Choose the $w_i>0$ such that $\sum_{i=1}^dw_ia_i$ is minimized and $\sum_{i=1}^dw_i=1$

Question

Say $w,a\in\mathbb R^d$. How can we choose the $w_i$ such that they minimize $$\sum_{i=1}^dw_ia_i\tag1$$ and satisfy $$w_i>0\tag2$$ $$\sum_{i=1}^dw_i=1?\tag3$$ Usually I would solve $$\frac{\rm d}{{\rm d}w_i}\left[\sum_{i=1}^dw_ia_i+\alpha\left(\sum_{i=1}^dw_i-1\right)\right]=0\tag4$$ for that (Lagrange multiplier), but this doesn't help here, since the highest order to $w_i$ is $1$ ... So, how do we find the answer here?

Answer 1

If we find the smallest $a_i$ and set that $w_i$'s weight to 1, that should satisfy the constraint and minimize $\sum_{i=1}^dw_ia_i$. There is no reason to give weight to any $a_i$ than the minimal under this formulation, as this would only increase the sum.

Mathematically, I believe this is because $\sum_{i=1}^dw_ia_i$ is a convex function under the constraints, so the minimizer will be one of the extreme points where only one $a_i$ has weight.

Answer 2

I have an answer to a very similar question, assuming also that all the $a_i\geq 0.$

Say $w,a\in\mathbb R^d$. How can we choose the $w_i$ such that they minimize $$\sum_{i=1}^dw_ia_i\tag1$$ and satisfy $$w_i\geq 0\tag2$$ $$\sum_{i=1}^dw_i=1?\tag3$$

Suppose WLOG that $a_1 \leq a_2 \leq \ldots \leq a_d.$

Since $ a_j\geq a_1,\ \forall j\in \{2,3,\ldots,d\},\ $ it follows that, for any $w_i$ satisfying $(2)$ and $(3)$,

$$ a_1w_1 + a_2 w_2 + \ldots + a_d w_d \geq a_1 w_1 + a_1 w_2 + \ldots + a_1 w_d = a_1 \sum_{i=1}^{d} w_i$$

$$= a_1\cdot 1 = a_1\cdot 1 + a_2 \cdot 0 + a_3\cdot 0 + \ldots + a_d\cdot 0. $$

So the minimum is $a_1.$

I think $a_1$ will be the infimum of $\displaystyle\sum_{i=1}^dw_ia_i$ if we change $(2)$ to $w_i > 0.$