We have a simple problem, namely minimizing: $$f(x) = x_1^2 + x_2^2 - x_1.$$

The gradient is $$\nabla f(x) = \begin{bmatrix} 2x_1 - 1 \\ 2x_2 \end{bmatrix},$$ hence the unique stationary point is: $x_* = (\frac{1}{2}, 0)$. Now, I have a simple introduction to OR level question asking me why this is globally optimal.

My typical answer is to show that $f(x)$ is convex, e.g.,: $$ \nabla^2 f(x) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \succ 0,$$ therefore a local minimizer is a global minimizer.

However, in some engineering course-notes, the proof of global optimality is:

It is also a global minimizer since the function $f$ is $C^1$ and radially unbounded, therefore the global minimizer is a stationary point.

  1. What is $C^1$?

  2. What does radial unboundedness imply here?

Can anyone, please, enlighten me?

Edit: I think they are using the fact that $\lim\limits_{k \to \infty} \|x_k\|=\infty \implies \lim\limits_{k \to \infty} f(x_k) = \infty$. Hence, all level sets of $f$ are compact. Hence the global minimum exists in $\mathbb{R}^2$. And the unique stationary point is also globally optimum. But the main question is,

  1. Why not show convexity and go in this way?
  1. The notation $C^1$ means $f'$ is continuous (on $\Bbb R$ as the interval is not stated). In general $C^k(a,b]$ means that all of $f',f'',\cdots,f^{(k)}$ are continuous on $(a,b]$.

  2. You are correct that radial unboundedness means that $f\to\infty$ as $\|x\|\to\infty$. This method is essentially that for Lyapunov stability.

  3. Ahmadi and Jungers (2018)1 proved that if a function $f$ satisfies $f(0)=0$ and is positive for all $x_i\ne0$ then its convexity implies its radial unboundedness. However, notice that $f$ is not always positive for all $x_1,x_2\ne0$ (choose $x_1=1/2$ and any $|x_2|<x_1$) so the second criterion is not satisfied.


[1] Ahmadi, A. A., Jungers, R. M. (2018). SOS-Convex Lyapunov Functions and Stability of Difference Inclusions. CoRR abs/1803.02070.

  • $\begingroup$ Thanks for your answer. Do you see any reason why this condition is preferred over convexity? $\endgroup$ – independentvariable Apr 17 '20 at 21:15
  • $\begingroup$ I mean in general, we can say that the function is convex so (0.5, 0) is a global minimizer, right? $\endgroup$ – independentvariable Apr 17 '20 at 21:27
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    $\begingroup$ Yes, I suppose the solution provided is alternative. Of course the other condition required is that the domains of $\boldsymbol x$ are convex which is clearly the case here. $\endgroup$ – TheSimpliFire Apr 18 '20 at 9:53

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