zeebek
Sep1-10, 05:34 PM
I am reading the definition in wiki ( nothing better at the moment)
http://en.wikipedia.org/wiki/Lorentz_space
It seems too vague for me, namely what they call "rearrangement function" f^{*}:
f^{*}: [0, \infty) \rightarrow [0, \infty]; \\
f^{*}(t) = \inf\{\alpha \in \mathbb{R}^{+}: d_f(\alpha) \leq t\}; \\
d_f(\alpha) = \mu(\{x \in X : |f(x)| > \alpha\}).
I am trying to put in words what is written. Is it right:
first for a given twe are looking for all \alpha's, so that d_f(\alpha) \leq t, where d_f(\alpha) is basically a size of the area where |f(x)| > \alpha ? Then we take infinum via \alpha , so as a result there will be the smallest d?
Still I cannot imagine "geometrically" how is it?
At last, I need just simpler difinition for the case when f is real.
thanks!
http://en.wikipedia.org/wiki/Lorentz_space
It seems too vague for me, namely what they call "rearrangement function" f^{*}:
f^{*}: [0, \infty) \rightarrow [0, \infty]; \\
f^{*}(t) = \inf\{\alpha \in \mathbb{R}^{+}: d_f(\alpha) \leq t\}; \\
d_f(\alpha) = \mu(\{x \in X : |f(x)| > \alpha\}).
I am trying to put in words what is written. Is it right:
first for a given twe are looking for all \alpha's, so that d_f(\alpha) \leq t, where d_f(\alpha) is basically a size of the area where |f(x)| > \alpha ? Then we take infinum via \alpha , so as a result there will be the smallest d?
Still I cannot imagine "geometrically" how is it?
At last, I need just simpler difinition for the case when f is real.
thanks!