# Are minimax and maximin condition interchangable?

I've came across a classic problem in my field where

$$\min_h \max_{\delta,\omega} |h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$$ where $h$ ( a set of coefficients), $\omega$, and $\delta$ are independent variables.

My question here is that whether or not this 'minimax' problem can be rewritten/reformulated as a 'maximin' problem?

Such that $$\max_{\delta,\omega} \min_h |h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$$

Will that make sense? Sorry if this is a silly question, but I tried googling it and wiki it. Doesn't really answer this general question!

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Please: Don't write "\underset{\delta,\omega}{min}\:" Instead, write \min_{\delta,\omega}. When in "displayed" rather than "inline" mode that automatically puts the subscript directly under "min". It also does two other things, regardless of whether it's inline or displayed: it causes proper spacing to appear between "min" and what follows or precedes it, and it prevents italicization. It's standard usage. – Michael Hardy Feb 20 '12 at 18:26
Thank you Michael! Noted. Still quite noob with latex maths format. – JuniorEngie Feb 20 '12 at 23:59
See this question: math.stackexchange.com/questions/840299/… – EthanAlvaree Apr 3 at 21:50

There is an approach that addresses the problem in terms of interchange of limits on double sequences. The sufficient condition is the uniform convergence with respect to a parameter of the limits.

Fixed $h$ exists a sequence $(\delta_k,\omega_k)\to (\delta^*,\omega^*)$ shout that $$|h^TP{(\delta^{*},\omega^{*})}-R_d(\delta^*,\omega^*)| = \max_{\delta,\omega} |h^TP{(\delta,\omega)}-R_d(\delta,\omega)|.$$ Now, there is a sequence $h_k\to h_*$ shout that

$$|h_*^TP{(\delta^*,\omega^*)}-R_d(\delta^*,\omega^*)| = \min_h \max_{\delta,\omega}|h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$$

Let's $F_h(\delta,\omega)=|h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$. Now note that

$$\lim_{h\to h^*}\lim_{(\delta,\omega)\to(\delta^*,\omega^*)}F_h(\delta,\omega)= \min_h \max_{\delta,\omega} |h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$$

and

$$\lim_{(\delta,\omega)\to(\delta^*,\omega^*)}\lim_{h\to h^*}F_h(\delta,\omega)= \max_{\delta,\omega} \min_h |h^TP{(\delta,\omega)}-R_d(\delta,\omega)|$$

Then it applies the following theorem with $t = h$ and $x = (\delta, \omega)$

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality

Proof. See Zoric. P. 381.

Within the limits of the above theorem replace $\lim_{\mathcal{B}_{T}}$ by $\lim_{h\to h^*}$ and $\lim_{\mathcal{B}_{X}}$ by $\lim_{(\delta,\omega)\to(\delta_*,\omega_*)}$.

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Please: Don't write "\underset{\delta,\omega}{min}\:" Instead, write \min_{\delta,\omega}. When in "displayed" rather than "inline" mode that automatically puts the subscript directly under "min". It also does two other things, regardless of whether it's inline or displayed: it causes proper spacing to appear between "min" and what follows or precedes it, and it prevents italicization. It's standard usage. – Michael Hardy Feb 20 '12 at 18:30
Thank you Elias! The explanation was simple and easily understood. However just wondering if the above condition have any restrictions on the properties of $P(\delta,\omega)$ and $R_d(\delta,\omega)$? As I want to see if I am have a solid prove that this limiting evaluation can indeed work for the exchange between max and min, vice-versa. Thanks! – JuniorEngie Feb 21 '12 at 8:17
The theorem only requires the convergence of the uniform limit with respect to the first argument of the second limit. Or uniform convergence second limit with respect to the argument of the first limit. However the more regularly you have but it is easy to prove the uniform convergence. – MathOverview Feb 21 '12 at 12:29
If $R$ and $P$ are continuous try to prove that $F(\delta, \omega)$ is a family of functions uniformimente equicontinuas with respect to the parameter $h$ for the convertgencia uniform. If $R$ and $Q$ have continuous partial derivatives continuous attempt to prove, using the mean value theorem, which $F_h (\delta, \omega)$ is a family of functions uniformimente Lipchtiz with respect to the parameter $h$. That is, the constant Lipchtiz does not depend on $h$. – MathOverview Feb 21 '12 at 12:29