I've got this exercise for uni that I've been absolutely racking my brain over for the past couple hours, the problem goes as follows:
Let $\theta_1, \cdots, \theta_d \in (0, \infty)$. Consider the optimization problem $$\max_{w \in \mathbb{R}^d}\sum_{i=1}^d \log{(\theta_i + w_i)} \ \ \text{ s.t. } \ w_i \geq 0, \text{for }i=1, \cdots, d, \ \sum_{i=1}^dw_i = 1$$ a) show that the problem is (equivalent to) a convex optimization problem
b) Show that the solution $w^*$ of this problem is given by $$w^*_i = \max\bigg\{0, \frac{1}{\lambda^* - \theta_i}\bigg\}$$ where $\lambda^*$ is uniquely determined by $$\sum_{i=1}^d \max\bigg\{0, \frac{1}{\lambda^* - \theta_i}\bigg\} = 1$$
The answer to (a) seemed quite simple to me, as $\log$ is concave, the new problem would be minimising the negative objective function from above. But (b) is where I am struggling severely.
It would seem logical to me to use the KKT-theorem here. Writing out the KKT-conditions for the problem resulting from (a) problem gives us, for all $i=1, \cdots, d$:
$$-\frac{1}{\theta_i + w^*_i} - \lambda^* - \mu_i^* = 0$$ $$\mu^*_iw^*_i = 0$$
Multiplying both sides of the first equation with $(\theta_i + w^*_i)$ gives $$-1 - (\theta_i + w^*_i)\lambda^* - (\theta_i + w^*_i)\mu_i^* = 0$$ $$\mu^*_iw^*_i = 0$$ Filling in our definition of $w_i^*$, and applying our second condition gives $$-1 - \max\Big\{\theta_i, \frac{1}{\lambda^*}\Big\}\lambda^* - \theta_i\mu_i^* = 0$$ $$\mu^*_i\max\Big\{0, \frac{1}{\lambda^* - \theta_i}\Big\} = 0$$
Which is in turn equivalent to: $$-1 - \max\Big\{\theta_i\lambda^*, 1\Big\} - \theta_i\mu_i^* = 0$$ $$\mu^*_i\max\Big\{0, \frac{1}{\lambda^* - \theta_i}\Big\} = 0$$
But now, looking at the first condition, we have two negative numbers being subtraced by a nonnegative number, which should combine to equal zero. I just don't see where I am mistaken. Any help with this would be greatly appreciated!! Thanks in advance.
We have $\mu^*_i \geq 0$, but no restriction on $\lambda^*$. So if $\lambda^*$ is negative, then $\max\Big\{\theta_i, \frac{1}{\lambda^*}\Big\}= \theta_i$, and therefore $\max\Big\{\theta_i, \frac{1}{\lambda^*}\Big\}\lambda^*=\theta_i \lambda^*$