Hahn Banach Theorem: Clarification on meaning of extending a functional?

Question

Hahn Banach Theorem:

Given linear (vector) space $\mathbb{X}$, define $u \in \mathbb{L} \subset \mathbb{X}$, $A,B,C$ functionals, A sublinear.

$A:\mathbb{L} \to \mathbb{R}, B:\mathbb{L} \to \mathbb{R}, A(u) \leq B(u)$ and $C:\mathbb{X} \to \mathbb{R}$.

Then $B$ can be extended to $C$, $C(u) \leq A(u)$

I am not familiar with what it means precisely to extend one function to another.

Can someone illustrate with a simple example what extending mean in this context? What does it mean precisely to extend a function

Answer 1

First off, I think your statement of the theorem is a bit off. It should go something like:

Given $A:\mathbb{X}\rightarrow\mathbb{R}$ (not $\mathbb{L}\rightarrow\mathbb{R}$) sublinear and $B:\mathbb{L}\rightarrow\mathbb{R}$ linear such that $B(u)\le A(u)$ for all $u\in\mathbb{L}$, there exists a linear extension of $B$ (call it $C$) from $\mathbb{X}\rightarrow\mathbb{R}$ such that $C(u)\le A(u)$ for all $u\in\mathbb{X}$.

In this case, $C$ being an extension of $B$ means that $C$ restricted to $\mathbb{L}$ is equal to $B$, i.e. $C(u) = B(u)$ for all $u\in\mathbb{L}$.

For a very easy example, let $\mathbb{L} = \mathbb{R}\times\{0\}$, $\mathbb{X} = \mathbb{R}^2$. Let $B(x,0) = x$. If $A(x,y) = \sqrt{x^2+y^2}$, then it is clear that $B(x,0)\le A(x,0)$ for all $x$, and we can let $C(x,y) = x$, and in fact this is the only possible linear extension which satisifies $C\le A$. If we let $A(x,y) = 2\sqrt{x^2+y^2}$ instead, then we still have $B\le A$, and $C(x,y) = x$ still works as an extension, but $C(x,y) = x+y$ would also work (you can check this for yourself).

Answer 2

A function is said to be an extension of another one if, given the original function $f_1 : A \to X$, there is a function $f_2:B \to X$ such that $A\subset B$ and $f_1(x) = f_2(x) \,\,\,\forall x \in A$. Clearly, an extension doesn't have to be unique, it just has to be defined on a larger domain that contains the original one and agree with the first function where both are defined.