary metric $d$. Unless the $d$ is somehow recording the fact that the elements of $C$ are functions, there is no reason to think that there is a relationship between $d$ and pointwise convergence. $\endgroup$
    – Hayden
    Oct 3, 2016 at 21:00
  • $\begingroup$ It doesn't hold for completely arbitrary metrics on $C$. The metric must be such that convergence in the metric implies pointwise convergence. But all metrics on $C$ one is actually interested in do have that property. $\endgroup$ Oct 3, 2016 at 21:01
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    $\begingroup$ @DanielFischer Some don't. With $\Omega=[0,1]$, $f_n(x)=x^n$, $f_n\to 0$ in $\|\cdot\|_1$, but not pointwise. $\endgroup$ Oct 3, 2016 at 21:05
  • $\begingroup$ What did you mean by this "I assume the metric property $d(f,g)=0$ iff ⟺ $f=g$ is also defined up to this pointwise equivalence"? $\endgroup$
    – BigbearZzz
    Oct 3, 2016 at 21:09

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No.

Let $\Omega=[0,1]$ and $d(f,g)=\int_0^1|f(x)-g(x)|\,\mathrm dx$. Let $f_n(x)=x^n$. Then pointwise $f_n\to f_*$ with $$f_*(x)=\begin{cases}1&x=1\\0&x\ne 1\end{cases} $$ (which is $\notin C$), whereas $f_n\to 0$ with respect to $d$.

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  • $\begingroup$ This is the exact counter example of $f_* = g_*$ ($g_* = 0$), but $f_* = g_*$ almost everywhere. Then, can we say that $f_∗=g_*$ almost everywhere for the general case with any metric $d$? $\endgroup$ Oct 3, 2016 at 22:12

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