Simultaneous optimization

Question

Consider a continuous and differentiable function $f(x,y,z)$ with $x,y,z\in \mathbb{R}$. For each $(y,z)$, there exists a unique $x_1(y,z)$ that maximizes $f(x,y,z)$. For each $(x,z)$, there exists a unique $y_1(x,z)$ that maximizes $f(x,y,z)$. $f(x,y,z)$ increases in $z$.

Hypothesis: $\frac{\partial x_1}{\partial z}>0$

I am pretty sure that this Hypothesis is a standard analytical result but somehow I don't see it. Can you hint me towards the Theorem that shows this or sketch a proof?

Do I need $\frac{\partial^2 f(x,y,z)}{\partial x \partial y}>0$ on top of this?

@Jean-ArmandMoroni! showed with below counterexample that the Hypothesis is not true with the above assumptions.

Would it be sufficient to limit $x,y\in \mathbb{R}^+$? Are there any other easy conditions that make my hypotheses correct?

Jean-Armand Moroni · Accepted Answer · 2022-07-28T21:53:01.567

0

EDIT No conclusion can be made on $x_1$. Here is a counterexample.

If $f$ is a function which satisfies the conclusion, i.e.
$\frac {\partial x_1} {\partial z} \gt 0$,
then define $g(x, y, z) = f(-x, y, z)$.

$g$ is continuous and differentiable, and for each $(y, z)$ there exists a unique $x'_1(y, z)$ which maximizes $g(x, y, z)$: this is $x'_1(y, z) = - x_1(y, z)$.
$g(x, y, z)$ increases in $z$.

So $g$ verifies all hypotheses. But $\frac {\partial x'_1} {\partial z} = -\frac {\partial x_1} {\partial z}$, so
$\frac {\partial x'_1} {\partial z} \lt 0$

Below is what I first wrote.

Actually $y$ plays no role in the problem: the reasoning can take place by fixing $y$, and reasoning on $f_y$ which is the restriction of $f$ to a fixed value of $y$:

$x'_{1,y}(z)$, which maximizes $f_y(x, z)$, is the same as $x_1(y, z)$ which maximizes $f(x, y, z)$.
And so $\frac {\partial x'_{1,y}} {\partial z} (x, z) = \frac {\partial x_1} {\partial z} (x, y, z)$.

Then if
$\forall y, x, z_a, z_b$ we have $z_a \lt z_b \Rightarrow f_y(x, z_a) \lt f_y(x, z_b)$,
$\sup_x f_y(x, z_a) \lt \sup_x f_y(x, z_b)$
(the inequality stays strict because the $\sup$ is attained on some $x$)

The rest of the proof was wrong.

edited Jul 28 '22 at 21:53

answered Jul 28 '22 at 20:59

Jean-Armand Moroni

5,632

So $x'_{1,y}$ is increasing. But is it continuous and differentiable? – aschepler Jul 28 '22 at 21:27
@aschepler You are right. And more than that, my proof above is wrong. So I'll delete it. – Jean-Armand Moroni Jul 28 '22 at 21:33
@aschepler Instead of deleting it, I wrote a counterexample; and left the beginning of the previous post, to give some context to your comment. – Jean-Armand Moroni Jul 28 '22 at 21:55
Thank you very much @Jean-ArmandMoroni! Given your counterexample, would it be sufficient to limit $x,y\in \mathbb{R}^+? Do you see any other easy condition that make my hypotheses correct? – Paul Jul 29 '22 at 07:43
Thanks again. I created a follow up question. – Paul Jul 29 '22 at 17:03

Simultaneous optimization

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