Let $f(x,y)$ be a real function of the variables $x,y$ (which can be real vectors). Under what conditions do we have the following equality:

$$\min_x \max_yf(x,y) = \min_y \max_x f(x,y)$$

For example, this equality is true if $f(x,y) = xy$ and $x,y$ are real scalars.

Note that this is not the same as Von Neumann's minimax theorem (https://en.wikipedia.org/wiki/Minimax_theorem), because here the role of the variables is exchanged (e.g., $x$ is minimized on the left-hand side, but it is maximized on the right-hand side).

Though I do not know if convexity/concavity of $f(x,y)$ with respect to either arguments plays a role here (like it does for Von Neumann's minimax), I am using the convex-related tags here since that's the context where I've seen related questions. Similarly I am tagging game-theory, though I'm not sure it's directly applicable.

I also expect that the a condition for this equality to be true is that the saddle-points of both sides of the equation be attained at points where the gradient of $f$ vanishes (see my answer below).

  • $\begingroup$ Also posted here: math.stackexchange.com/q/3592868/10063 $\endgroup$
    – becko
    Mar 24, 2020 at 13:35
  • $\begingroup$ Why the close vote? $\endgroup$
    – becko
    Mar 24, 2020 at 19:39
  • $\begingroup$ It's a slightly strange question, since for example the equality wouldn't even hold for $ax+by$ in $[0,1]\times [0,1]$ (unlike the Minimax Theorem). Any particular reason why you are interested in this? $\endgroup$ Mar 26, 2020 at 12:59
  • $\begingroup$ @YaakovBaruch I have a variational formulation of a certain statistical mechanics problem, where I derive an optimization like one side of this equation, but where it would make a lot of sense physically that the optimization were like the other side. I need to understand when both sides are equal, because if they are not it would be potentially interesting. Numerically I find they typically are equal (but this could be because I am only looking at the points where the derivative are zero). $\endgroup$
    – becko
    Mar 26, 2020 at 14:01
  • $\begingroup$ In your example $ax+by$, the saddle occurs at the boundary of the domain, where the gradient of $f$ need not be zero. I do not expect my equality to hold in that case. $\endgroup$
    – becko
    Mar 26, 2020 at 14:03

1 Answer 1


Here is a tentative proof, under some assumptions. Would like to see some additional arguments (or counterexamples) to make this clearer.


  1. We suppose that $f(x,y)$ is concave in $y$ for all $x$.
  2. That the equation $\partial f/\partial y=0$ has a unique solution $x$ for every $y$.
  3. That the solution to both saddle-point optimizations is attained at a point where $\partial f/\partial x = \partial f/\partial y = 0$.


Then finding $\max_y f(x,y)$ is equivalent to $\partial f/\partial y = 0$. It follows that the original problem is equivalent to a constrained minimization over all variables:

$$\begin{aligned} \min_x \max_y f(x,y) &= \min_{x,y} f(x,y) \quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right) \\ &= \min_y \min_x f(x,y) \quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right) \end{aligned}$$

where we simply changed the order of the minimizations. Since there is a unique $x$ that makes $\partial f/\partial y=0$ for any $y$, the inner minimization on $x$ is trivial, and can be formally changed to a maximization:

$$\begin{aligned} \min_x \max_y f(x,y) &= \min_y \max_x f(x,y) \quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right) \end{aligned}$$

Finally, if the unconstrained version of the right-hand side optimization is solved at a point where $\partial f/\partial y=0$ (this assumption 3 above), then the constrain can be ignored:

$$\min_y \max_x f(x,y) = \min_y \max_x f(x,y) \quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right)$$

In this case, we obtain:

$$\min_x \max_y f(x,y) = \min_y \max_x f(x,y)$$

as desired.


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