Multivariate convex optimization problem involving logarithms 2

Question

This is an extension to previous question in Multivariate convex optimization problem involving logarithms. Thanks a lot to David M. for the answer to there. Now, I like to extend the question a little more to include solutions with $a_i=0$ or $b_i=0$.

$$\min_{a, b} \sum_{i=1}^K b_i I(a_i>0, b_i>0)f(\frac{a_i}{b_i}) $$

s.t. $$ f(x) = (1+x) \log(1+x) -\log(x) - (1+x) \log(2)$$

$$ \sum_{i=1}^K a_i = 1.$$

$$ \sum_{i=1}^K b_i = 1.$$

$$ a,b\geq 0. $$

$I(a_i>0, b_i>0)$ is indicator function i.e. it is 1 when ($a_i>0$ and $b_i>0$) and $0$ otherwise.

I believe the answer to the question is the set of (a,b) s.t. $\sum_{i=1}^K a_i = \sum_{i=1}^K b_i = 1$ and $a_i=b_i$ for all $i$. I just wanted to make sure if I am correct.

Answer 1

Yes I think you're correct. Note that your function $f$ satisfies $f(t)\geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)\geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $a\equiv{b}$.

You can just eliminate all the edge cases explicitly:

1. If $a_i=0$ and $b_i>0$, then $b_i\cdot f(a_i/b_i)=b_i\cdot f(0)$, which isn't defined. However, $\lim_{t\to0}f(t)=\infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=\infty$, which obviously isn't the minimizer.
2. If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that $$ 0\cdot f(a_i/0)=\lim_{t\to0}t\cdot f\big(\frac{a_i}{t}\big)=\infty, $$ again obviously not the minimizer.
3. Finally, if $a_i=b_i=0$, then $b_i\cdot f(a_i/b_i)=0\cdot f(0/0)$. Again, doing the sensible thing, $$ 0\cdot f(0/0)=\lim_{t\to0}t\cdot{f\big(\frac{t}{t}\big)}=f(1)\cdot\lim_{t\to0}t=0. $$

Hence, the optimal set is indeed $\{a,b\geq0\;|\;a\equiv{b},{1}^\text{T}a=1\}$.