Suppose that $f$ is a Lipschitz continuous function on a compact set, $K\in\mathbb{R}^d$. Suppose that it attains its minimum uniquely at some point $x_0$. Given $\delta>0$, can we show that for $|x-x_0|>\delta$, $f(x)-f(x_0)>t(\delta)$, where $t(\delta)$ is a positive number depending on $\delta$?
One way I can do this is to define the set $$\{f(x):|x-x_0|\geq\delta/2\}.$$ This is a compact set since $f$ is defined on a compact set. Also, the singleton $f(x_0)$ is a compact set. Therefore, there is a positive distance between the two sets.
However, I want to estimate this distance. Is this possible? Can we use the Lipschitz property?
Given the information you have provided, we may not derive such a bound because $f$ might be arbitrarily flat in an arbitrarily large neighbourhood of its unique minimiser. It would be useful if, for example, we knew that $f$ is strongly convex with a known strong convexity modulus so that we know that $f$ grows faster than some quadratic function close to its minimiser.
Additionally, even if we know that $L$ is the best Lipschitz constant of the function, it may be arbitrarily flat close to the minimizer and steeper aways from it.