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Consider the function: $f(x,y) = \dfrac{ax+by+c}{x+y}$

Given that $x,y,a,b,c > 0$ and $a \neq b \neq c$, we can show that this function is non-convex (by taking the Hessian of $f$ wrt $(x,y)$ and checking for its positive semidefiniteness).

Let us say we now have another function:

$$ g(x_{1},x_{2} \cdots x_{n},y_{1},y_{2} \cdots y_{n}) = \sum\limits_{i=1}^{n} f(x_{i},y_{i}) $$

Given that $f$ is non-convex, will the function $g$ be non-convex as well? Investigating its Hessian with respect to $2n$ variables seems to be tedious.

Another approach would be to get the Hessian of $g$ as a function of the Hessian of $f$. Is there an expression for the Hessian of a sum of functions?

V-Red
  • 193

1 Answers1

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Consider the Hessian $\mathbf{H_{g}}$ of $g$ with respect to $(x_{1},y_{1},x_{2},y_{2} \cdots x_{n},y_{n})$.

\begin{eqnarray} \mathbf{H_{g}} &=& \left( {\begin{array}{cc} \dfrac{\partial^{2} g}{\partial x_{1}^{2}} & \dfrac{\partial^{2} g}{\partial x_{1}\partial y_{1}} & \dfrac{\partial^{2} g}{\partial x_{1}\partial x_{2}} & \dfrac{\partial^{2} g}{\partial x_{1}\partial y_{2}} & \cdots & \dfrac{\partial^{2} g}{\partial x_{1}\partial x_{n}} & \dfrac{\partial^{2} g}{\partial x_{1}\partial y_{n}} \\ \dfrac{\partial^{2} g}{\partial y_{1}\partial x_{1}} & \dfrac{\partial^{2} g}{\partial y_{1}^{2}} & \dfrac{\partial^{2} g}{\partial y_{1}\partial x_{2}} & \dfrac{\partial^{2} g}{\partial y_{1}\partial y_{2}} & \cdots & \dfrac{\partial^{2} g}{\partial y_{1}\partial x_{n}} & \dfrac{\partial^{2} g}{\partial y_{1}\partial y_{n}} \\[5pt] \vdots & & \ddots \\[5pt] \vdots & & & \ddots \\[5pt] \dfrac{\partial^{2} g}{\partial x_{n}\partial x_{1}} & \dfrac{\partial^{2} g}{\partial x_{n}\partial y_{1}} & \dfrac{\partial^{2} g}{\partial x_{n}\partial x_{2}} & \dfrac{\partial^{2} g}{\partial x_{n}\partial y_{2}} & \cdots & \dfrac{\partial^{2} g}{\partial x_{n}^{2}} & \dfrac{\partial^{2} g}{\partial x_{n}\partial y_{n}} \\ \dfrac{\partial^{2} g}{\partial y_{n}\partial x_{1}} & \dfrac{\partial^{2} g}{\partial y_{n}\partial y_{1}} & \dfrac{\partial^{2} g}{\partial y_{n}\partial x_{2}} & \dfrac{\partial^{2} g}{\partial y_{n}\partial y_{2}} & \cdots & \dfrac{\partial^{2} g}{\partial y_{n}\partial x_{n}} & \dfrac{\partial^{2} g}{\partial y_{n}^{2}} \end{array} } \right) \\[15pt] &=& \left( {\begin{array}{cc} \mathbf{H_{f}(x_{1},y_{1})} & 0 & \cdots & 0 \\ 0 &\mathbf{H_{f}(x_{2},y_{2})} & \cdots & 0 \\ \vdots & \ddots & \cdots & 0 \\ 0 & 0 & \cdots & \mathbf{H_{f}(x_{n},y_{n})} \end{array} } \right) \end{eqnarray}

$\mathbf{H_{g}}$ is positive semidefinite iff $\mathbf{H_{g}}$ is symmetric and its eigenvalues are non-negative. Given that $\mathbf{H_{f}(x_{i},y_{i})}~~\forall i \in [1,n]$, is symmetric and its eigenvalues are negative for $x_{i}, y_{i},a,b,c >0$ and $a \neq b \neq c$, it follows that $\mathbf{H_{g}}$ is also not positive semidefinite (since the eigenvalues of $\mathbf{H_{g}}$ is the list of eigenvalues of $\mathbf{H_{f}(x_{i},y_{i})}~~\forall~i \in[1,n]$).

Hence, $g$ is non-convex.

V-Red
  • 193