Consider the optimization problem
f(x) = infimum $(−x^2) s.t. 0 ≤ x ≤ 1$.
What will be its dual in simplified closed form with no 'inf'.
I know its Lagrangian function will be $ L(x,) = -x^2 +x , ≥0$. But I am not sure wether its $$ minimize L(x,) s.t ≥0, (dL/d)=0 is its dual.
The minimum of the Lagrangian of the given problem with respect to $x$ is $-\infty$. I guess then the point behind this question is that the dual of a nonlinear problem which attains its optimal value is not necessarily even well-defined. Note that this does not happen in the case of linear programs and if one of the problems attains its optimal, the other one also does. But in case you have: $$\min\ x^2 \quad \text{ subject to } \quad {0\le x\le1}.$$ The Lagrangian is $L(x;\lambda_1,\lambda_2)=x^2-\lambda_1x+\lambda_2(x-1)=x^2-x(\lambda_1-\lambda_2)-\lambda_2$. Thus, we have $$\min_{x\in\mathbb R} L(x;\lambda_1,\lambda_2)= \left\{ \begin{array}{ll} -1/4\lambda_1^2-1/4\lambda_2^2+1/2\lambda_1\lambda_2-\lambda_2 & \mbox{if $2x-\lambda_1+\lambda_2=0$};\\ -\infty & \mbox{if $2x-\lambda_1+\lambda_2\ne 0$}.\end{array} \right. $$ Hence, the dual is $$ \max \ -1/4\lambda_1^2-1/4\lambda_2^2+1/2\lambda_1\lambda_2-\lambda_2 \quad \text{ subject to } \quad {\lambda_1,\lambda_2\ge 0}, $$ which is a concave program.