Home > Uncategorized > Hahn-Banach Theorem (real version)

Hahn-Banach Theorem (real version)


Suppose X is a vector space over \mathbb{R}, p:\mathbb{X} \to \mathbb{R} has the following properties: p(\lambda x)=\lambda p(x),\ \forall x \in X, \lambda \in\mathbb{R}_+ and p(x+y)\leq p(x)+p(y),\ \forall x,y \in X.
Let X_0 be a subspace of X and u: X_0\to \mathbb{R} a linear functional such that u(x) \leq p(x),\ \forall x \in X_0.
Then we can find f:X \to \mathbb{R} a linear functional such that f| _{X_0} =u and f(x) \leq u(x),\ \forall x \in X.
Solution: Consider the set M= \{ (Y,g) : Y is a subspace of X,\ X_0 \subset Y,\ g: Y \to \mathbb{R} is a linear functional which extends u and g \leq p on Y \}.
Define an order relation on M like this (Y_1,g_1)< (Y_2,g_2) if Y_1 \subset Y_2 and g_2 is an extension for g_1.
We show that in M every chain has an upper bound. Suppose M_0 is a totally ordered subset of M. Then define Y_0=\displaystyle \bigcup_{(Y,g) \in M_0} Y and g: Y_0 \to \mathbb{R},\ g(y)= g_0(y) if y \in Y_0 and (Y_0,g) \in M_0. This function is well defined, and Y_0 is a subspace of X because the set M_0 is totally ordered. Furthermore, from the definition for g_0, we have that g_0 \leq p. Therefore (Y_0,g_0) \in M, and is obviously an upper bound for M_0. By Zorn’s Lemma, we find that M has at least one maximal element (Z,h).
Suppose X \neq Z. Then we can find x_0 \in X\setminus Z. Define W=Span\{ Z,x_0\} = \mathbb{R} \cdot x_0 \oplus Z. Therefore, W is a linear subspace in X.
Let y,z \in Z. Then h(y)+h(z)= h(y+z)\leq p(y+z) = p(y-x_0+x_0+z) \leq p(x_0+y)+p(-x_0+z).
Therefore, we have h(z)-p(-x_0+z)\leq -h(y)+ p(x_0+y),\ \forall y,z \in Z. Therefore, we can say a= \sup_{z \in Z} (h(z)-p(-x_0+z)) \leq \leq \inf_{y \in Z} (-g(y)+ p(x_0+y))=b. Pick one c \in [a,b] and define h_1(z)=\lambda c+h(y), where z=\lambda x_0+y (unique representation). h_1 is linear, and extends h on W, which means that it extends u on X_0. We can check that (W,h_1)\in M and is an extension for the maximal element, which is a contradiction. Therefore Z=X, and the maximal element h is the requested functional

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