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So I was trying to prove that the sum of two log convex functions is also log convex. In my attemp I was able to conclude that if

$$f(x)^{1-t} f(y)^t + g(x)^{1-t} g(y)^t \leq (f(x)+g(x))^{1-t} (f(y)+g(y))^t $$

holds for any $t \in [0,1]$ then the proof is completed. I found that another user has asked how to prove the same question as me. This is the question How to prove that the sum of two log-convex functions is log-convex?

The first answer by the user Xiang Yu gets to the same point as me, and then he assume that $f(x)+g(x)=f(y)+g(y)=1$ and then continues the proof. Now I can´t figure out why we can assume this. I was trying to prove that this is indeed the worst case for the inequality to hold, but I am not sure this is true.

Can anybody explain me why we can assume $f(x)+g(x)=f(y)+g(y)=1$ , or maybe tell me what other options are to conclude the proof.

Thank you

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    $\begingroup$ because the problem is homogeneous, so you can always multiply a positive number to f(x) and f(y) without changing the problem's nature. $\endgroup$
    – Yimin
    Mar 1, 2019 at 1:19

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I believe the relevant part of the answer in the other question the OP is referring to is

Set $a=f(x),b=f(y),c=g(x),d=g(y)$, then we need to show $$a^\theta b^{1-\theta}+c^\theta d^{1-\theta}\leq (a+c)^\theta(b+d)^{1-\theta}.$$ By dividing $(a+c)^\theta$ and $(b+d)^{1-\theta}$ on both sides, we may assume that $a+c=b+d=1$.

As suggested, dividing both sides by $(a+c)^\theta (b+d)^{1-\theta}$ gives

$$\cfrac{a^\theta b^{1-\theta}+c^\theta d^{1-\theta}}{(a+c)^\theta (b+d)^{1-\theta}} \leq 1 \tag{1}\label{eq1}$$

The question's comment by Yimin refers to "the problem is homogeneous". I believe this means that if, using the question variables, consider $a + c = k$ for some $k$, then let $a = ka_1$ and $c = kc_1$ to get, if $k \neq 0$, that $a_1 + c_1 = 1$. Substituting the redefined $a$ and $c$ values into the LHS of \eqref{eq1} gives

\begin{align} \cfrac{a^\theta b^{1-\theta}+c^\theta d^{1-\theta}}{(a+c)^\theta (b+d)^{1-\theta}} & = \cfrac{\left(ka_1\right)^\theta b^{1-\theta}+\left(kc_1\right)^\theta d^{1-\theta}}{({ka_1}+{kc_1})^\theta (b+d)^{1-\theta}} \\ & = \cfrac{k^\theta a_1^\theta b^{1-\theta}+ k^\theta c_1^\theta d^{1-\theta}}{k^\theta(a_1 + c_1)^\theta (b+d)^{1-\theta}} \\ & = \cfrac{a_1^\theta b^{1-\theta}+ c_1^\theta d^{1-\theta}}{(a_1 + c_1)^\theta (b+d)^{1-\theta}} \tag{2}\label{eq2} \end{align}

As such, $a_1 + c_1 = 1$ also works in the inequality where there is just a relabeling of $a$ to $a_1$ and $c$ to $c_1$. You can do a similar thing for $b + d$.

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  • $\begingroup$ Thank you very much, nice explanation!! $\endgroup$
    – JoseSquare
    Mar 1, 2019 at 1:43
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    $\begingroup$ You are welcome. Note this is not my area of expertise, so it wasn't immediately obvious to me either & took me a while to figure it out. $\endgroup$ Mar 1, 2019 at 1:44
  • $\begingroup$ I appreciate it. I had no clue that the reason was the homogeneity. Thanks so much for illuminating me :-) $\endgroup$
    – JoseSquare
    Mar 1, 2019 at 9:58

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