Let $(\mathbb R^2, \|\cdot\|_1)$ be normed space and $E$ be a 1-dimensional subspace. If $f:E \to \mathbb R$ is linear functional with induced 1-norm one, i.e $\|f\|_{1,E}=1$ on $E$ extend it to linear functional $f:\mathbb R^2 \to \mathbb R$, with the same norm, $\|f\|_1 = 1$.
Some definitions:
- For a vector $x \in \mathbb R^2$ the norm is defined as $\|x\|_1 := |x_1| + |x_2|$.
- For a linear operator the norm $\displaystyle\|f\|_1 := \sup_{\|x\|_1 = 1} | f(x) |$
- The induced norm of $f$ is defined as $\displaystyle\|f\|_{1,E} := \sup_{x \in E,\ \|x\|_1 = 1} | f(x) |$
Attempt 1:
A vector $x \in E$ would have components related as $x_1 a_1 + x_2 a_2 = 0$ for some non-vanishig $a = (a_1,a_2)$. If $\|f\|_{1,E} = 1$ then
\begin{align}
1 = \sup_{x \in E,\ \|x\|_1 = 1} | f(x) |
&= \sup_{x_1 a_1 + x_2 a_2 = 0,\ \|x\|_1 = 1}\left| x_1 f_1 + x_2 f_2 \right| \\
&= \sup_{\|x\|_1 = 1}\left| x_1 f_1 -\frac{a_1}{a_2} x_1 f_2 \right| \\
&= \left| f_1 -\frac{a_1}{a_2} f_2 \right|\sup_{\|x\|_1 = 1} |x_1| \\
&= \left| f_1 -\frac{a_1}{a_2} f_2 \right| \frac{|a_2|}{|a_1| + |a_2|}
= \frac{ |a_2 f_1 – a_1 f_2| }{|a_1| + |a_2|} = \frac{|\det(f,a)|}{|a_1| + |a_2|}
\end{align}
or more concisely
$$
|\det(f,a)| = \| a \|_1
$$
But that doesn't determine $f$.
Attempt 2:
I wanted to find a linear functional such that $f(v) = v_1 f_1 + v_2 f_2 = 1$, for $v\in E$ and $f(u) = u_1 f_1 + u_2 f_2 = 0$ for linearly independent vector, where $\|v\|_1 = \|u\|_1 = 1$. Then we get the solutions
\begin{align}
f_1 &= \frac{u_2}{\det(v,u)} & f_2 &= -\frac{u_1}{\det(v,u)}
\end{align}
But that stll doesn't determine $u$ nor $f$.
By the way, I found the solutions geometrically, i.e. without computations as above, but I don't know how to justify them with computation. Any hints would be helpful. It will be great to learn something for the infinite dimensional case from this exercise.
Best Answer
There are no many options for a subspace, so lets assume that $E=\{(x_1, x_2)\in \mathbb{R}^2: x_2=ax_1\}$ for some $a\in \mathbb{R}$, then we also may assume that the linear functional $f: E\to \mathbb{R}$ must be of the form \begin{align*} f(x_1, x_2)=\eta x_1, \end{align*} for some $\eta\in \mathbb{R}$. Now we observe that \begin{align*} |f(x_1, x_2)|=|\eta||x_1|=\dfrac{|\eta|}{1+|a|}\|(x_1, x_2)\|_1 \end{align*} hence by assumption the relation between the data $\eta$ and $a$ must be \begin{align*} \dfrac{|\eta|}{1+|a|}=1. \end{align*} Now, by Hahn-Banach, $f$ admits an extension to the whole $\mathbb{R}^2$, denoted by $F$. By letting $\beta_1=F(1, 0)$ and $\beta_2=F(0, 1)$, we can describe $F$ by \begin{align*} F(x_1, x_2)=\beta_1x_1+\beta_2x_2,\ \forall x_1, x_2\in\mathbb{R}. \end{align*} Since $F$ extends $f$, over $E$ we must have \begin{align*} \eta x_1=f(x_1, ax_1)=F(x_1, ax_1)=(\beta_1+a\beta_2)x_1, \end{align*} which results on the first condition for the extension: $\beta_1+a\beta_2=\eta$. The induced $1$-norm of $F$ is given by \begin{align*} \|F\|_1=\max\{|\beta_1|, |\beta_2|\}, \end{align*} therefore the second conditions for the extension is \begin{align*} \max\{|\beta_1|, |\beta_2|\}=1. \end{align*} since $F$ has norm $1$. Hence by solving the system \begin{align*} \begin{cases}\beta_1+a\beta_2=\eta,\\ \max\{|\beta_1|, |\beta_2|\}=1,\end{cases} \end{align*} you can determine $F$.
Observe that this system does not have unique solutions. For instance, $(\beta_1, \beta_2)=(0, 1)$ and $(\beta_1, \beta_2)=(1, 0)$ solves the system, and each solution is associated to a different linear extension. The condition to have a unique Hahn-Banach extension (preserving the norm) for a linear functional $f: M\leq X\to \mathbb{R}$, is that the dual space $X^*$ is strictly convex.