# Function linear in one variable and quadratic concave on another, is concave in jointly?

I have a function $f(x,y,z)$. $f$ is convex with respect to $x$, and linear with respect to $y$. Also, it is quadratic concave with respect to $z$. I want to solve the following optimization problem: \begin{align} \min_{x\in C} \max_{y\in D_y} \max_z f(x,y,z) \end{align} My question is can I assume that this problem is concave with respect to $(y,z)$? Let $g(y,z) = \min_{x\in C} f(x,y,z)$. Function $g(u)$, where $u=(y,z)$ is concave? What if $f(x,y,z) = f_1(x,y) + f_2(x,z)$?

No, you may not. Suppose $f(x,y,z) = -yz^2$, with $D_y = \mathbb{R_+}$, then the Hessian is: $$\begin{pmatrix}0 & -2z \\ -2z & -2y\end{pmatrix}$$ which is indefinite for $z \neq 0$, since the eigenvalues are $-y \pm \sqrt{y² + 4z²}$
• What if we know that $f(x,y,z)=f_1(x,y) + f_2(x,z)$, that is we don't have any multiplication with the variables $y$ and $z$. – user85361 Jan 10 '17 at 16:06
• Then $f$ is jointly concave, since the hessian w.r.t. $y$ and $z$ is a diagonal matrix with elements $d^2 f / dy^2$ and $d^2 f / dz^2$, which are both negative due to the concavity assumption. – LinAlg Jan 10 '17 at 16:15
• I don't understand, how you assume from hessian of f with respect to $y$ and $z$, concavity of function $g(u)$, in the question? – user85361 Jan 10 '17 at 16:27