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I was given a question to apply the following corollary to the function $f(x)=x^3$ where $f: \mathbb{R}^1 \rightarrow \mathbb{R}^1$ and S is the set of nonnegative numbers $\{ x \mid x \geq 0 \}$.

The corollary is: Suppose the feasible region is the nonnegative orthant, that is $ S = \{ x \in \mathbb{R}^n \mid x \geq 0\}$ and f is continuouosly differentiable.

(a) If $\bar{x}$ is a global solution, then $\nabla f(\bar{x}) \geq 0, \bar{x} \geq 0$ and $\nabla f(\bar{x})'(\bar{x}) = 0$

(b) If f is convex and $\nabla(\bar{x}) \geq 0, x \geq 0, \text{ and } \nabla f(\bar{x}))\bar{x}) = 0$, $\bar{x}$ is a global solution.

In this case, what do I pick for $\bar{x}$ in order to prove that the corollary applies to $f(x) = x^3$? As for (b), I'm guessing I would have to prove convexity of $f(x) = x^3$ in addition to picking $\bar{x}$ such that $f(\bar{x}) \leq f(x)$?

Michael Hardy
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Nikitau
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1 Answers1

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Hints:

(a) Since $x \geq 0, x^3 \geq 0$, can you think of an $\bar{x}$ such that $\bar{x}^3=0$? That must be the global solution that you want to pick.

(b) Compute $\nabla f$ and verify it satisfies those conditions. Also, compute the second derivative of $f$. If the second derivative is nonnegative, what can you conclude?

Siong Thye Goh
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  • Ah I'm feeling silly now. So a complete proof would be: Let $\bar{x} =0$. Then clearly $\nabla f(\bar{x}) = 3\bar{x}^2 = 0$, satisfying the first condition in (a). And trivially, $\nabla f(\bar{x})^{T} \bar{x} = 0$. Naturally, this also proves (b). And so the proposition holds. – Nikitau Apr 23 '17 at 21:36