Extend the linear functional on $\mathbb R^2$ 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.
 A: 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.
