More likely, you misquoted that guy. :)

You don't even understand the definition.

:)

所有跟帖： 

• Your memory of constructing positive definite funciton is failin -weston- ♀ (101 bytes) () 03/07/2015 postreply 09:41:18

• You forgot any integral over a zero-valued set yields zero. :) -career- ♂ (62 bytes) () 03/07/2015 postreply 11:19:23

• No. But the integral takes place over the entire space. Silly -weston- ♀ (0 bytes) () 03/07/2015 postreply 11:22:38

• Then look at your insisting on including the zero-valued subspac -career- ♂ (125 bytes) () 03/07/2015 postreply 11:29:43

• You learned Fourier by skipping all non-zero values of a functio -weston- ♀ (0 bytes) () 03/07/2015 postreply 11:42:21

• Show me where I did that? :) -career- ♂ (0 bytes) () 03/07/2015 postreply 11:48:47

• Out of (-.5, 0.5), rect's Lebesgue measure is zero:) -career- ♂ (200 bytes) () 03/07/2015 postreply 12:16:05

• Repeat one more time: Go and study Bochner Theorem (actually -weston- ♀ (35 bytes) () 03/07/2015 postreply 12:26:42

• Bochner/ Lebesgue won't CYA :) -career- ♂ (813 bytes) () 03/07/2015 postreply 12:40:48

• Gave you the math definition already. Your comment adds nothing -weston- ♀ (0 bytes) () 03/07/2015 postreply 13:07:01

• Got your "Say it the 3rd time, rec funcition is a step function. -career- ♂ (80 bytes) () 03/07/2015 postreply 13:10:25

• Too bad, you can't read math -weston- ♀ (0 bytes) () 03/07/2015 postreply 14:22:16