Let be a normed space.
Show the function is a metric.
Hi biddoggy!
a metric satisfies:
(1) d(x,x) = 0
and , if
So: ||x-x|| = ||0|| = 0
OKAY
||x-y|| > 0, because , thus
OKAY
(2) d(x,y) = d(y,x)
||x-y|| =||y-x||
||x-y|| = ||-(x-y)||
||x-y|| = |-1|*||x-y|
||x-y|| = ||x-y||
OKAY
(3)
||x+z - (y+z)|| \le ||x+z|| + ||-(y+z)|| = ||x+z|| + ||y+z||
OKAY
Are there any questions on that?
Yours
Rapha