# Show the taxican norm is lipschitz equivalent to the Euclidean norm

• Nov 7th 2009, 02:50 AM
phillips101
Show the taxican norm is lipschitz equivalent to the Euclidean norm
What it says in the title. (Aside from the fact 'taxican' should be read as 'taxicab').

It's obvious that the taxicab norm is always larger than the Euclidean norm, but I can't find a constant such that h*taxicab norm is always less than the Euclidean norm.

Help? :(

Thanks
• Nov 7th 2009, 11:50 AM
Abu-Khalil
By taxicab you mean $\sup x_i$?
• Nov 7th 2009, 12:13 PM
Opalg
Quote:

Originally Posted by phillips101
What it says in the title. (Aside from the fact 'taxican' should be read as 'taxicab').

It's obvious that the taxicab norm is always larger than the Euclidean norm, but I can't find a constant such that h*taxicab norm is always less than the Euclidean norm.

Assuming that this is in 2-dimensional space, if $x = (x_1,x_2)$ then the taxicab norm of x is $\|x\|_1 = |x_1|+|x_2|$. If $\|x\|_2 = \sqrt{x_1^2+x_2^2}$ is the euclidean norm then $\|x\|_1^2 = \bigl(|x_1| + |x_2|\bigr)^2\leqslant2(|x_1|^2 + |x_2|^2) = 2\|x\|_2^2$ (Cauchy–Schwarz inequality). So $\|x\|_1\leqslant\sqrt2\|x\|_2$. (In n-dimensional space, the √2 becomes √n.)