Say we are in a (bounded) metric space $(X,d)$ and we consider a (bounded) Lipschitz function $f:X\to\mathbb{R}$ with optimal Lipschitz constant
$$L(f)=\sup\{\frac{f(x)-f(y)}{d(x,y)}\mid x,y\in X, x\neq y\}.$$
That means that for any $\varepsilon>0$, there are $x,y\in X$ such that $d(x,y)>0$ and
$$L(f)-\frac{f(x)-f(y)}{d(x,y)}<\varepsilon.$$
Would there be a way to explicitly determine a function $\delta(\varepsilon)$ such that for any $\varepsilon>0$, there exist $x,y\in X$ such that $d(x,y)\geq\delta(\varepsilon)$ and
$$L(f)-\frac{f(x)-f(y)}{d(x,y)}<\varepsilon$$
and could such a function $\delta$ be effective in terms of $\varepsilon$? Lastly, in what way would $\delta$ depend on $f$?
What conditions on $(X,d)$ could be assumed to make such a $\delta$ of a simple form?
Just to clarify, my motivation comes from the following: in the norm of the dual of a linear space $X$, the norm can be equivalently expressed by
$$\sup\{f(x)/\|x\|\mid x\in X,x\neq 0\}$$
as well as
$$\sup\{f(x)\mid x\in X, \|x\|\leq 1\}$$
where one uses the linearity of $f\in X^*$. I wonder if $L(f)$ can have a representation where division is avoided and so I am exploring similar questions.
Generally, it may not be possible to determine an explicit function $\delta(\varepsilon)$ that satisfies the condition you described for all Lipschitz functions on a given metric space. The reason for this is that Lipschitz functions may behave in a variety of different ways depending on the space and function involved.
There are cases where such a $\delta(\varepsilon)$ can be determined or bounded. One approach is to consider spaces with additional structural properties that allow for more precise analysis of Lipschitz functions.
For example, if $(X,d)$ is a compact metric space, then the Lipschitz constant $L(f)$ of any Lipschitz function $f:X\to\mathbb{R}$ is attained on $X$, i.e., there exist $x_0, y_0 \in X$ such that $L(f) = \frac{f(x_0) - f(y_0)}{d(x_0, y_0)}$. In this case, you can take $\delta(\varepsilon)$ to be any positive value smaller than the minimum non-zero distance between any two distinct points in $X$. Then, for any $\varepsilon > 0$, you can find $x, y \in X$ such that $d(x, y) \geq \delta(\varepsilon)$ and $L(f) - \frac{f(x) - f(y)}{d(x, y)} < \varepsilon$.
In terms of the dependence of $\delta(\varepsilon)$ on $f$, it will generally vary depending on the specific function $f$ and the properties of the metric space. It may be possible to establish relationships or bounds between $\delta(\varepsilon)$ and certain properties of $f$, such as its Lipschitz constant or its behavior on certain subsets of $X$. However, a general form of $\delta(\varepsilon)$ that applies to all Lipschitz functions on all metric spaces is unlikely.