Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c$ and $|f(b)| = c$ and $\varepsilon > 0.$ It is easy to see that if $\|f\|_{\infty}< \frac{\varepsilon}{2} =: \delta (\varepsilon)$ then we can find $g$ with followning properties: 1. $$\|f-g\|_{\infty}< \varepsilon$$ 2. $$g(a)=f(a), \,\, g(b)=f(b)$$ 3. $$|g| \geq c$$ Indeed, it is enough to take $g$ with the given values in $a$ and $b$, such that $c \leq |g| < \frac{\varepsilon}{2}.$ Is it possible if we replace $\| \cdot \|_{\infty}$ by Lipschitz norm? There should be some simple counterexample.