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.

  • $\begingroup$ @ChristianRemling it’s not possible, because |f| is less than $\varepsilon / 2$ and $ |f(a)| \geq c$ $\endgroup$
    – Hpela
    Nov 9, 2022 at 18:02

1 Answer 1


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 $L(f - g) \leq \frac{1 + \pi}{2}$.

  • $\begingroup$ Thank you! Could you please explain more precisely where this estimate came from: $|g(t + \delta) - g(t)| \approx \frac{\pi \delta}{4}$? $\endgroup$
    – Hpela
    Nov 10, 2022 at 21:48
  • 2
    $\begingroup$ If $L(f) < \frac{1}{2}$ then $|b - a| > 4c$ since diametrically opposed points are $2c$ units apart. Now $g$ has to travel $\pi c$ units because that is the arc length between diametrically opposed points, so this yields $L(g)<\frac{\pi c}{4c} = \frac{\pi}{4}$. $\endgroup$
    – Nik Weaver
    Nov 10, 2022 at 21:56
  • $\begingroup$ (So it should be $\lessapprox$, not $\approx$.) $\endgroup$
    – Nik Weaver
    Nov 10, 2022 at 21:57
  • $\begingroup$ I don't understand why if we have a function g which is an arc then the Lipschitz constant is the ratio of the arc length to the segment length. Do we treat g as a function of distance from time and the Lipschitz constant is then velocity (derivative of g w.r.t time)? Could this be how we compute the Lipschitz constant for any function that encircles the circle? $\endgroup$
    – Hpela
    Nov 11, 2022 at 13:21
  • $\begingroup$ Yes, you can think of it that way. Any Lipschitz function on $[a,b]$ is differentiable almost everywhere and its Lipschitz number is the sup norm of its derivative. $\endgroup$
    – Nik Weaver
    Nov 11, 2022 at 14:38

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