# Construction of the Lipschitz function with a given Lipschitz constant, given two values and with small Lipschitz norm

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.

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

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

• Thank you! Could you please explain more precisely where this estimate came from: $|g(t + \delta) - g(t)| \approx \frac{\pi \delta}{4}$? Nov 10, 2022 at 21:48
• 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 $L(g)<\frac{\pi c}{4c} = \frac{\pi}{4}$. Nov 10, 2022 at 21:56
• (So it should be $\lessapprox$, not $\approx$.) Nov 10, 2022 at 21:57
• 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? Nov 11, 2022 at 13:21
• 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. Nov 11, 2022 at 14:38