{p(x)<1} is a Convex Set

This theorem can be considered a converse of a previous theorem.

Theorem: Let p denote a positive homogenous, subadditive function defined on a linear space X over the reals.

(i) The set of points x satisfying p(x)<1 is a convex subset of X, and 0 is an interior point of it.

(ii) The set of points x satisfying p(x)\leq 1 is a convex subset of X.

Proof: (i) Let K=\{x\in X\mid p(x)<1\}. Let x_1,x_2\in K. For 0\leq\alpha\leq 1,

\begin{aligned}p(\alpha x_1+(1-\alpha)x_2)&\leq \alpha p(x_1)+(1-\alpha)p(x_2)\\    &<\alpha+(1-\alpha)\\    &=1    \end{aligned}

Therefore K is convex. We also have p(0)=0\in K.

The proof of (ii) is similar.

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.