Are Minkowski functionals (a.k.a. gauges) *strictly* convex?

Question

Take a strictly convex subset $C$ of say, $\mathbb{R}^d$. For the sake of simplicity, assume that $C$ is compact and $0$ belongs to its interior.

Define the Minkowski functional \begin{align*} f \colon \mathbb{R}^d & \to [0,+\infty)\\ x &\mapsto \min\{\tau\ge 0 : x \in \tau C\} \end{align*}

It is known that $f$ is convex. Is its square $f^2$ strictly convex?

At least in two dimensions, it feels like it should be. It looks like a paraboloid whose sections are scaled $\partial C$'s (rather than ellipses). However, I have no idea how to prove it I and have no geometric intuition whatsoever for $d\ge 3$.

Answer 1

It is true, what follows is a proof.

If $x,y \in \mathbb{R}^d \backslash \{0\}$ are such that $\forall \alpha>0, \alpha x \neq y$, then $f(x)>0, f(y)>0$ and $\frac{x}{f(x)},\frac{y}{f(y)} \in \partial C$ and $ \frac{x}{f(x)}\neq\frac{y}{f(y)}$ so, since $C$ is strictly convex, for each $\lambda \in (0,1)$ we have that $$(1-\lambda) \frac{x}{f(x)} + \lambda\frac{y}{f(y)} \in \operatorname{int}(C)$$ and then, in particular, getting $\lambda=\frac{f(y)}{f(x)+f(y)}$ we get $$\frac{x+y}{f(x)+f(y)}\in \operatorname{int}(C)$$ and so $$f(x+y)<f(x)+f(y).$$ So if $x,y \in \mathbb{R}^d \backslash\{0\}$ are such that $\forall \alpha>0, \alpha x \neq y$, then also $$\forall \gamma \in (0,1), \forall \alpha>0, (1-\gamma)x \neq \alpha \gamma y$$ and so \begin{align*} \forall \gamma \in (0,1), \Bigl(f\bigl((1-\gamma)x+\gamma y\bigr)\Bigr)^2 &< \Bigl(f\bigl((1-\gamma)x\bigr)+f\bigl(\gamma y\bigr)\Bigr)^2 \\ &= \Bigl((1-\gamma)f(x)+\gamma f(y)\Bigr)^2 \\ &\le (1-\gamma)\bigl(f(x)\bigr)^2+\gamma\bigl(f(y)\bigr)^2. \end{align*} On the other hand, if $x,y \in \mathbb{R}^d \backslash \{0\}$ are such that there exists $\alpha \in (0,+\infty) \backslash \{1\}$ such that $\alpha x = y$, then \begin{align*} \forall \gamma \in (0,1), \Bigl(f\bigl((1-\gamma)x+\gamma y\bigr)\Bigr)^2 &= \Bigl(f\bigl((1-\gamma)x+\gamma \alpha x\bigr)\Bigr)^2 \\ &= \Bigl(f\bigl((1-\gamma+\gamma \alpha) x\bigr)\Bigr)^2 \\ &= (1-\gamma+\gamma \alpha)^2\bigl(f(x)\bigr)^2 \\ &= \bigl((1-\gamma)\cdot1+\gamma \alpha\bigr)^2\bigl(f(x)\bigr)^2 \\ &< \bigl( (1-\gamma)\cdot 1^2 + \gamma \alpha^2 \bigr) \bigl(f(x)\bigr)^2 \\ &= (1-\gamma)\bigl(f(x)\bigr)^2+\gamma\bigl(f(\alpha x)\bigr)^2 \\ &= (1-\gamma)\bigl(f(x)\bigr)^2+\gamma\bigl(f(y)\bigr)^2. \end{align*} Finally, if $x,y\in \mathbb{R}^d$ and $x=0\neq y$ then \begin{align*} \forall \gamma \in (0,1), \Bigl(f\bigl((1-\gamma)x+\gamma y\bigr)\Bigr)^2 &= \bigl(f(\gamma y)\bigr)^2 \\ &= \gamma ^2 \bigl(f(y)\bigr)^2 \\ &< \gamma \bigl(f(y)\bigr)^2 \\ &= (1-\gamma) \bigl(f(x)\bigr)^2 + \gamma \bigl(f(y)\bigr)^2 \end{align*} and obviously the same argument apply to the case $x,y\in \mathbb{R}^d$ and $x\neq 0 =y$ switching $x$ and $y$.

Answer 2

Here is another approach that pushes the technicalities under the rug of existing results.

It is straightforward to check that $f$ is positive homogeneous, that is $f(tx) = tf(x)$ for $t \ge 0$.

Note that the gauge $f$ is the largest positive homogenous convex function majorized by $x \mapsto \delta_C(x) + 1$ (where $\delta_C$ is the 'convex' indicator function, that is $0$ for $x \in C$ and $+\infty$ for $x \notin C$. See Section 5 of Rockafellar's "Convex Analysis" for example). In particular, $f$ is convex.

In order to establish strictness, we just need to show that $f$ is strict in a particular case: If $f(x_1) = f(x_2) \neq 0$ and $x_1 \neq x_2$, we see that $t {x_1 \over f(x_1)} + (1-t) {x_2 \over f(x_2)} \in C^\circ$ for $t \in [0,1]$ and so $f(t {x_1 \over f(x_1)} + (1-t) {x_2 \over f(x_2)}) < 1$, or $f(t x_1 + (1-t) x_2) < f(x_1) = t f(x_1) + (1-t)f(x_2)$.

Now let $q(x) = x^2$ and note that $q$ is strictly convex and strictly increasing for $x \ge 0$. Let $s=f^2$.

Suppose $t \in (0,1)$ and $x_1 \neq x_2$.

If $f(x_1) = f(x_2)$ then $s(t x_1 + (1-t) x_2) < s(x_1) = t s(x_1) + (1-t)s(x_2)$ since $q$ is strictly increasing.

Otherwise, $s(t x_1 + (1-t) x_2) = q(f( t x_1 + (1-t) x_2)) \le q(t f(x_1) + (1-t)f(x_2)) < t s(x_1) + (1-t)s(x_2)$.