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 true if we replace supremum norm by Lipschitz norm? I.e. can we find $\delta (\varepsilon)$ and $g$ such that the above three conditions hold if we replace $\| \cdot \|_{\infty}$ by Lipschitz norm everywhere above?

There should be some simple counterexample.

Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c,$ $|f(b)| = c$ 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 true if we replace supremum norm by Lipschitz norm? I.e. can we find $\delta (\varepsilon)$ and $g$ such that the above three conditions hold if we replace $\| \cdot \|_{\infty}$ by Lipschitz norm everywhere above?

There should be some simple counterexample.