I have a polynomial that I know to be convex. If I homogenize the polynomial, is the resulting homogeneous polynomial also convex? I know that the perspective of a convex function is convex, but cannot find a citation for the homogenization of a convex function.


The function whose convexity I would like to show in the variables $(a,p\_1,\dots,p\_N)$ for $a > 0$ and under some bound on the magnitudes of the $p\_n$'s (I suspect the bound will be $\vert p\_n \vert < a$) is


$a^2 \prod\_{n=1}^N \left(1 + \vert p\_n \vert^2/a^2\right)$.


The function $\prod\_{n=1}^N \left(1 + \vert p\_n \vert^2\right)$ is log-convex and therefore convex in the $p\_n$'s, but I am having difficulty showing that the original function is convex. However, the original function is a homogenized version of this convex function.


Thank you!



Convex polynomial homogenization and convexity
Amir - thanks for your nice answer. I did end up figuring this one out:


The function $a\_0^2 \prod\_{j=1}^{N\_t} \left( 1 + \vert p\_j \vert^2/a\_0^2 \right)$ is convex under certain bounds on $a\_0$ and the $p\_j$.
Consider the function 
\begin{equation}
\label{eq:sqrtabmag}
\sqrt{\prod\_{j=1}^{N\_t} \left( 1 + \vert p\_j \vert^2 \right)},
\end{equation}
whose logarithm is:
\begin{equation}
\label{eq:logsqrtabmag}
\frac{1}{2}\sum\_{j=1}^{N\_t} \log\left(1 + \vert p\_j \vert^2 \right).
\end{equation}
The logarithm's Hessian matrix is diagonal with entries
\begin{equation}
\begin{array}{ll} \frac{1 - \vert p\_j \vert^2}{( 1 + \vert p\_j \vert^2 )^2}, & j=1,\dots,N\_t, \end{array}
\end{equation}
which are greater than or equal to zero if $\vert p\_j \vert \leq 1$, so the logarithm is convex if $\vert p\_j \vert \leq 1$.
Because log-convex functions are themselves convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.5.1), $\sqrt{\prod\_{j=1}^{N\_t} \left( 1 + \vert p\_j \vert^2 \right)}$ is also convex if $\vert p\_j \vert \leq 1$.
Convexity of the function $a\_0 \sqrt{\prod\_{j=1}^{N\_t} \left( 1 + \vert p\_j \vert^2 / a\_0^2 \right)}$ for $a\_0 > 0$ and $\vert p\_j \vert \leq a\_0$ follows from the fact that the perspective of a convex function is convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.2.6). 
Finally, because a nonnegative convex function raised to a power $b \geq 1$ is convex (Ref. \cite{Boyd:ConvexOpt}, \textsection 3.2.4), the original function $a\_0^2 \prod\_{j=1}^{N\_t} \left( 1 + \vert p\_j \vert^2 / a\_0^2 \right)$ is convex.


