The Hahn-Banach theorem and two applications November 28, 2009Posted by Akhil Mathew in analysis, functional analysis, MaBloWriMo.
Tags: convex sets, Hahn-Banach theorem, hyperplane separation theorem, linear functionals, Muntz approximation theorem
I have been finishing my MaBloWriMo series on differential geometry with a proof of the Myers comparison theorem, which right now has only an outline, but will rely on the second variation formula for the energy integral. After that, it looks like I’ll be posting somewhat more randomly. Here I will try something different.
The Hahn-Banach theorem is a basic result in functional analysis, which simply states that one can extend a linear function from a subspace while preserving certain bounds, but whose applications are quite manifold.
Edit (12/5): This material doesn’t look so great on WordPress. So, here’s the PDF version. Note that the figure is omitted in the file.
The Hahn-Banach theorem
Theorem 1 (Hahn-Banach) Let be a vector space, a positive homogeneous (i.e. ) and sublinear (i.e. ) function.
Suppose is a subspace and is a linear function with for all .
Then there is an extension of to a functional with
Corollary 2 Let be a normed linear space, a subspace. Then if and only if: Every bounded linear functional vanishing on vanishes on .
One implication is immediate; linear functionals are continuous. For the other, note that the condition is equivalent to “any bounded linear functional on satisfies , where denotes the image of .”In particular, we are reduced to showing that if is a normed linear space is nonzero, there is a bounded linear functional that does not annihilate . To do this, we apply the Hahn-Banach theorem to and the subspace , with as some multiple of the norm. Choose any nonzero linear functional on this subspace, and then extend to all of .
An approximation result for powers of A spectacular application of this is the following:
Theorem 3 (Muntz) If is a sequence of positive real numbers with , then the linear combinations of the functions are dense in , the space of real continuous functions on the unit interval with the sup norm.
The proof starts by employing the above corollary: for a sequence with the powers not dense in , there is a nonzero continouus functional with The elegant idea in this proof is to consider the complex function defined in the right half-plane . This is an analytic function in , because When , the function tends uniformly in (in the unit interval) to .These are the basic properties of :
- . Indeed, is a bounded functional and the function has norm 1.
- . Indeed, otherwise for , implying vanishes on polynomials. But the Weierstrass theorem implies these are dense in , so , contradiction.
There are general theorems characterizing the roots of a bounded analytic function in some disk (in some, boundedness can be weakened to a condition). Since the right half-plane is conformally equivalent to the disk via the transformation we can consider the bounded analytic function on the unit disk, which has roots at Now I quote the following theorem:
Theorem 4 Let be a bounded analytic function. If the zeros of in are , then
This can be proved using Blaschke products; see Rudin’s Real and Complex Analysis.So for our function , we find which shows that , and which proves the theorem.It’s in fact true that the condition is necessary, and the can be made complex. Cf. Rudin.
Geometric applications Let be a closed convex plane region, and :Then there exists a line through such that lies on one side of —i.e., a support line. This can be proven using a bit of elementary geometry, but it also follows from fancier arguments as well.Basically, the assertion of the theorem can be rephrased as follows: the line corresponds to a linear functional , such that and .
Theorem 5 Let , be disjoint convex sets in a topological vector space such that contains an interior point. Then there is a nonzero continuous linear functional with
The geometric fact above follows from taking , .For a convex set , we define the Minkowski functional Then:
- is positive homogeneous—this is evident.
- . Indeed, if , then by convexity. So , which proves subadditivity.
If has an interior point at zero, is always finite.Now consider . This is a convex set, and I will show that there is a continuous functional with , which will prove the theorem.Choose and consider the set ; if is chosen appropriately, this will contain a neighborhood of . Now (since otherwise and ). There is a linear functional with bounded by (this is seen for with by positive homogeneity; for with , is negative), which we can extend to all of with the same bound. Then where I have abused notation by writing for when every element of is greater than or equal to every element of .
Corollary 6 If are closed convex subsets of a locally convex space , we can get a nonzero continuous as above with even if .
Indeed, we can find a small convex neighborhood of the origin such that are disjoint. Then apply the above result to and note that .