Hahn-Banach theorem with convex majorant

Question

At least 99% of books on functional analysis state and prove the Hahn-Banach theorem in the following form: Let $p:X\to \mathbb R$ be sublinear on a real vector space, $L$ a subspace of $X$, and $f:L\to \mathbb R$ linear with $f\le p|_L$. Then there is a linear $F:X\to\mathbb R$ with $F\le p$ and $F|_L=f$.

However the theorem is true if the majorant $p$ is merely convex. This version has a very similar proof as the classical statement and several advantages. For instance, there is no need to introduce the new notion of sublinearity and the result is even interesting for $X=\mathbb R$.

The only reference I know is the book of Barbu und Precupanu Convexity and Optimization in Banach Spaces.

Two questions:

1. Who first observed that sublinearity can be replaced by convexity?

2. Is there any (e.g. pedagocial) reason to prefer the sublinear version?

Answer 1

If $p$ is convex, then $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.

Answer 2

Other References Containing the Convex Hahn-Banach

The convex majorant version of HBT (Hahn-Banach Theorem) also appears in

• "Methods of Modern Mathematical Physics Vol. 1 Functional Analysis" by Reed and Simon (p.75-76)

• "Handbook of Analysis and Its Foundations" by Eric Schechter (statement p.318; proof p.321-322).

"Convex" Hahn-Banach for Complex Vector Spaces

Reed and Simon also have a "convex" majorant HBT for complex vector spaces. The standard HBT for complex spaces has a semi-norm $p$ and a complex-linear functional $f$ with $|f| \leq p$. In the ``convex" majorant HBT for complex spaces, $p$ need not be a semi-norm, but is only required to satisfy the weaker condition \begin{align}\label{1}\tag{1} p(\alpha x + \beta y) \leq |\alpha|p(x) + |\beta|p(y) \quad \text{ when } |\alpha|+|\beta|=1. \end{align}

Reed and Simon deduce the "convex" complex HBT from the convex real HBT in the same way that the standard semi-norm complex HBT is deduced from the standard sublinear real HBT (that is, by exploiting the relationship between a complex-linear functional and its real part). And, of course, the proof of the convex real HBT is basically identical to the proof of the sublinear real HBT.

Alternatively, the "convex" complex HBT can be deduced from the semi-norm complex HBT by changing Fedja's $P(x)$ to $P(x)=\inf_{0 \neq s \in \mathbb{C}} |s|^{-1}p(sx)$. Indeed, if $p$ satisfies \eqref{1}, then $P$ is a semi-norm; if a linear functional is dominated by $p$, it is also dominated by $P$; and $P \leq p$.

Who First Noted the Convex HBT

Since the proofs of the convex and sublinear HBT are so similar and in light of Fedja's argument, I would guess that the convex HBT was known basically as soon as the sublinear HBT was. That being said....

Schechter on page 318 points out that the convex HBT is not mentioned much in the literature but was known at least as early as in

• Nakano, Hidegorô. On an extension theorem. Proc. Japan Acad. 35 (1959), no. 3, 127. doi:10.3792/pja/1195524402. https://projecteuclid.org/euclid.pja/1195524402

See also

• Nakano, Hidegorô. An extension theorem. Proc. Japan Acad. 33 (1957), no. 10, 603--604. doi:10.3792/pja/1195524883. https://projecteuclid.org/euclid.pja/1195524883

According to the last reference, the convex real HBT may also appear in the following two references, but I haven't been able to check.

• H. Nakano: Modulared linear spaces, Jour. Fac. Sci. Univ. Tokyo, I, 6, 85 (1951).

• H. Nakano: Topology and Linear Topological Spaces, Tokyo (1951).

Answer 3

Just two remarks on the above answer but too long for a comment. Firstly, there is no reason why the above sublinear function should be finite everywhere, indeed it can be the constant function $-\infty$. I don't think that this invalidates the argument but it does make it a tad messier.

Secondly, I think that there is a very good reason for using sublinear functions. There is a perfect duality between such functionals on a vector space and between the algebraically closed, convex, absorbent subsets which contain zero. This supplies a transparent link between the analytic and the geometric versions of the result in question (a fact which has been mentioned several times in the above comments). Of course, the sublevel sets of a a convex function are also convex, but, in contrast to the case of a sublinear functional where these are all dilations of its unit ball, these form an infinite family of such sets, typically of different form. Of course, a convincing example of an interesting application of the version involving domination by a convex function would change the whole ball game.

Answer 4

Yet another reference is Theorem A on page 105 in:

A. Wayne Roberts, Dale E. Varberg, Convex functions. Pure and Applied Mathematics, Vol. 57. Academic Press , New York-London, 1973.

This reference provides also an answer to a question in one of the comments:

What are some applications which can not be proved with the "sublinear" version?

If $p:\mathbb{R}^n\to\mathbb{R}$ is convex, then the subdifferential $\partial p(x)$ is the set of all $v\in\mathbb{R}^n$ such that $$ (*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p(y)\geq p(x)+\langle v,y-x\rangle \ \quad \text{for all $y\in\mathbb{R}^n$.} $$ It is easy to prove that $\partial p(x)\neq\emptyset$ for all $x$. The result mentioned by OP has the following application:

Theorem. $p$ is differentiable at $x$ if and only if $\partial p(x)$ consists of one vector.

Sketch of the proof. If $p$ is differentiable, then it easily follows that $\partial p(x)=\{\nabla p(x)\}$. Suppose now that $\partial p(x)=\{ v\}$. We want to show that $p$ is differentiable at $x$. It easily follows from the convexity that if partial derivatives exist at $x$, then $p$ is differebtiable. Indeed, if $A=[\frac{\partial p}{\partial x_1}(x),\ldots,\frac{\partial p}{\partial x_n}(x)]$, then $y\mapsto p(x+h)-p(x)-Ah$ is convex and one can use Jensen's inequality to estimate this remiander in terms of partial derivatives (very nice exercise, cf. p. 101 in the above book). Thus it remains to show that $p$ has partial derivatives at $x$. One-sided partial derivatives exist (due to monotonicity of difference quotients) and it suffices to show that they are equal. For if not, they would give two different one dimensional linear functions satisfying $(*)$ in the direction on the partial derivative. Then the "convex" Hahn-Banach theorem mentioned by OP would give two different vectors $v$ satisfying $(*)$ for all $y\in\mathbb{R}^{n}$ which would contradict the assumption that $\partial p(x)=\{ v\}$.