 Solution

For the same multiplier convention, the Lagrangian dual problem isFor every feasible $x$ and every $\lambda$, the equality constraint implies $q(\lambda )\le L(x,\lambda )=f(x)$. Taking first the infimum over feasible $x$ and then the supremum over $\lambda$ proves weak duality:The inequalities remain meaningful with the usual extended-value conventions, including $q(\lambda )=-\infty$ when the Lagrangian is unbounded below. Equality of optimum values, or their attainment, does not follow from weak duality. If the sufficient pair in part (a) exists, it attains equality, since $q({\lambda }^{\ast })=L(x^{\ast },{\lambda }^{\ast })=f(x^{\ast })$.