# Why compactness w.r.t strong convergence can only be used in finite-dimensional spaces

I am reading the following book:

Optimization by Vector Space Methods written by D. G. Luenberger

My problem comes from the Example 1. in p.127:

The strong convergence means the convergence in norm. However, we can still define a norm for infinite-dimensional space such as the $l_p$ norm. Therefore, we can still define compactness w.r.t strong convergence in the infinite dimensional spaces.

So I am confused about what did the author try to say.

• For a start, in infinite dimensional spaces, the unit ball is never (strongly) compact. (Compactness of a set is still defined, of course.) – David Mitra Sep 15 '17 at 11:36
• What is meant is that compactness with respect to strong topology is so strong requirement that it is hard to find interesting applications. So, weaker notions are used. – Felix Klein Sep 15 '17 at 12:26

Yes, of course, you can define compactness w.r.t. strong topology. The problem is that it is not useful in optimization problems in infinite dimensional spaces. Let us first recall the usual argument in finding the minimizers of a linear functional $F$ over an space $X$. The argument goes as follows: Find a sequence $\{u_n\}$ such that $F(u_n)\to \inf F$ as $n\to \infty$. If this sequences is sequentially compact in a topology $\tau$ and $F$ is sequentially lower semicontinuous, then we can find $F$ admits a minimizer.
The problem is that in most cases we can only prove that $\{u_n\}$ is bounded. So we hope that boundedness implies so kind of compactness. However, only in finite dimensional space the boundedness can ensure that pre-compactness in strong topology. In infinite dimensional space (say reflexive Banach space), weak topology is applied, since boundedness can imply sequential compactness in weak topology.