convex unit ball

Asked Dec 10 '22 at 12:27

Active Dec 10 '22 at 13:11

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Let $x ∈ R^n $ and define $||x||p=$ $(\sum_{i=1}^n |x_i|^p)^(1/p) $ where $p>=1$ what is need to show is that for p ≥ 1 the unit ball ${∥x∥_p<=1}$ is convex.

I understand that to show this i need to show that for every two pionts $x_1$ and $x_2$ and scalar a : $0<=a<=1$ the line $a*x_1+(1-a)*x_2$ is in the unit ball, i tried using the norm properties :

$(\sum_{i=1}^n |a*x_1+(1-a)*x_2|^p)^(1/p) <= |a|^(1/p)*(\sum_{i=1}^n|x_1|^p)^(1/p) + |(1-a)|^(1/p)*(\sum_{i=1}^n|x_2|^p)^(1/p) $

but after that im stuck :(

edited Dec 10 '22 at 13:11

asked Dec 10 '22 at 12:27

Eliza Romanski

Any norm is convex. Do you the properties of a norm on a vector space? – geetha290krm Dec 10 '22 at 12:28
@geetha290krm i understand this intuitively but not how to proof that and yes i now the properties of norm on vector space – Eliza Romanski Dec 10 '22 at 12:30
I think you mean $\vert \vert x \vert \vert_p \leq 1$, otherwise it's not convex. So what you need to prove is $\vert \vert ax_1+(1-a)x_2\vert \vert_p \leq 1$. Use the properties of the norm to do so. – CyclotomicField Dec 10 '22 at 12:52
Does this answer your question? Confusion related to the convexity of norms. Btw, your "try using the norm properties" was wrong: your $|a|^{1/p}$ and $|1-a|^{1/p}$ should be replaced by $|a|$ and $|1-a|$ (or simply $a$ and $(1-a)$ since $a\in[0,1]$). – Anne Bauval Dec 10 '22 at 14:50

convex unit ball

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