# Why we can assume $f(x)+g(x)=f(y)+g(y)=1$ in sum of log convex functions is log convex proof?

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

• 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. Mar 1, 2019 at 1:19

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 $$b + d$$.

• Thank you very much, nice explanation!! Mar 1, 2019 at 1:43
• 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. Mar 1, 2019 at 1:44
• I appreciate it. I had no clue that the reason was the homogeneity. Thanks so much for illuminating me :-) Mar 1, 2019 at 9:58