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Suppose

(i) $G:\mathbb{R}^n\to\mathbb{R}$ is a $C^{1}$ function; and that

(ii) $\frac{G(x)}{|x|}\to\infty$ as $|x|\to\infty$; and that

(iii) $\nabla G = 0 $ has precisely one solution (called $x_{0}$).

My question is, do (i) - (iii) imply that $x_{0}$ is a strict global minimum of $G$?

I think this is true, but how do I prove it?

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1 Answer 1

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I think hypothesis (ii) can be weakened as $G(x)\to+\infty$ as $|x|\to+\infty$. In fact this implies that for every $M\in\mathbb R$ the function $G(x)$ is greater than $M$ out of a certain ball $B(0,r)$ with $r=r(M)$.

Let $M$ be big enough so that $G(x)\leq M$ has non-empty solution, and consider $r=r(M)$. Then $G(x)>M$ out of $B(0,r)$; since $\overline{B(0,r)}$ is a compact set, $G$ restricted to this closed ball has a global maximum and a global minimum. Note that the global minimum is at most $M$, therefore it is a global minimum of $G(x)$ on $\mathbb R^n$ too (remember that $G(x)$ is greater than $M$ out of the ball).

Since $\mathbb R^n$ is open, a global minimum is also a local minimum and consequently - by hypothesis (i) - the gradient is the null vector on it. By uniqueness stated in (iii), it follows that $G(x)$ has a global minimum in $x_0$ and nowhere else.

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