Does this Lipschitz approximation of a continuous, monotone function keep monotonicity intact?

Question

Consider a continuous, monotone function $f: [0, 1] \rightarrow \mathbb{R}$ and a constant $K > 0$. Further, define $h: [0, 1] \rightarrow \mathbb{R}, x \mapsto \inf_{y \in [0, 1]} f(y) + K|x-y|$. It has been shown in other posts that $h$ is $K$-Lipschitz continuous and approximates $h$ uniformly for $K \rightarrow \infty$. What I want to know, is whether $h$ is still monotone. This should be easy to show, but I just can't work it out.

Answer 1

Yes, $h$ is monotone.*

We will assume $f$ is nondecreasing, with the nonincreasing case following from similar logic.

Let $x_1<x_2$, and let $y=y_2$ realize the infimum in the definition of $h(x_2)$. If $x_1<y_2$, then we certainly have

$$h(x_1)\leq f(x_1)\leq f(y_2)\leq h(x_2),$$

since we always have $h(x)\leq f(x)$ (take $y=x$ in the definition of $h$).

On the other hand, if $y_2\leq x_1$, then $|x_1-y_2|<|x_2-y_2|$, so we have $$h(x_1)\leq f(y_2)+K|x_1-y_2|< f(y_2) + K|x_2-y_2|=h(x_2).$$

*Remark

Note that I am answering the question in the post itself. The title at the moment currently asks a different question, "Does the Lipschitz extension of a continuous, monotone function keep monotonicity intact?", which is generally not true - see Jose27's answer for a counterexample.

Moreover, one can't speak of "the" Lipschitz extension of a function in the first place, as when a function is Lipschitz, there can be many different Lipschitz extensions to a larger domain.

Answer 2

No, for example if $f\equiv 0$ in $[0,1]$, viewed as a $K$-Lipschitz function, then $$h(x)= \begin{cases} -Kx, & x<0\\ f(x), & 0\leq x\leq 1\\ K(x-1), & x>1\end{cases}.$$

Another example is $f(x)=x$, for which $h(x)=|x|$.