If is a normed vector space over or , then its dual space is a Banach space.
Proof Let be a normed vector space. The dual space is the space of all bounded linear functionals on , equipped with the operator norm:
To show that is a Banach space, we need to prove it is complete. Let be a Cauchy sequence in . This means that for every , there exists an such that for all , we have:
By the definition of the operator norm, this implies that for any with :
This shows that for each fixed , the sequence of scalars is a Cauchy sequence in or . Since and are complete, this sequence converges. We can therefore define a function (or ) by the pointwise limit:
The function is linear because each is linear. We must now show two things:
is bounded (i.e., )
in the norm of
First, let’s show in . From the inequality for and , we can take the limit as :
This holds for all and for all with . Taking the supremum over all such gives:
This is precisely the definition of convergence of to in the norm of .
Second, we need to verify that our limit function is actually in . We know is linear; we just need to show it is bounded. We use the triangle inequality for any :
From our previous step, we know . For instance, we can take . Since is in , its norm is finite. Therefore:
This shows that is a bounded linear functional, so .
Since every Cauchy sequence in converges to an element in , the space is complete and is therefore a Banach space.
The Hahn-Banach Theorem
Hahn-Banach Theorem (Real Vector Spaces)
Let be a vector space over , be a vector subspace of . Suppose is a sublinear functional, meaning it satisfies:
Positive homogeneity: for all and .
Subadditivity: for all .
If is a linear functional that is dominated by on , i.e., for all , then there exists a linear extension of such that:
for all ( is an extension).
for all (the extension is also dominated by ).
Remark
In the setup of the theorem:
is a vector space, not necessarily normed or complete.
is a subspace, not necessarily closed.
A typical example of a sublinear functional is for some constant .
Corollary
Let be a subspace of a normed vector space over . Every bounded linear functional can be extended to a bounded linear functional such that the extension has the same norm:
Proof Let us denote and . Let . We define a function by . This function is sublinear, as it is positively homogeneous and satisfies the triangle inequality.
For any , we have .
The conditions for the Hahn-Banach theorem are met. Therefore, there exists a linear extension of such that for all . This gives us an upper bound for : For a lower bound, we consider : Combining these two inequalities, we get for all . This shows that is a bounded linear functional (so ) and that . Since is an extension of , its norm must be at least as large as the norm of . That is, Thus, we conclude that .
Corollary
If is a normed vector space over , then for any non-zero , there exists a functional such that:
Remark
This corollary is significant because it guarantees that the dual space is non-trivial (if is non-trivial) and is “large enough” to separate points. It also shows that the supremum in the dual definition of the norm of an element (i.e., ) is actually achieved.
Proof Sketch of the Hahn-Banach Theorem
The full proof requires Zorn’s Lemma to extend the functional from a subspace to the whole space by considering a partially ordered set of all valid extensions. The key step, however, is to show that a functional can be extended by just one dimension.
Let be the functional satisfying the theorem’s hypotheses. Pick an element . We want to extend to the subspace . Any element in can be uniquely written as for and .
For the extension, which we’ll also call , to be linear, it must have the form:
We need to define the value . Let’s call this value . Our task is to choose a such that the extension condition holds for all .
By considering cases for and and rearranging, we find that must satisfy for all :
Such a exists if and only if the supremum of the left side is less than or equal to the infimum of the right side. This is equivalent to showing that for all :
We know that , so . Also, by the subadditivity of :
Putting these together:
The inequality holds, so a valid choice for exists. This completes the one-step extension. Zorn’s Lemma is then invoked to show this process can be continued to cover the entire space .