# Geometry of vector spaces: if a subset is convex and satisfies the line condition, then it induces a norm

I'm trying to prove Theorem 3 on page 8 of these lecture notes.

Recall:

• A subset $B$ of some vector space V is convex if $(\forall v_1, v_2 \in B)(\forall t \in [0, 1])(tv_1 + (1-t)v_2 \in B)$. (In words: given any two points in B, the line segment between them must also be in B.)

• A subset $B$ of some vector space $V$ satisfies the line condition if, given any non-zero vector $v \in V$, there exists an associated line condition constant $a$ such that $(\forall t \in \mathbb{R}) (tv \in B \Leftrightarrow | t | \leq a)$.

If we have a subset $B$ which satisfies both conditions above, we can use it to define a norm $\| v \| = a^{-1}$, where $a$ is the line condition constant for $v$.

I am trying to prove that $\| \cdot \|$ is indeed a norm, but my lecture notes don't go in full depth on how to prove subadditivity.

The proof begins with the following (trivial) equality:

$v_1 + v_2 = (\| v_1 \| + \| v_2 \|)(\frac{\| v_1 \|}{\| v_1 \| + \| v_2 \|} \frac{v_1}{\| v_1 \|} + \frac{\| v_2 \|}{\| v_1 \| + \| v_2 \|} \frac{v_2}{\| v_2 \|})$

We then define $v = \frac{\| v_1 \|}{\| v_1 \| + \| v_2 \|} \frac{v_1}{\| v_1 \|} + \frac{\| v_2 \|}{\| v_1 \| + \| v_2 \|} \frac{v_2}{\| v_2 \|}$, and note that $v \in B$ (by convexity of $B$).

Hence, $v_1 + v_2 = v(\| v_1 \| + \| v_2 \|)$. From this, we somehow conclude that $\| v_1 + v_2 \| \leq \| v_1 \| + \| v_2 \|$.

How is that last step justified?

Notice that, by your definition of $\|\cdot\|$, then $v\in B\implies \|v\|\le 1$, so the last step is
$$\| v_1 + v_2 \| = \|v\| (\| v_1 \| + \| v_2 \|) \leq \| v_1 \| + \| v_2 \|$$