Consider the problem: minimize $ x^3$ subject to $x^2 \le 0, x \in \Bbb R$
I want to show the Lagrange Dual Problem is infeasible (i.e. $d^* = -\infty$)
We have $ L (x, \lambda) = x^3 + \lambda x^2 $
Then $g(\lambda) = \inf(L(x,\lambda)) = \inf( x^3 + \lambda x^2)$
$\frac{d}{dx}(x^3 + \lambda x^2) = 0 \Leftrightarrow 3x^2 + 2 \lambda x =0 \Leftrightarrow x = \{ 0, -\frac23 \lambda \}$
Hence, plugging in the values we found into $g$ and varying $x$ we have that $g(\lambda) = - \infty $ if $\lambda =0$ and $g(\lambda) = \frac{4 \lambda ^3}{27} $ if $\lambda >0$
The dual is then: $ \max g(\lambda)$ subject to $\lambda \ge 0$
So the problem is infeasible because when $\lambda = 0 $ we have $d^* = -\infty $ and so constraint cannot be attained
Is this approach correct?