# The $p$-th Root of a Convex Homogeneous Function of Degree $p$

## Backgroud

I was stuck at a question from a convex analysis book for more than a week. I couldn’t directly find its solution.

Let $K$ be a convex cone in $\R^n$, $p \ge 1$. The function $f: K \to [0,\infty)$ is homogeneous of degree $p$, i.e.

$f(\lambda x) = \lambda^p f(x),\quad \forall x \in K, \forall \lambda \ge 0.$

Suppose also that $f(x) > 0 \,\forall x \in K \setminus \{x\}$. Show that $g(x) := [f(x)]^{\frac1p}$ is convex on K.

## Resolution

I saw a proof for the Minkowski inequality using the convexity of the function $x \mapsto x^p$ on the webpage for a real analysis course. In fact, by replacing $\norm{\cdot}$ with $g(\cdot)$ in the proof of Theorem 7, we’re done.

First, observe that

\begin{equation} g(\lambda x) = [f(\lambda x)]^{\frac1p} = \lambda g(x),\quad \forall x \in K, \forall \lambda \ge 0 \label{homo} \end{equation}

Therefore, it suffices to show that

\begin{equation} g(x + y) \le g(x) + g(y). \label{goal} \end{equation}

The assumption $\norm{\vect{u}}_p = \norm{\vect{v}}_p = 1$ can be changed to $g(x) = g(y) = 1$. Then we’ve $f(x) = f(y) = 1$. By the convexity of $f$ on $K$, $\forall x,y \in K, \forall \lambda \in [0,1]$,

\begin{equation} f(\lambda x + (1 - \lambda) y) \le \lambda f(x) + (1 - \lambda) f(y) \le 1. \end{equation}

Since it’s given that $f(x) \ge 0 \,\forall x \in K$, we can say that

\begin{equation} g(\lambda x + (1 - \lambda) y) \le 1. \label{trick} \end{equation}

Finally, for any $u,v \in K$, we let

\begin{equation} \begin{alignedat}{2} \lambda &{}= \frac{g(u)}{g(u) + g(v)} &{} \qquad x &{}= \frac u{g(u)} \\ 1 - \lambda &{}= \frac{g(v)}{g(u) + g(v)}&{} \qquad y &{}= \frac v{g(v)}. \end{alignedat} \label{coeff} \end{equation}

Making use of \eqref{homo}, we substitute \eqref{coeff} into \eqref{trick}.

\begin{align*} g \left( \frac{g(u)}{g(u) + g(v)} \frac u{g(v)} + \frac{g(v)}{g(u) + g(v)} \frac v{g(v)} \right) &\le 1 \\ g \left(\frac u{g(u)+g(v)} + \frac v{g(u)+g(v)} \right) &\le 1 \\ g \left(\frac{u + v}{g(u) + g(v)} \right) &\le 1 \\ g(u + v) &\le g(u) + g(v) \end{align*}

We get \eqref{goal} at last.