# Sequential maximization $\max_{x} \max_{y} f(x,y)$ vs. simultaneous maximization $\max_{x,y} f(x,y)$

Let $f(x,y)$ be a non-separable, non-negative real-valued function, that is jointly concave in $x$ and $y$. We want to maximize $f(x,y)$ over $x$ and $y$.

Is the sequential maximization $$\max_{x} \max_{y} f(x,y)$$ always equal to the simultaneous maximization $$\max_{x,\ y} f(x,y)$$ Or is there a simple counter-example for this?

I know that if $f(x,y)$ is separable, then the sequential maximization and simultaneous maximization are equal. Are there any conditions on a non-separable function such that this result still holds?

Any help is much appreciated!

• – A.Γ. Jun 19 '18 at 17:33
• @A.Γ. That's not a duplicate. Technically, this implies the statement you're linking to. – Yanko Jun 19 '18 at 18:46

## 1 Answer

Let $(x_0,y_0)$ be any two elements. Then clearly we have that $f(x_0 ,y_0) \leq\max_y f(x_0 ,y)$

Moreover we notice that $\max_x \max_y f(x,y)$ is the maximum over elements of the form $\max_y f(x',y)$ for some $x'$. In particular this implies that $f(x_0 ,y_0)\leq \max_y f(x_0 ,y)\leq \max _x \max _y f(x,y)$. As $x_0,y_0$ were chosen arbitrarily we may conclude that $\max_{x,y} f(x,y)\leq \max_x\max_y f(x,y)$ and the inequality in the other direction is obvious.

• Thanks! That's very helpful – Wav Jun 20 '18 at 10:53