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.


1 Answer 1


$\newcommand\ep\varepsilon$Yes, it is easy to construct a counterexample here.

Indeed, if $g_\ep$ is such a function for each given real $\ep>0$ (so that $|g_\ep| \geq c$, $g_\ep(a)=f(a)$, $g_\ep(b)=f(b)$, and the Lipschitz constant of $f-g_\ep$ is less than $\ep$), then $g_\ep\to f$ pointwise (as $\ep\downarrow0$). So, it would follow that $|f|\ge c$ everywhere.

However, it is very easy to construct a Lipschitz function $f\colon [a,b] \to\mathbb{C}$ such that $|f(a)|\geq c$ and $|f(b)| = c$, but $|f(x)|<c$ for some $x\in(a,b)$.

  • 1
    $\begingroup$ Thank you! Sorry, but I asked the question wrong and had to update. $\endgroup$
    – Hpela
    Nov 8, 2022 at 23:30
  • 1
    $\begingroup$ @Hpela : After the "update", the question looks quite incomprehensible to me. $\endgroup$ Nov 9, 2022 at 1:55
  • $\begingroup$ So I updated again. $\endgroup$
    – Hpela
    Nov 9, 2022 at 7:32
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    $\begingroup$ @Hpela it is considered a little rude to modify a question after the original version was correctly answered. Next time please accept Iosif's answer and post the new version separately. $\endgroup$
    – Nik Weaver
    Nov 9, 2022 at 11:43
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    $\begingroup$ @NikWeaver Sure, thank you for pointing that out $\endgroup$
    – Hpela
    Nov 9, 2022 at 12:34

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