Let $f$ and $g$ be two strictly increasing functions such that $%
f:[0,1]\rightarrow \mathbb{R}_{+}$ (with $f(0)=0$) and $g:[2,\infty )\rightarrow \mathbb{R}_{+}$.
Can anyone come up with sufficient conditions that will ensure (i) and (ii)
always hold whenever $0<x_{1}$, $0<x_{2}$, and
$x_{1}+x_{2}\leq 1$, respectively?
\begin{equation*}
\text{(i) }f(x_{1})g(y_{1})+f(x_{2})g(y_{2})\leq f(x_{1}+x_{2})g(y_{1}+y_{2})%
\text{.}
\end{equation*}
\begin{equation*}
\text{(ii) }f(x_{1})g(y_{1})+f(x_{2})g(y_{2})\geq
f(x_{1}+x_{2})g(y_{1}+y_{2})\text{.}
\end{equation*}
One thing I can see is that since both functions are strictly increasing,
(i) is always true if
\begin{equation*}
\text{(a) }f(x_{1})+f(x_{2})\leq f(x_{1}+x_{2})
\end{equation*}
or
\begin{equation*}
\text{(b) }g(y_{1})+g(y_{2})\leq g(y_{1}+y_{2})
\end{equation*}
Is always true. But, this begs the question of what conditions guarantee (a)
and (b)?
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MathGuy
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Do you have any guarantee that $x_1+x_2$ is less than $1$? – Andrei Feb 28 '22 at 21:38
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Your first observation is wrong. Let $f(x)=x-0.5$. Then $f(x_1)+f(x_2)=x_1+x_2-1<x_1+x_2-0.5=f(x_1+x_2)$. I will choose $x_1=0.2, x_2=0.4$ so I have $f(x_1)=-0.3, f(x_2)=-0.1, f(x_1+x_2)=0.1$. Now let me choose $g(y)$ strictly increasing, but negative for certain $y$ values, say $g(y)=y-100$. $g(3)=-97, g(5)=-95, g(3+5)=-92$. Notice that the left hand side is positive, the right hand side is negative – Andrei Feb 28 '22 at 21:43
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@Andrei First of all, thank you so much for your comments. Yes, indeed you are right. I have now revised my question: (1) f(x)>=0 and g(x)>=0 for all x in their domains. (2) x1+x2<=1. (3) it can be assumed that f(0)=0. I will appreciate any further comments you might have. – MathGuy Mar 01 '22 at 09:53
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There is no guarantee, even with these restrictions. I don't think (I did not try to prove it) that there is any condition that you can put on $f$ or $g$ only, to guarantee one of the inequality. For example, choose $f(x)=x$. Then, for $x_1=x_2$, you get $g(y-1)+g(y_2)\le g(y_1+y_2)$. If (a) and(b) are simultaneously satisfied, then (i) is true. – Andrei Mar 01 '22 at 15:38
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@Andrei Thank you again, Andrei. – MathGuy Mar 01 '22 at 18:47