I spend a lot time with this problem, and I need your help,
Let $(X,d)$ be a metric space and $A$ be a dense subset of $X$, suppose that there is a bounded distance $d’$ on $A$ topologically equivalent to $d$ and $(X,d’)$ is complete metric space, define for all $z$ in $X$:
$$f(z)= \sup \{ \limsup_{n\to\infty} d’(x_n , x_{n+1}) : (x_n)_{n \in \mathbb{N}} \subset A \text{ converges to }z\},$$ the task is to show that $f$ is not continuous outside $A$.
For my approach, we can easily show that $f(x)=0$ for any $x \in A$, we take $z \in X$ and suppose that $f$ is continuous at $z$, since $A$ is dense then there is $(x_n)$ a sequence of $A$ that converges to $z$, by continuity of $f$ at $z$, we deduce that $f(x_n)$ converges to $f(z)$ and then $f(z) =0$, hence for any sequence of points $(x_n)$ of $A$ that converges to $z$, we have $\lim_{n\to\infty} d’(x_n,x_{n+1}) = 0$. I arrived here and I didn’t make an advance. I would like to see ideas how to continue or how to solve it.
Take a sequence $x_n$ such that $\lim_{n\to\infty} x_n = z$, where $z\in X$ is such that $f(z) = 0$. Suppose that $x_n$ is not a $d'$-Cauchy sequence, so that there is $r > 0$ such that for any $N$ there are $n, m\geq N$ with $d'(x_n, x_m) \geq r > 0$.
Thus we can recursively define a subsequence $n_k$ such that $d'(x_{n_{k+1}}, x_{n_k}) \geq r > 0$ by taking $N = n_k$ in the above. But then $\lim_{k\to\infty} x_{n_k} = z$ yet $\limsup_{n\to\infty} d'(x_{n_k}, x_{n_{k+1}}) \geq r > 0$.
The obtained contradiction shows that $x_n$ is a $d'$-Cauchy sequence, so it converges to some $y\in A$, of course we need to have $y = z$, so $z\in A$.