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Given \begin{align} s(\theta)= &\text{arg min}( g( \boldsymbol{x}) ) \\ \text{subject to }& \boldsymbol{A}(\theta) \boldsymbol{x} = \boldsymbol{b}(\theta) \\ &c_1 \le x_i \le c_2 , i \in \{1,2,\cdots, n\} \end{align}.

where $ \theta\in \Theta$ , $ \boldsymbol{A} \in \mathcal{R} ^{ m \times n}$ is full row rank, with $m<n $, $ c_1 $ and $ c_2 $ are constants, both $ \boldsymbol{A}(\theta) $ and $ \boldsymbol{b}(\theta) $ is continuous and the problem is feasible for all $ \theta \in \Theta $. Further $g(\boldsymbol{x})$ is continuous and strictly convex.

Does it follow from the Maximum theorem that $s(\theta)$ is a continuous vector valued function?


The theorem is short and seems rather straight forward, however, I am in need a second opinion verifying that what I believe to hold is valid. (I still somewhat struggle with the concepts).

Below follows my simple draft


Let $\boldsymbol{x} \in \mathbb{R}^n$, and $\theta \in \Theta \in \mathbb{R}^1$

Let the constraint set be denoted

$ C(\theta)= \{ \boldsymbol{x}: \boldsymbol{A}(\theta) \boldsymbol{x} = \boldsymbol{b}(\theta) , c_1 \le x_i \le c_2 , i \in \{1,2,\cdots, n\} \} $

Now:

$ \theta \mapsto C(\theta) $ is compact (affine constraints) such that $ C(\theta) \neq \emptyset$ (assumption of feasibility) $\forall \theta \in \Theta $. Therefore $ C:\mathbb{R}^1 \rightrightarrows \mathbb{R}^n$ $ is a compact valued correspondence.

With $\boldsymbol{A}(\theta)$ full row rank, given continous parameters, $ C(\theta) $ is continous (implicit function theorem). Since also $g( \boldsymbol{x}) $ is continuous, by the maximum theorem (in this case we use the opposite version, i.e., minimum) , $ s(\theta) $ is continuous (i.e. both upper and lower hemicontinuous). Now due to strict convexity any solution is single-(vector) valued and $ s(\theta) $ is a continuous vector valued function.

Am I missing something?


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  • $\begingroup$ what is $\Theta$ here? $\endgroup$ Jul 28, 2019 at 8:01
  • $\begingroup$ It is the (topological) space of the scalar parameter $\theta$. Could to my understanding be replaced with forexample $\Theta =\{ \theta : 0 \le \theta \le k \} $. (i tried to write it similar as it is written in the theorem) $\endgroup$
    – Einar U
    Jul 28, 2019 at 13:31

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