I have a minimization problem with some function $f(x)$ and constraints:
$x_1^2+x_2^2+x_3^2\leq1$
$x_1\geq0$
$x_2\geq0$
$x_3\geq0$
I need to find the set of feasible directions $P$ at $a=(0,1,0)$
Ok, so from the definition of feasible directions set I know that a point $(a+\delta p)$ where $\delta$ is small and positive must be in the set of constraints. From the first inequality:
$(\delta p_1)^2+(1+\delta p_2)^2+(\delta p_3)^2\leq1$
$(\delta p_1)^2+1+2\delta p_2+(\delta p_2)^2+(\delta p_3)^2\leq1$
$(\delta p_1)^2+2\delta p_2+(\delta p_2)^2+(\delta p_3)^2\leq0$
All squared elements are always greater than 0 so:
$\delta p_2\leq0$
$p_2\leq0$
From another inequalities it's clear that $p_1\geq0$ and $p_3\geq0$
$1+\delta p_2\geq0$
$\delta p_2\geq-1$
$p_2\geq\frac{-1}{\delta}$
And now I'm stuck. According to the solution $p_2<0$ but I can't figure out how to achieve this. Did I do anything wrong?
What you did is correct. But there is more you can get from the inequality $$ (\delta p_1)^2 + 2\delta p_2 + (\delta p_2)^2 + (\delta p_3)^2 \leq 0 $$ This is equivalent to $$ \delta^2 p_1^2 + 2\delta p_2 + \delta^2 p_2^2 + \delta^2 p_3^2 \leq 0 $$ or $$ 2p_2 + \delta (p_1^2 + p_2^2 + p_3^2) \leq 0 $$ or $$ p_2 \leq -\frac{1}{2} \delta (p_1^2 + p_2^2 + p_3^2) $$ We know that $\delta$ is positive. $p_1^2 + p_2^2 + p_3^2$ is also positive, because the direction $p$ must not be the $0$ vector. (If $p$ was $0,$ then $a+\delta p$ wouldn't take us away from $a.$) Therefore $p_2<0.$ If $p$ was meant to be a unit vector anyway, then the equation simplifies to $p_2\leq -\frac{1}{2}\delta$.