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Reading a paper, they use the fact that the function $x \mapsto \log |\det Df(x)|$ is $v$-Hölder, where $Df$ is the derivative of some map. Then, they state that the function $f$ is $C^{1+v}$ and continue the paper with this assumption. Does the condition of $f$ being $C^{1+v}$ imply the Hölder continuity of $\log|\det Df(x)|$?

Thanks

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    You might want to recall how $C^{1+v}$ regularity is defined (presumably for non integer $v$).2012-02-17

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Since $f$ is $C^{1+v}$, the components of $Df$ are $v$-Hölder. Then $\det Df$, being a sum of products of $v$-Hölder functions, is also $v$-Hölder, and so is $|\det Df|$. Finally, if $\det Df(x)\ne0$, then also $\log|\det Df|$ is $v$-Hölder.

This shows local Hölder continuity in the set where $\det Df$ does not vanish. For global Hölder continuity some additional conditions are required:

  • Boundedness of the derivatives of $f$.
  • A uniform in $x$ lower bound on $|\det Df|$, as indicated in Didier's comment.
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    It seems the missing condition is not only that $\det Df(x)\ne0$ for every $x$ but that $|\det Df(x)|\geqslant c$, with $c\gt0$, uniformly on $x$.2012-02-17
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    @Didier My answer was about local Hölder continuity; I will edit it.2012-02-17
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    Understood. $ $2012-02-17
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    Given the question, I would bet that the underlying space is compact, so that local Hölder continuity and global Hölder continuity are the same.2012-02-17