Show the taxican norm is lipschitz equivalent to the Euclidean norm

**phillips101** · November 7th 2009, 02:50 AM

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

**Abu-Khalil** · November 7th 2009, 11:50 AM

By taxicab you mean $\sup x_i$ ?

**Opalg** · November 7th 2009, 12:13 PM

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.)

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