There already exist several posts on similar questions, but I could not find any, which really gives a precise answer to my problem.
Let $A \subseteq C([0,T],E)$, where $E$ is a topological space and C$([0,T],E)$ denotes the set of all continuous functions from $[0,T]$ to $E$. Assume I am given a metric on $E$, let me call it $d$, which induces the prescribed topology on $E$.
Then, equicontinuity of $A$ means $sup_{f \in A}d(f_{t_n},f_t) \to 0$ for each $t \in [0,T], (t_n)_{n \in \mathbb{N}}$ converging to $t$. One usually does not find this as the definition of equicontinuity, but I think this is an equivalent way to express it, right?
Now, suppose someone considers a topologically (i.e. "weakly") equivalent metric, $\rho$, on $E$. Is equicontinuity of $A$ invariant under switching between such equivalent metrics? Intuitively, this should be the case, but I would very much like to have clarification for it.
Thanks in advance!
The answer is negative. Here is a very simple example: For $a\in [0,\infty)$ consider $f_a:[0,1]\to \mathbb [0,\infty)$, $x\mapsto a+x$. Of course, the family $A=\{f_a: a\ge 0\}$ is equicontinuous with respect to the usual (translation-invariant) metric $d(x,y)=|x-y|$ on $[0,\infty)$. Now, consider the metric $\delta(x,y)=|x^2-y^2|$. It is easily seen that this is a metric which gives the same topology as $d$ (continuity of the square function and the square root). But $A$ is not equicontinuous with respect to $\delta$: $$\sup\{\delta(f_a(x),f_a(0)):a\ge 0\} = \sup\{|2ax+x^2|:a\ge 0\}=\infty$$ for each $x\neq 0$.