Uniqueness of a Hahn Banach extension Let $M = \{(x_n) \in l_1: x_1-3x_2=0\}$ and $f: M \to \mathbb{R}$ given by $f((x_n)) = x_1$. Let $g: l_1 \to \mathbb{R}$ be the Hahn Banach extension of $f$. 
Show that  $g$ is unique.
I know that $\|g\|=\|f\|$. So, I tried to compute $\|f\|$. 
It's easy to see $\|f\| \leq 1$. But I couldn't prove $\|f\|=1$.
And I don't know what to do with the equality $\|g\|=\|f\|$ in order to prove that  $g$ is
unique.
I would really appreciate your help
 A: First note that $M=\left\{(3y,y,x_3,x_4,x_5,\ldots)\in l_1:y,x_i\in\mathbb{R}\right\}$, so $l_1=M\oplus\mathbb{R}(1,0,0,0,\ldots)$.
Now note that $|f(3,1,0,0,\ldots)|=3$, so $\Vert f\Vert\geq 3/4$. Show that in fact for every $x\in M$, $|f(x)|\leq\frac{3}{4}\Vert x\Vert$, and then conclude that $\Vert f\Vert=3/4$. This seems weird, but it happens because we are working in a subspace of $l_1$.
A Hahn-Banach extension has the property that $\Vert g\Vert=\Vert f\Vert=3/4$, and $g$ is determined by $\alpha=g(1,0,0,0,\ldots)$. Now note that
\begin{align*}
g(x_1,x_2,x_3,\ldots)&=g(3x_2,x_2,x_3,\ldots)+g(x_1-3x_2,0,0,\ldots)\\
&=f(3x_2,x_2,x_3,\ldots)+g(x_1-3x_2,0,0,\ldots)\\
&=3x_2+(x_1-3x_2)\alpha
\end{align*}
So we want to find $\alpha$ such that
$$|3x_2+(x_1-3x_2)\alpha|\leq\frac{3}{4}(|x_1|+|x_2|)$$
for all $x_1$ and $x_2$. Letting $x_1=tx_2$, this is equivalent to finding $\alpha$ for which
$$|3+(t-3)\alpha|\leq\frac{3}{4}(|t|+1)$$
for all $t\in\mathbb{R}$.
Maybe you can solve the rest.
HINT: Do not try to solve this by taking squares to get rid of the absolute values, because I think you will get something of order $8$ on $\alpha$ in the end. Instead, substitute for some values of $t$ which make the expression simpler, or see how it behaves when $t\to\infty$, etc...
