# Thread: how did they do this transition..

1. ## how did they do this transition..

http://i36.tinypic.com/lw3uh.jpg

i would have transform each norm into inner product form
and use the distributive law.

but here they just do some magic
put out thr Re part

i dont know whats the Re part in that context

what method they use here?

as if they say
||f+g||^2 =(||f||+||g||)^2

thats not true its a wrong way of the triangle innequality

2. Originally Posted by transgalactic
http://i36.tinypic.com/lw3uh.jpg

i would have transform each norm into inner product form
and use the distributive law.

That's what they did but in a long, sloppy way:

$||f+if||-||f-if||=|1+i|\cdot||f||-|1-i|\cdot||f||=0\,,\,\,since\,\,||kf||=|k|\cdot||f|| \,,\,\,k\in \mathbb{C}$[/color]

but here they just do some magic
put out thr Re part

i dont know whats the Re part in that context

$||f+if||=-i+i+$ $since\,\,=\overline{k}$ , but for some reason I can't understand, instead of cancelling the

two central summands, they thought it'd be a good idea to write

$i-i=--\overline{}=-2Re()$ , which seems a pretty goofy and weird thing to do, though the final result still is right.

Tonio

what method they use here?

as if they say
||f+g||^2 =(||f||+||g||)^2

thats not true its a wrong way of the triangle innequality
.

$
i-i=--\overline{}=-2Re()
$

$
i-i=0
$

i do understand this step
$
--\overline{}
$

but why
$
--\overline{}=-2Re()
$

?

4. Originally Posted by transgalactic
$
i-i=--\overline{}=-2Re()
$

$
i-i=0
$

i do understand this step
$
--\overline{}
$

but why
$
--\overline{}=-2Re()
$

?

Hmmm...I think you're rushing to write back without properly thinking things over: $\forall z\in \mathbb{C}\,,\,\,z+\overline{z}=2Re(z)$ , as you can easily check if, for example, you write $z=x+iy\,,\,\,z\,,\,y\in \mathbb{R}$ , and then put $z=-$

Tonio

5. yes i understand now thanks