How can I prove that for 0<p<1 the defined function norm(x)_p = (sum_(i=1)^n |x_i|^p)^(1/p) for n >= 2 does not fulfill the triangle inequality anymore?

beefypy 2022-10-19 Answered
I need to prove that for 0 x p = ( i = 1 n | x i | p ) 1 / p for n 2
no longer satisfies the triangle inequality?
I know that in one dimension, the norm is just an absolute value, and the triangle inequality holds. I could take two unit vectors.
For example, a = (0,1,0) and b = (2/3.2/3.1/3)
But I don't really know how I can put the unit vectors in the sum above to show that
a + b a + b does not hold.
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Answers (2)

giosgi5
Answered 2022-10-20 Author has 15 answers
You cannot add vectors in different dimensions. Your a is in R 2 and b is in R 3 so a+b has no meaning. What you are asked to show is that if you fix n 2 then we can have vectors a and b in R 3 such that a + b > a + b . This is very easy: take the vectors (1,0,0,...,0) and (0,1,0,...,0) and verify that a + b = 2 1 / p , a + b = 2. Since p<1 we have 2 1 / p > 2
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djo57bgfrqn
Answered 2022-10-21 Author has 4 answers
Well, actually, this function is concave when you restrict your study to positive vectors (i.e., vectors with positive coordinates). If 2 f ( x ) denotes the Hessian matrix of your function f ( x ) = ( i = 1 n x i p ) 1 / p then, for any vector u you can show that u T 2 f ( x ) u 0 , hence the Hessian of -f is positive semi-definite, so, by a known theorem, -f convex, so f is concave.
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