# 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?

I need to prove that for 0${‖x‖}_{p}$ = ($\sum _{i=1}^{n}|{x}_{i}{|}^{p}$)${}^{1/p}$ for $n\ge 2$
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, $\stackrel{\to }{a}$ = (0,1,0) and $\stackrel{\to }{b}$ = (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
$‖a+b‖$ $\le$ $‖a‖$ + $‖b‖$ does not hold.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

giosgi5
You cannot add vectors in different dimensions. Your a is in ${\mathbb{R}}^{2}$ and b is in ${\mathbb{R}}^{3}$ so a+b has no meaning. What you are asked to show is that if you fix $n\ge 2$ then we can have vectors a and b in ${\mathbb{R}}^{3}$ 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 ${2}^{1/p}>2$
###### Did you like this example?
djo57bgfrqn
Well, actually, this function is concave when you restrict your study to positive vectors (i.e., vectors with positive coordinates). If ${\mathrm{\nabla }}^{2}f\left(x\right)$ denotes the Hessian matrix of your function $f\left(x\right)={\left(\sum _{i=1}^{n}{x}_{i}^{p}\right)}^{1/p}$ then, for any vector $\stackrel{\to }{u}$ you can show that ${\stackrel{\to }{u}}^{T}{\mathrm{\nabla }}^{2}f\left(x\right)\stackrel{\to }{u}\le 0,$ hence the Hessian of -f is positive semi-definite, so, by a known theorem, -f convex, so f is concave.
###### Did you like this example?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee