Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c$ and $|f(b)| = c$. Is there a Lipschitz function $g$ such that $|g| \geq c,$ $g(a)=f(a),$ $ g(b)=f(b)$ and Lipschitz constant of $f-g$ is less than epsilon for any positive epsilon?


There should be some simple counterexample.



Construction of the Lipschitz function with a given Lipschitz constant and given two values
$\newcommand\ep\varepsilon$Yes, it is easy to construct a counterexample here.


Indeed, if $g\_\ep$ is such a function for each given real $\ep>0$ (so that $|g\_\ep| \geq c$, $g\_\ep(a)=f(a)$, $g\_\ep(b)=f(b)$, and the Lipschitz constant of $f-g\_\ep$ is less than $\ep$), then $g\_\ep\to f$ pointwise (as $\ep\downarrow0$). So, it would follow that $|f|\ge c$ everywhere.


However, it is very easy to construct a Lipschitz function $f\colon [a,b] \to\mathbb{C}$ such that $|f(a)|\geq c$ and $|f(b)| = c$, but $|f(x)|<c$ for some $x\in(a,b)$.


