Supposing you have uniformly continuous function between metric space, is the epsilon delta relationship continuous. More precisely, for $X,Y$ metric, given $f :X \rightarrow Y$, uniformly continuous, if for a given $\varepsilon>0$ we define the $\delta(\varepsilon)$ to be the sup of all $\delta>0$ s.t $ d_X(x,y)< \delta \Rightarrow d_Y(f(x),f(y))< \varepsilon $ , is  $g:\mathbb{R}^+ \rightarrow\mathbb{R}^+$  given by $g(\varepsilon)= \sup${ $\delta$ >0 that work }, a continuous a function?
Edit: It occurs to me that the sup defined here way me infinity, as for constant functions. If we exclude these cases, is the above true?

Consider a sawtooth wave from the reals to the reals...but let's make the rising slope be 1, and the falling slope be 2. That's uniformly continuous, with $\delta = \epsilon/2$ working everywhere. But on the rising segments, your $\delta_{max}$ will be $1$, and on the falling segments, it'll be $1/2$, and hence it must be discontinuous at each "breakpoint". Actully, the same argument works for
$$
f(x) = \begin{cases}
x & -1 \le x \le 0 \\
-2x & - \le x \le \frac12
\end{cases}
$$
which is a uniformly continuous function from $[-1, 0.5]$ to $[-1, 0]$.
