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I'm reading a book on ODE, and find $|\cdot|$ is confusing. It says:

Consider a function $g:\Omega \rightarrow \mathbb{R}^n$. For every compact $K\subset \Omega$, there exist constants $C$ and $L$ such that $$ |g(t,x)| \leq C, \quad |g(t,x)-g(t,y)| \leq L |x-y| \qquad \forall (t,x),(t,y)\in K$$

What is $|g(t,x)|$? $g$ is $n$-dimensional. It seems neither just the absolute value with $C$ being a constant vector, nor the norm.

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  • $\begingroup$ Maybe $|\cdot|$ is defined somewhere before. If not, my intuition is, that it is most often the 1-norm, the 2-norm oder the $\infty$-norm. $\endgroup$
    – Thomas
    Feb 22, 2013 at 16:30
  • $\begingroup$ What makes you think it's not the absolute value, $C$ and $L$ are real numbers. $\endgroup$
    – JSchlather
    Feb 22, 2013 at 16:42
  • $\begingroup$ @JSchlather $C$ is a real number, but $g\in \mathbb{R}^n$. That's why I thought it is not the absolute value. $\endgroup$
    – Chang
    Feb 22, 2013 at 16:50
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    $\begingroup$ Sorry, I tend to use absolute value and norm interchangeably. What I meant to say is that $|\cdot|$ is likely the euclidean norm. That is $|g(t,x)|=\sqrt{g_1(t,x)^2+\cdots g_n(t,x)^2}$. $\endgroup$
    – JSchlather
    Feb 22, 2013 at 16:52

1 Answer 1

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Take any norm of $\mathbb{R}^n$, and the result will be fine (up to changing of the constants $C$ and $L$), since all norms on a fixed finite dimensional vector space are comparable.

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  • $\begingroup$ Thanks. For another statement, the author, for $f\in\mathbb{R}^n$, wrote, $|f|\leq C, \ ||D_x f|| \leq L$. What would be the author's motivation to use $|\cdot|$ and $||\cdot||$ here? $\endgroup$
    – Chang
    Feb 22, 2013 at 16:53
  • $\begingroup$ "What would be the author's motivation to use ..." Why are you asking me? I'm not the author, and you conveniently didn't tell us which book this is! :-p Either the author explained his/her notation in the book (in which case you should read it more thoroughly), or the author didn't (in which case if you really want to know the "motivation" you have to contact the author and ask). $\endgroup$
    – Willie Wong
    Feb 24, 2013 at 7:59

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