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Be a concave mutivariate function $f(\textbf{x})=\textbf{y}$. I observed the following conjecture: the maximum value of $f$ is achievable when all entries of $\textbf{x}$ are equal.

How to prove such property or what are the tools used to prove such propoerties?

Thank you in advance

John
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This doesn't work in general, but it is a very useful observation in the particular cases where it does hold.

Starting with why it doesn't work, notice that concavity does not guarantee that a maximum exists - for instance $$f(x,y)=\begin{cases}x &&\text{if }x\leq 0 \\\log(x+1) &&\text{if }x>0\end{cases}$$ is a concave function, but has no maximum. Also, if we take a function like: $$f(x,y)=-x^2-(y-1)^2$$ which is concave, it has a maximum at $(x,y)=(0,1)$, which is not on the line $x=y$ as you conjecture.

However, what you are probably running into is a very important idea in mathematics that will make your life a lot easier. In particular, I suspect that examples you are solving are symmetric in their arguments- that is, the order of the variables doesn't matter, so, for $3$ variables: $$f(x,y,z)=f(y,x,z)=f(x,z,y)=f(z,x,y)=f(y,z,x)=f(z,y,x)$$ or similar expressions in other dimensions. This, alongside concavity, guarantees that every maximum (and local maximum) will have every coordinate equal. This follows from the fact that, for instance in $2$ dimensions, if you thought $(a,b)$ for $a\neq b$ was a maximum, concavity would hold that since $f(a,b)=f(b,a)$, if we took the average of the positions $(a,b)$ and $(b,a)$ to get $\left(\frac{a+b}2,\frac{a+b}2\right)$, the function $f$ would obtain at least the same value as it did at $f(a,b)$ here, thus $(a,b)$ cannot be a strict maximum.

This is a very handy tool to have - if you notice your function is concave and it has this symmetry, then you can immediately know that, to find the maximum, if one exists, you only need to check the case where all coordinates are equal. There are all sorts of other symmetries you might notice that can yield similar simplifications (for instance, reflections over planes like $f(x,y,z)=f(-x,y,z)$, which, alongside concavity, would imply $x=0$ for any maximum).

Milo Brandt
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    Thank you for your anwser. Actually I have an opimization problem and after some simulations I noticed that the max occurs when all components are equal. My objective is actually symmetric in the way you mentionned, but taken into account the equality contraints (Ax=b), the variables are no more symmetric – John Dec 10 '14 at 00:53
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    If the constraints have a similar symmetry it still could work; for instance, maximizing $x+y+z$ over the unit sphere (i.e. $x^2+y^2+z^2=1$) yields $x=y=z$ since if $(x,y,z)$ is on the sphere so is any permutation of the coordinates - but you say that's not the case. I can't think of any other general reason why the coordinates would be equal (though there might be some symmetry you haven't noticed). What exactly is the optimization problem you're thinking on? – Milo Brandt Dec 10 '14 at 00:56
  • $max \ f(x) subject \ to: Ax=b, a<x<b $.

    The product $Ax=b$ makes the components of $x$ not symmetric since for a solution $x_0$, inverting components in $x_0$ will not satisfy necessarily the equality constraint...

     – John Dec 10 '14 at 01:02