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Liad
Liad
07:12
nice, im trying to generalize it for $R \ ^ d $
 
8 hours later…
Liad
Liad
15:18
@hardmath i think the "/2" in your def. of $||||_1$ should be on both x and y , dont you?
 
2 hours later…
hardmath
hardmath
17:22
@Liad: I don't think it would hurt to do it that way, but it isn't necessary. Check the work.
Once you have it working in two dimensions, you can extend it to higher dimensions in various ways.
Liad
Liad
@hardmath i said it because you wrote $||T(x,y)|| < sqrt(0.5)x \ ^ 2 $ , i think $T(x,y) = sqrt(x \ ^ 2) $ , in your current def. am i missing something?
hardmath
hardmath
I actually think we need the custom norm the way I set it up,, so that ||x||_1 = 1 when we take x = (1,0)'.
Maybe I got the inequality wrong. I'll go check.
No, what I wrote was that ||T(x,y)'||_1 < 0.8 ||(x,y)'||_1, showing that T is a contraction in this norm.
Liad
Liad
17:38
Huh, i will go see it again.
hardmath
hardmath
17:52
@Liad: Keep in mind that $T(x,y)' = (0,x)'$, so that $||T(x,y)||_1 = ||(0,x)||_1 = \sqrt{x^2/2} = \sqrt{2} \cdot |x|$. This is less than $0.8 \cdot ||(x,y)||_1$.
Liad
Liad
yea, i thought that $T(x,y) = (x,0) $ thanks :)
@hardmath Are you a math. student?
hardmath
hardmath
18:10
@Liad: Aren't we all? But sadly I am no longer allowed to go back to school in the fall, because I'm too old and I already got a Ph.D. in math.
Liad
Liad
Nice ! :)

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