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## 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