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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.

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.

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 $\delta (\varepsilon)$ with $\|f\|_{\infty}< \delta (\varepsilon) $ such that we can find 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 any $\varepsilon$?

There should be some simple counterexample.

It is possible that $\|f-g\|_{\infty}< \varepsilon$. Indeed, let $\delta= \frac{\varepsilon}{2}$. Then we can take $g$ with $\|g\|_{\infty}< \delta$.

Is it possible if we replace condition $\|f\|_{\infty}< \delta (\varepsilon)$ by: Lipschitz constant of $f$ is less than $\delta (\varepsilon) $?

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.