# Fundamentals: Extreme Value Theorems—Part 1

In the last post, we proved that the gradient descent algorithm can be used to computationally find a global minimizer of the least-squares cost functional, i.e., it converges to a vector $\mathbf{x}^\ast \in \mathbb{R}^N$ such that

for all $\mathbf{x} \in \mathbb{R}^N$, where $\mathcal{C}(\mathbf{x}) \triangleq \frac{1}{2}\|\mathbf{y}-\mathbf{H}\mathbf{x}\|_{\ell_2}^2.$

Today, we want to be more general and ask the following question: given an arbitrary continuous function $\mathcal{C}(\mathbf{x})$, with $\mathbf{x} \in \mathbb{R}^N$, how to know that the function has a global minimizer?

Specifically, we will discuss two famous theorems that establish existence of global minimizers.

Theorem 1 [Extreme Value Theorem]: If $\mathcal{X} \subseteq \mathbb{R}^N$ is a compact set and $\mathcal{C}$ is a continuous function on $\mathcal{X}$, then $\mathcal{C}$ has a global minimizer on $\mathcal{X}$

Theorem 2 [Extreme Value Theorem]: If $\mathcal{C}$ is a continuous coercive function defined on all of $\mathbb{R}^N$, then $\mathcal{C}$ has a global minimizer.

Those theorems provide sufficient conditions for the existence of global minimizers. The rest of the post will be about defining the terms compact and coercive; in Part 2 of the post we shall prove both theorems.

We start by establishing two fundamental concepts of open and closed sets in $\mathbb{R}^N$.

Definition 1 [open set]: A subset $\mathcal{X} \subset \mathbb{R}^N$ is open if and only if for every $\mathbf{x} \in \mathcal{X}$  there exist $\epsilon > 0$ such that the open ball $\mathcal{B}_{\epsilon}(\mathbf{x})$ of center $\mathbf{x}$ and radius $\epsilon$ remains in $\mathcal{X}$, i.e., $\mathcal{B}_{\epsilon}(\mathbf{x}) \subset \mathcal{X}$.

Remark that the radius $\epsilon$ may depend on $\mathbf{x}$, and also we remind the definition $\mathcal{B}_{\epsilon}(\mathbf{x}) \triangleq \{\mathbf{y} \in \mathbb{R}^N : \|\mathbf{x}-\mathbf{y}\|_{\ell_2} < \epsilon\}$.

Definition 2 [closed set]: A subset $\mathcal{X} \subset \mathbb{R}^N$ is closed if and only if its complement $\mathcal{X}^\complement = \mathbb{R}^N \setminus \mathcal{X}$ is open.

More intuitively, a closed set is a set that includes its boundary (if there is one), while an open set does not. To establish the term compact set, we introduce one more intermediate definition.

Definition 3 [bounded set]: A subset $\mathcal{X} \subseteq \mathbb{R}^N$ is bounded if there exists a constant $C > 0$ such that $\|\mathbf{x}\|_{\ell_2} < C$ for all $\mathbf{x} \in \mathcal{X}$.

Finally, we now define the compact set by combining Definitions 2 and 3.

Definition 4 [compact set]: A subset $\mathcal{X} \subseteq \mathbb{R}^N$ is compact if it is closed and bounded.

From the perspective of Definition 4, Theorem 1 simply states that if we are optimizing $\mathcal{C}$ over some compact subset of $\mathcal{X} \subseteq \mathbb{R}^N$, then we are sure that there exists a global minimizer $\mathbf{x}^\ast$ that we may hope to find computationally.

Now, what happens if we are interested in optimizing over the whole $\mathbb{R}^N$ rather than on some subset?

This is where the notion of coercive functions becomes useful.

Definition 5 [coercive function]: A continuous function $\mathcal{C}$ that is defined on all of $\mathbb{R}^N$ is coercive if

This means that for any constant $C > 0$ there exists a constant $X > 0$ (that may depend on $C$) such that $\mathcal{C}(\mathbf{x}) > C$ whenever $\|\mathbf{x}\|_{\ell_2} > X$.

Intuitively, for a function to be coercive, it must approach $+\infty$ along any path within $\mathbb{R}^N$ on which $\|\mathbf{x}\|_{\ell_2}$ becomes infinite.

Going back to the example of the least-squares cost functional, we have from the reverse triangular inequality that

where the constant $C = \sqrt{\lambda_{\text{min}}(\mathbf{H}^\mathrm{T}\mathbf{H})}$ with $\lambda_{\text{min}}(\mathbf{H}^\mathrm{T}\mathbf{H})$ denoting the smallest eigenvalue of the matrix $\mathbf{H}^\mathrm{T}\mathbf{H}$.

Thus, when the matrix $\mathbf{H}$ is non-singular, i.e., $C > 0$, and since

we see that least-squares is indeed a coercive function.

## 5 thoughts on “Fundamentals: Extreme Value Theorems—Part 1”

1. Bo Zhao says:

Thank you for posting this well-written note on the existence of a global minimizer . It is really a great pleasure to read. Additionally, I have a comment on your coerciveness argument of the least-square cost function. To prove that it blows up as x goes to infinity, I think you should use a lower bound on ||y - Hx||, rather than a upper bound. More specifically, you may want to use the reverse triangle inequality, i.e., ||y - Hx|| >= ||H||*||x|| - ||y||. Please correct me if I am wrong.

Thanks,
Bo Zhao

1. kamilov says:

Dear Bo Zhao,
Thank you for your thoughtful comment. I have now fixed and updated the post.
Kind Regards,
- Ulugbek

2. Very concise and clear.

1. kamilov says:

Thanks!