Consider the following function
\begin{equation}
f(x,y)=g(x)\times y,
\end{equation}
where all derivatives exist.
I am wondering whether it would be possible to pick a $g(x)$ function (other than the constant function $g(x)=k$ for some $k\in R$) such that $f(x,y)$ is jointly convex in $(x,y).$ Denoting the first and the second derivatives of $g(x)$ as $g^{'}$ and $g^{''},$ we obtain the Hessian of $f(x,y)$ as
\begin{equation}
H=\left[\begin{array}{cc}
g^{''}y & g^{'}\\
g^{'} & 0
\end{array}\right].
\end{equation}
Given that the second leading principal minor is $-[g^{'}]^{2}<0,$ $f(x,y)$ is neither convex nor concave in $(x,y).$
So, it seems to me that no $g(x)$ can make $f(x,y)$ convex or concave. Am I right? Is there a known-result that says that a "multiplicatively separable" function cannot be convex?
Best Answer
You could also let $g(x)=ax$ to get a convex/concave function. I do not think that the reason for $f(x,y)=g(x)\times y$ to be rarely convex/concave is because it is separable but rather because of the linear function in $y$. A general separable function $f(x,y) = g(x)\times h(y)$ can be either convex or concave. For instance $f(x,y) = e^x e^y$ is convex.