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One proof that the $p$-norm $\| x\|_p = (|x_1|^p + \ldots + |x_n|^p)^\frac{1}{p}$ satisfies the triangle inequality exploits the fact that $ x \mapsto |x_1|^p + \ldots + |x_n|^p$ is a convex function on $\mathbb{R}^n$. I abstracted the ideas of this proof into the following proposition.

Proposition: Let $X$ be vector space over $\mathbb{R}$. Let $\| \cdot \|$ be a function on $X$ satisfying all the axioms for a norm, except possibly the triangle inequality. To be precise:

  • $\| t x\| = |t| \|x\|$ for all $t \in \mathbb{R}$, $x \in X$ (positive homogeneity).
  • $\|x\| \geq 0$ for all $x \in X$, with equality if and only if $x=0$ (positive definiteness).

Then, the following are equivalent:

  1. $\|\cdot\|$ is a norm.
  2. $\| \cdot \|$ is a convex function.
  3. There exists a strictly increasing function $f : [0,\infty) \to [0,\infty)$ such that $x \mapsto f(\|x\|)$ is a convex function.

Proof: It's easy to check that even a seminorm is convex, so (1) implies (2). If (2) holds, then (3) holds with $f(t) = t$. Suppose, then, that $f$ is as in (3). Let $x,y \in X$. We want to establish the triangle inequality, $\|x+y\| \leq \|x\| + \|y\|$. If either of $x,y$ is zero, this holds trivially, so assume $x$ and $y$ are both nonzero. Define $u = \frac{x}{\|x\|}$, $v = \frac{y}{\|y\|}$ and observe $\|u\| = \|v\| =1$. Define $s = \frac{\|x\|}{\|x\| + \|y\|}$, $t = \frac{\|y\|}{\|x\| + \|y\|}$ and observe $s,t \in [0,1]$ with $s+t=1$. Observe $su+tv = \frac{x+y}{\|x\|+\|y\|}$. Thus, from $f(\|su+tv\|) \leq sf(\|u\|) + tf(\|v\|)$, we get $$f(\frac{\|x+y\|}{\|x\|+\|y\|}) \leq s f(1)+tf(1) = f(1).$$ Since $f$ is strictly increasing, this shows that $\frac{\|x+y\|}{\|x\|+\|y\|} \leq 1$, which is the triangle inequality. QED.

To apply this proposition to the $p$-norm, simply note that $f(t) = t^\frac{1}{p}$ is a strictly increasing function such that $x \mapsto f(\|x\|_p) = |x_1|^p + \ldots + |x_n|^p$ is a convex function.

Question: Is the naïve anlalog for seminorms of this result valid? To be precise, suppose $X$ is a vector space over $\mathbb{R}$. Let $\| \cdot \| : X \to [0,\infty)$ be any positive homogeneous function. Is it true that the following are equivalent?

  1. $\|\cdot\|$ is a seminorm.
  2. $\| \cdot \|$ is a convex function.
  3. There exists a strictly increasing function $f : [0,\infty) \to [0,\infty)$ such that $x \mapsto f(\|x\|)$ is convex.

Some observations:

  • The proofs that (1) implies (2) implies (3) go through unchanged. So, we just need to assume that an $f$ as in (3) exists and check that $\|x+y\| \leq \|x\| + \|y\|$ for all $x,y \in X$.
  • If $\|x+y\| = 0$, nothing needs to be said.
  • If $\|x\|$ and $\|y\|$ are both nonzero, then the proof given above works.
  • If $\|x\| = \|y\|=0$, then we can argue as follows: $$ f(\|\frac{x+y}{2}\|) \leq \frac{f(\|x\|) + f(\|y\|)}{2} = f(0)$$ so, since $f$ is strictly increasing, we get $\frac{x+y}{2} \leq 0$. Thus, in this case, both sides of the triangle inequality are zero.

The remaining case to consider is when, say, $\|x\|$ and $\|x+y\|$ are both nonzero, but $\|y\|=0$. Thanks for reading!

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OK I managed to figure this out. I'll post my argument here.

Proposition: Let $X$ be a vector space over a field $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$ and let $\| \cdot \| : X \to [0,\infty)$ satisfy $\|\lambda x\| = |\lambda| \|x\|$ for all $\lambda \in \mathbb{F}$ and all $x \in X$. Thus, $\| \cdot \|$ is a seminorm if and only if it satisfies the triangle inequality. Then, the following are equivalent:

  1. $\| \cdot \|$ is a seminorm.
  2. $\|\cdot\|$ is a convex function.
  3. There exists a strictly increasing function $f:[0,\infty) \to [0,\infty)$ such that $x \mapsto f(\|x\|)$ is convex.

Proof: A simple calculation shows that seminorms are convex, so (1) implies (2). Trivially, (2) implies (1), since we may take $f$ to be the identity map. Thus, we assume that (3) holds. Let $f : [0,\infty) \to [0,\infty)$ be a strictly increasing map such that $x \mapsto f(\|x\|)$ is convex. The goal is to prove that $\|x+y\| \leq \|x\| + \|y\|$ where $x,y \in X$.

Case 1: $\|x\| >0$ and $\|y\| >0$. Then, we may proceed exactly as in the question statement above.

Case 2: $\|x\| = \|y\| =0$. Then, we have $$ f( \frac{\|x+y\|}{2}) = f( \| \frac{x+y}{2} \|) \leq \frac{ f(\|x\|) + f(\|y\|)}{2} = f(0)$$ which implies $\frac{\|x+y\|}{2} \leq 0$ so that $\|x+y\| =0$, and the triangle inequality holds.

Case 3: $\|x\|=0$, but $\|y\| > 0$. Let us note that there is no loss of generality to assume that $f(0)=0$ (since, otherwise, we can replace $f$ with $x \mapsto f(x) - f(0)$, which is still strictly increasing and whose composition with $\|\cdot\|$ is still convex). Let $a>0$ be arbitrary. Define \begin{align*} u = \frac{x}{a} && v = \frac{y}{\|y\|} && s = \frac{a}{a+\|y\|} && t = \frac{\|y\|}{a+\|y\|} \end{align*} and observe that $su+tv = \frac{x+y}{a+\|y\}}$. We have $$ f( \frac{\|x+y\|}{a+\|y\|}) = f(s u + tv) \leq s f(u) + t f(v) = 0 + t f(1) \leq f(1)$$ so that, since $f$ is strictly increasing, $$\frac{\|x+y\|}{a+\|y\|} \leq 1.$$ Letting $a \to 0$, we get $\|x+y\| \leq \|y\|$, so the triangle inequality holds in this last case.

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