TITLE: Hahn-Banach theorem with convex majorant
QUESTION [32 upvotes]: 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:

Who first observed that sublinearity can be replaced by convexity?
Is there any (e.g. pedagocial) reason to prefer the sublinear version?

REPLY [2 votes]: 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 $\partial p(x)=\{ v\}$.