Explicit example of Hahn-Banach theorem on the finite dimensional space $\mathbb{R}^2$? The Hahn-Banach theorem allows us to extend linear functionals defined on a subspace of some vector space $V$ to the entire space. Is it possible to construct an explicit example of this in the finite dimensional case? For example, suppose $V = \mathbb{R}^2$ and $U=\mathbb{R} \subset V$. What would be a simple explicit example of the Hahn-Banach theorem, i.e. what are explicit expressions for


*

*$p:V \to \mathbb{R}$ is a sublinear function

*$\varphi: U \to \mathbb{R}$ is a linear functional on the linear subspace $U \subset V$ which is dominated by $p$ on $U$.

*The linear extension $\psi:V \to \mathbb{R}$ of $\varphi$ to the whole space such that
\begin{align}
\psi(x) = \varphi(x) \quad \forall x \in U, \\
\psi(x) = p(x) \quad \forall x \in V.
\end{align}

 A: In case of $(\mathbb{R}^2, \|\cdot\|_2)$, things are quite simple.
Let $U \le \mathbb{R}^2$ be a subspace and $\phi : U \to \mathbb{R}$ a linear functional. By the Riesz representation theorem, there exists $a \in U$ such that $\phi(x) = \langle x, a\rangle, \forall x \in U$. Then the linear functional $\psi : \mathbb{R}^2 \to \mathbb{R}$ given by the same formula $\psi(x) = \langle x, a\rangle, \forall x \in \mathbb{R}^2$ is the unique Hahn-Banach extension of $\phi$.
Namely, clearly $\psi$ extends $\phi$ and $\|\phi\| = \|a\|_2 = \|\psi\|$ so $\psi$ is a Hahn-Banach extension of $\phi$.
Let $\zeta : \mathbb{R}^2 \to \mathbb{R}$ be another Hahn-Banach extension of $\phi$. By the Riesz representation theorem, there exists $b \in \mathbb{R}^2$ such that $\zeta(x) = \langle x, b\rangle,\forall x \in \mathbb{R}^2$. Since $\zeta$ extends $\phi$, we have
$$\langle x, a\rangle = \phi(x) = \zeta(x) = \langle x, b\rangle, \forall x \in U \implies \langle x, a - b\rangle = 0, \forall x \in U \implies a - b \perp U$$
Since $b = \underbrace{a}_{\in U} + \underbrace{(b - a)}_{\in U^\perp}$, the Pythagorean theorem gives
$$\|a\|_2^2 + \|b - a\|_2^2 = \|b\|_2^2 = \|\zeta\|^2 = \|\phi\|^2 = \|a\|_2^2 \implies b - a = 0 \implies a = b$$
Therefore $\zeta = \phi$.
This explicit construction is in fact always possible when dealing with a Hilbert space, which $(\mathbb{R}^2, \|\cdot\|_2)$ is an example of.

For an explicit example of the above discussion, consider the subspace $Y = \{(x,2x) \in \mathbb{R}^2 : x \in \mathbb{R}\} \le \mathbb{R}^2$ and the linear functional $\phi :Y \to \mathbb{R}$ given by $\phi(x,y) = x$.
An orthonormal basis for $Y$ is $\left\{\frac1{\sqrt{5}}(1,2)\right\}$ so the orthogonal projection $P_Y$ onto $Y$ is given by 
$$P_Y(x,y) = \left\langle (x,y),\frac1{\sqrt{5}}(1,2)\right\rangle \frac1{\sqrt{5}}(1,2) = \left(\frac{x+2y}5, \frac{2x+4y}5\right)$$
Now notice that for all $(x,y) \in Y$ we have
$$\phi(x,y) = x = \langle (x,y), (1,0)\rangle = \langle (x,y), P_Y(1,0)\rangle = \left\langle (x,y), \left(\frac15, \frac25\right)\right\rangle$$
Now the above discussion implies that the unique Hahn-Banach extension of $\phi$ is given by $\psi : \mathbb{R}^2 \to \mathbb{R}$ defined as
$$\psi(x,y) = \left\langle (x,y), \left(\frac15, \frac25\right)\right\rangle = \frac{x}2 + \frac{2y}5, \quad\forall (x,y)\in\mathbb{R}^2$$
A: You could take
$$
 p(x) = \| x\|_2,
 \quad
 \phi(u) = \frac12 u
$$
Then a possible extension for $\psi:\mathbb R^2\to\mathbb R$ would be
$$
 \psi(x) = \frac12 x_1.
$$
Another valid extensions would be
$$
 \psi(x) = \frac12 x_1 + \frac12 x_2,
$$
because this functional is also bounded by $p$.
