I have some problem with problem 3.23 in Luenberger book "Optimization by vector space methods".
Consider the problem of finding the vector $x$ of minimum norm satisfying
$(x|y_i) \geq c_i, i=1,2,\dots n$
where the $y_i$'s are linearly independent.
(a) Show that this problem has a unique solution.
This part has already been proved but I have more issues with.
(b) Show that a necessary and sufficient condition that
$x = \sum_{i=1}^n a_i y_i$
be the solution is that the vector $a$ with component $a_i$ satisfy
$G'a \geq c$ $a \geq 0$
and that $a_i = 0$ if $(x|y_i) > c_i$. $G$ is the Gram matrix of $\{y_i, y_2, \dots y_n\}$
I want to use the projection theorem but has problem how to solve it.
Thanks!
Here is a starting point (one direction):
if $x = \sum_{i=1}^n a_i y_i$, then it is straightforward that $G'a \geq c$. What remains to show is that if $x$ is the minimum then $a \geq 0$, and $a_i = 0$ if $(x|y_i) > c_i.$
Take $v \in \{y_2, \dots y_n\}^{\perp}$, $v \neq 0$. Now, $(x+\lambda v|y_1) = (x|y_1) +\lambda (v|y_1).$ WLOG, we can assume that $(v|y_1) >0.$ Then $(x+\lambda v|y_1) \geq c_1$ for $\lambda > 0$. Also, by definition of $v$, $(x+\lambda v|y_i) \geq c_i, i \geq 2.$ So $x+\lambda v$ is also a solution. Now, as $x$ is the minimum norm solution, $\|x+\lambda v\| \geq \|x\|,$ or $\lambda^2 \|v\|^2 + \lambda (x,v) \geq0.$ Divide by $\lambda$, and let $\lambda$ goes to $0$ to get $(x|v) \geq 0, $ which is $a_1(y_1|v) \geq 0, $ or $a_1 \geq 0$.
Now, if $(x|y_1) > c_1$, then for $\lambda <0$, and small enough, we would also have $x+\lambda v$ a solution and the final steps above would give us $a_1 \leq 0.$