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

1. ## 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

2. By taxicab you mean $\sup x_i$?

3. 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.)