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:
- $$\|f-g\|_{\infty}< \varepsilon$$
- $$g(a)=f(a), \,\, g(b)=f(b)$$
- $$|g| \geq c$$
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.
Best Answer
I can achieve $L(f - g) \leq (\frac{1}{2} + \frac{\pi}{4})\epsilon = (1.285\ldots)\epsilon$. Two reductions: (1) we can assume $|f(t)| < c$ for all $t \in (a,b)$ and (2) we can take $\epsilon = 1$. (1) because $C = \{t: |f(t)| \geq c\}$ is a closed subset of $[a,b]$, so its complement is a countable set of disjoint open intervals $[a_i, b_i]$ such that $|f(a_i)| = |f(b_i)| = c$ and $|f(t)| < c$ for all $c \in (a_i,b_i)$; then we can set $g = f$ on $C$ and handle each of these intervals separately. (2) just by scaling.
Assuming these reductions, define $g: [a,b] \to \mathbb{C}$ by letting $g(a) = f(a)$, $g(b) = f(b)$, and letting $g(t)$ move along the $|z| = c$ circle from $f(a)$ to $f(b)$ at uniform speed. The greatest possible discrepancy between $|f(b) - f(a)|$ and the length of the arc from $f(a)$ to $f(b)$ occurs when $f(a)$ and $f(b)$ are diametrically opposed and the arc length is $\frac{\pi}{2}$ times longer then the secant line. In that case, over any small interval $[t, t + \delta]$ we have $|f(t + \delta) - f(t)| < \frac{\delta}{2}$ since $L(f) < \frac{1}{2}$ by hypothesis, and $|g(t + \delta) - g(t)| \approx \frac{\pi \delta}{4}$ since small segments of a circle are approximately straight lines. Just from this we get $|(f - g)(t + \delta) - (f - g)(t)| \lessapprox \frac{\delta}{2} + \frac{\pi\delta}{2} = \frac{1 + \pi}{2}\delta$, so that $L(f - g) \leq \frac{1 + \pi}{2}$.