# Does convexity of a 'norm' imply the triangle inequality?

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:

1. $d(r \mathbf{v}) = |r|d(\mathbf{v})$,
2. $d(\mathbf{v})\ge 0$, with equality iff $\mathbf{v} = 0$, and
3. $d(\mathbf{u})+d(\mathbf{v}) \ge d(\mathbf{u}+\mathbf{v})$ (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!)

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Set $p = \frac{1}{2}$ and use property 1. –  Qiaochu Yuan May 29 '11 at 16:58
@Qiaochu Yuan: Thanks! Feeling a bit silly now! I even tried that at some point, and convinced myself it didn't work. :) –  duncanm May 29 '11 at 17:09
It's cool to know that a convex-(1)-(2) function is equivalent to being a norm... I didn't know the triangle inequality was equivalent to convexity under (1) and (2). Good to know. Thumbs up! –  Patrick Da Silva May 29 '11 at 17:14
@Patrick: Cheers! –  duncanm May 29 '11 at 19:23
@Patrick: The homogeneity condition $d(r\mathbf{v}) = |r| d(\mathbf{v})$ is of course crucial here. Translation invariance and the triangle inequality are not enough to infer homogeneity. Some time ago I wrote an answer trying to elucidate the relations between scalar products, norms and metrics, maybe you find that interesting. –  t.b. May 29 '11 at 21:06
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## 1 Answer

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.

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