TITLE: Does convexity of a 'norm' imply the triangle inequality?
QUESTION [18 upvotes]: Given a vector space $V$ (for convenience, defined over $\mathbb{r}$), we call $d:V\rightarrow\mathbb{R}$ a norm for $V$ if $\forall \mathbf{u}, \mathbf{v} \in V$ and $\forall r \in \mathbb{R}$ we have:

$d(r \mathbf{v}) = |r|d(\mathbf{v})$,
$d(\mathbf{v})\ge 0$, with equality iff $\mathbf{v} = 0$, and
$d(\mathbf{u})+d(\mathbf{v}) \ge d(\mathbf{u}+\mathbf{v})$ (triangle inequality)

I've read in a few places that an important property of a norm is that it is convex; that is, given $\mathbf{u},\mathbf{v} \in V$, and $p \in (0,1)$, we have $d(p \mathbf{u} + (1-p) \mathbf{v}) \le p d(\mathbf{u}) + (1-p) d(\mathbf{v})$.  This clearly follows from the triangle inequality.
My question is: Does the reverse also hold?  i.e. does a function satisfying (1) and (2) above which is convex necessarily satisfy the triangle inequality?  If not, what is an instructive counterexample?
Thanks! (btw: please feel free to suggest better tags / improvements to the question; I'm new to this!)

REPLY [10 votes]: Resolved in comments.
Setting $p=\frac12$ in the definition of convexity, we have
$$
d\Big( \frac{\mathbf u + \mathbf v}{2} \Big) \leqslant \frac12 d(\mathbf u) + \frac12 d(\mathbf v).
$$
By the scaling or homogeneity, the left hand side is simply $\frac12 d(\mathbf u + \mathbf v)$; plugging in this and simplifying, we get the triangle inequality.