Could lower semicontinuous functions have Lipschitz constant?

Question

For example, $f(x)=\left\{\begin{matrix} 1, & x\ne 0\\ 0, & 0 \end{matrix}\right.$ is lower semicontinuous. Does it have Lipschitz constant? Beased on defniation $|f(x_1)-f(x_2)|\leq K|x_1-x_2|$, it seems that there exist $K$ that makes $f(x)$ Lipschitz continuous.

Answer 1

I don't think so. Consider $|f(0)-f(x_0)| = 1$ for some $x_0\neq 0$. Then we have on the right side of the inequality $K|x_0|$. Choose your $K$, then I choose $0 < x_0 < (1/K)$. Then we have $K|x_0| < K|(1/K)| = 1 = |f(0)-f(x_0)|$.