Can You Explicitly Construct a Hahn Banach Extension?

[StatusX]

I'm trying to understand the Hahn Banach theorem, that every bounded linear functional f on some subspace M of a normed linear space X can be extended to a linear functional F on all of X with the same norm, and which agrees with f on M. But the proof is non-constructive, using zorn's lemma.

So I'm trying to come up with examples so I can understand it better. For example, let L be the space of all bounded infinite sequences with the sup norm, and let M be the subspace of L consisting of those sequences that converge to some finite limit. Then, on this subspace, the functional given by taking the limit of a sequence is clearly linear and bounded. So it must extend to a bounded linear functional on all of L. But I can't imagine what such a functional would look like. Is it possible to explicitly construct one?

 

[StatusX]

The functional would have to have the following properties:

1. Any two sequences whose difference converges to zero must get the same value (eg, if f(1,0,1,0,...)=a then f(1/2,1/2,2/3,1/3,3/4,1/4,...)=a).

2. If the sum of two sequence converges to some limit, the sum of the values for each sequence equals the limit of their sum. (eg, if f(1,0,1,0,...)=a, then f(0,1,0,1,...)=1-a).

But what is, say, a? Can it be anything?

 

[tacman]

First of all, it is great that you are trying to come up with examples to better understand the Hahn Banach theorem. It is a fundamental result in functional analysis and understanding it through examples can definitely help solidify your understanding.

In your example, you have correctly identified a subspace M of a normed linear space X and a bounded linear functional f on M. The Hahn Banach theorem guarantees the existence of a linear functional F on all of X that extends f and has the same norm. However, as you mentioned, the proof of this theorem is non-constructive and relies on the use of Zorn's lemma.

Unfortunately, it is not possible to explicitly construct such a functional F in this case. This is because the existence of such a functional relies on the axiom of choice, which is a fundamental principle in set theory. It states that for any collection of non-empty sets, it is possible to choose one element from each set. In the context of the Hahn Banach theorem, this means that we can choose a linear functional F that extends f on M, but we cannot explicitly construct it.

To better understand this, let's look at a simpler example. Consider the space of real numbers R and the subspace of rational numbers Q. We know that every real number can be approximated by a rational number, but we cannot explicitly construct a rational number that approximates a given real number. This is because the existence of such a rational number relies on the axiom of choice.

In the same way, the existence of a linear functional F that extends f on M in your example relies on the axiom of choice, and hence, cannot be explicitly constructed. However, we can still understand the properties of this functional F through the Hahn Banach theorem and its proof.

In conclusion, while it may be difficult to explicitly construct a functional F that extends f on M in your example, understanding the Hahn Banach theorem and its proof can help us understand its properties and significance in functional analysis.