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there is a Theorem If x is a local optimum to (G) then there is no feasible descent direction at x.

(G) min f(x)

I want to know if the reverse of theorem is true or not, that means if there is no feasible descent direction at x, then x is local optimum?

user509069
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  • well, its basically the definition of a local minimum $f(x^) \le f(y)$ $\forall y \in U_\epsilon(x^)$. – user160069 Dec 01 '17 at 18:49
  • If you have equality constraints then the set of feasible directions may be empty at every feasible point. – A.Γ. Dec 01 '17 at 22:26
  • Thank you @ A.Γ. That is true but I want to know about unconstrained problem. I asked from my instructor today and the answer is that converse of theorem does not hold. now I want an example. – user509069 Dec 02 '17 at 18:00
  • For unconstrained problem all directions are feasible, and the converse is true. How do you define a feasible descent direction? – A.Γ. Dec 04 '17 at 16:50

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