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This is the usual formulation of Hahn-Banach theorem (some books use sublinear function instead, but it probably does not make much difference):

Let $X$ be a vector space and let $p:X\to{\mathbb R}$ be any convex function. Let $M$ be a vector subspace of $X$ and let $f:M\to{\mathbb R}$ be a linear functional dominated by $p$ on $M$. Then there is a linear extension $\hat{f}$ of $f$ to $X$ that is dominated by $p$ on $X$.


However if you look into the proof, you find out that it shows that for a given $v\in X$, there exists an extension with the value $\hat{f}(v)=c$ for any choice of $c$ anywhere between

$$\sup_{x\in M,\lambda>0} \frac1\lambda [f(x)-p(x-\lambda{v})] \le c \le \inf_{y\in M,\mu>0} \frac1\mu [p(y+\mu{v})-f(y)].$$

If $p$ is positive homogeneous, i.e., $p(\lambda x)=\lambda p(x)$ for $\lambda>0$, than the expression for the range is simpler:

$$\sup_{x\in M} [f(x)-p(x-{v})] \le c \le \inf_{y\in M} [p(y+{v})-f(y)].$$

In case the $p$ and $f$ have the additional property that $$(\forall x\in X)(\forall y\in M) p(x+y)=p(x)+f(y)$$ then the above interval can be simplified to

$$-p(-v)\le c \le p(v).$$

(This situation happens quite often, take for example $f(x)=\lim x$ on the space $c$ of all convergent sequences and the function $p(x)=\limsup x$ on the space $\ell_\infty$ of all bounded sequences.)


On several occasions I found the above observations useful. (Since sometimes the function $p(x)$ can be chosen in a such way that the set of linear functionals which we want to study is precisely the set of all extensions of some given functional fulfilling $\hat{f}(x)\le p(x)$. Hence this might give some additional information about an interesting set of functionals.) So I wonder why I have not find a book where Hahn-Banach theorem was not formulated with the inclusion of the information about the possible range of those extensions.

Question 1 Do you know about a reference giving a formulation of Hahn-Banach theorem which includes the information about possible values of extensions?

Question 2 Were the extreme points of the set of Hahn-Banach extensions studied? (By the set of Hahn-Banach extensions I mean the set of all linear functionals $\hat{f}:X\to{\mathbb R}$ fulfilling $(\forall x\in X)\hat{f}(x)\le p(x)$ and $(\forall x\in M)\hat{f}(x)=f(x)$ for a given linear functional $f:M\to{\mathbb R}$ and a convex function $p:X\to{\mathbb R}$.)

EDIT: Perhaps I should mention that the formulation and the proof of Hahn-Banach theorem I followed is from Aliprantis, Border: Infinite-Dimensional Analysis. As far as I remember, all proofs of HBT I've seen have pretty much the same idea. Thus my guess is that the answer to the following question would be probably negative, but I'll ask it anyway.

Question 3 Have you seen a formulation (or a proof) of Hahn-Banach theorem that would yield substantially different expression for the bounds I've given above. (I believe its more-or-less obvious from the proof that these bounds are best possible in the sense that extension of $f$ dominated by $p$ cannot have value in the point $v$ outside the given intervals. So only change that could happen would be some different expression of the same value. Alternatively, some books might contain proof of HBT which does not obtain the optimal bounds, but again, I do not see any way how this would simplify the proof, so there's probably no reason to give such a proof.)

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I'd like to point out the paper by Bagget-Ramsay, A functional analytic proof of a selection lemma, Can. J. Math 32, No. 2 (1980), 441-448, proving some Borel selection theorems using your principal observation. Specifically, they prove as main theorem that for a subspace $E$ of a separable Banach space $F$ the restriction map $r: B_{\leq 1} F^{\ast} \to B_{\leq 1} E^\ast$ on the unit balls in the duals has a (weak$^{\ast}$)-Borel measurable section mapping extreme points to extreme points. "There is a Borel measurable way to choose Hahn-Banach extensions simultaneously." –  t.b. Jul 7 '11 at 19:25
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What is the advantage of having Hahn-Banach for functionals dominated by convex functions instead of sublinear functionals? Is there any serious gain in that, meaning: do Aliprantis and Border gain something using this formulation with a slightly more cumbersome proof? As far as I can see the important thing in applications will be continuity of the convex function anyway and hence the domination property immediately allows us to find a Minkowski functional dominating our functional and dominated by the original function, thus allowing us to use the usual formulation. Do I miss something? –  t.b. Jul 8 '11 at 11:03
    
@Theo I cannot think of any application where the formulation with convex function would bring some advantage. Of course, I cannot speak for the authors of the book. Nevertheless, I've edited the range of possible values in my question to reflect the fact, that if $p$ is sublinear, it can be simplified. –  Martin Sleziak Jul 8 '11 at 16:48
    
For positively homogeneous functions, convexity is equivalent to subadditivity. (The proof is easy, see here: books.google.com/… ) So if we want to find a situation, where the formulation with convex function would be in some sense superior to sublinear form of HBT, it would have to be a situation with $p(x)$ NOT positively homogeneous. –  Martin Sleziak Jul 9 '11 at 8:27

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