0
$\begingroup$

I am having a problem with one aspect of the following proof I came across in "An Easy Path to Convex Analysis and Applications" by Mordukhovich and Nam.

It is Proposition 3.9 and it is the line marked by * that is giving me troubles

Proposition 3.9 $\;\;\;\;\;\;\;$Let $a_1,\ldots,a_m$ be elements of $\mathbb{R}^n.$ Then the convex cone $K_{\Omega}$ generated by $\Omega:=\{a_1,\ldots,a_m\}$ is closed in $\mathbb{R}^n$.

Proof $\;\;\;\;\;$ Assume first that the elements $a_1,\ldots,a_m$ are linearly independent and take any sequence $\{x_k\}\subset K_{\Omega}$ converging to $x.$ By construction of $K_{\Omega}$ find numbers $\alpha_{ki}\geq0$ for each $i=1\ldots m$ and $k\in\mathbb{N}$ such that $$x_k=\sum_{i=1}^{m}\alpha_{ki}a_i$$ *Letting $\alpha_{k}:=(\alpha_{k1},\ldots,\alpha_{km})\in\mathbb{R}^n$ and arguing by contradiction, it is easy to check that the sequence $\{\alpha_{k}\}$ is bounded$\ldots$

I am not convinced to is easy to check that the sequence is bounded. My initial thoughts are as follows:

Assume the sequence $\{\alpha_k\}$ is not bounded. It then follows that $\{\alpha_k\}$ divergess to infinity. i.e. as $k\rightarrow \infty, \{\alpha_k\}\rightarrow\infty$. But then $\sum_{i=1}^{m}\alpha_{ki}a_i\rightarrow\infty$ which means $\{x_k\}\rightarrow\infty$. However, this contradicts our choice of $\{x_k\}$ which converges to $x$. Thus, $\{\alpha_k\}$ is bounded.

Does this seem like I am on the correct path?

$\endgroup$

1 Answer 1

0
$\begingroup$

I have to methods for proving the boundedness. The first one uses a contradiction, as proposed by the authors. However, the second method seems to be more elegant and even yields the convergence of the coefficients.

Method 1: Assume $\{\alpha_k\}$ is not bounded. Then, $$\frac{\sum_{i=1}^m \alpha_{ki} \, a_i}{\max_{i} \alpha_{ki}} = \sum_{i=1}^m \frac{\alpha_{ki}}{\max_j \alpha_{kj}} \, a_i \to 0.$$ Now, the sequences $\{\frac{\alpha_{ki}}{\max_j \alpha_{kj}}\}_k$ are bounded and you can extract convergent subsequences such that at least one of them does not converge to zero. Hence, $$\sum_{i=1}^m \alpha_i \, a_i = 0$$ and at least one $\alpha_i$ is not zero. This is a contradiction to the linear independence.

Method 2: Since the $a_i$ are linear independent, there are dual elements $b_j$ with $$b_j^\top a_i = \delta_{ij} = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \ne j.\end{cases}$$ Then, $$b_j^\top x \leftarrow b_j^\top x_k = b_j^\top \sum_{i=1}^m \alpha_{ki} \, a_i = \alpha_{kj}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.