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Positive real vectors $x,y,x',y'\in\mathbb R^2_+$; $f,g$ are multivariate functions: $f,g:\mathbb R^2\to\mathbb R^2$. Matrix $A\in\mathbb R^{2\times2}$.

Given that $y=Ax$, $y'=Ax'$, $A^Tg(y)=f(x)$, and $(x-x')\cdot(f(x)-f(x'))\leq0$

Prove that $(y-y')\cdot(g(y)-g(y'))\leq 0$.

The first step is obviously $(x-x')(f(x)-f(x'))\leq0$ implies $(x-x')\cdot A^T(g(y)-g(y'))\leq0.$

If we furhter assume the invertability of $A$, then$$A^{-1}(y-y')\cdot A^T(g(y)-g(y'))\leq0.$$

It seems that it cannot be further simplified.....

High GPA
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    The only thing left after the first step is to notice that $u\cdot A^Tv=Au\cdot v$ for any (real) vectors $u,v$ and (real) matrix $A$. – A.Γ. Jul 28 '19 at 16:17

1 Answers1

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Note that $$(A^{-1} a) \cdot (A^Tb) = (A^Tb)^T(A^{-1}a) = b^TAA^{-1}a = b^Ta = a \cdot b.$$ This completes your proof.

Theo Bendit
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    Readers new to matrices are advised there's a subtle abuse of notation in this calculation, which we've discussed before here. It makes the algebra more precise without really being harmful. – J.G. Jul 28 '19 at 15:54