Geometric interpretation of the norm of a linear functional So with $X=\Bbb R^2$ and using the standard Euclidean norm. For $x=(x_1,x_2)$ set the functional to be $f(x)=x_1+x_2$ and let $A=\{x\in \Bbb R^2: f(x)=1\}$
1) I want to find $\|f\|$
So $\|f\|=\displaystyle\sup_{\|x\|=1} |f(x)|$ and $\|x\|=1$ means we are mapping points from the unit circle. $x_1+x_2\leq \sqrt{2}$. So then $\|f\|=\sqrt{2}$?
2) I want to calculate $d=\inf \{\|x\|:x\in A\}$
So $x=(x_1,x_2)$ such that $x_1+x_2=1$ is the line $x_2=1-x_1$. The closest thing to the origin in $\Bbb R^2$ is at $\left(\frac12, \frac12\right)$ so then $d=\frac{1}{\sqrt{2}}$?
3) Compare 1) & 2) and make and prove a conjecture about $\|f\|$ and $d$ when $f(x) = a_1x_1+a_2x_2$.
In this case we have $\|f\| = \frac{1}{d}$ - so it seems this is what I am meant to conjecture about. 
So we have $\|f\|$ is given by $\sup|x_1+x_2|$ where $\sqrt{x_1^2+ x_2^2}=1$
and $d$ is given by $\inf \sqrt{x_1^2 + x_2^2}$ where $x_1+x_2=1$
$\|f\|$ is the maximum sum that can be achieved by the sum of coordinates on the unit circle.
$d$ is the minimum distance from the origin a point can be when it lies on the line $x_2=1-x_1$.
I mean this is the same line geometrically, we just do different things to them. 
Now I am lost, any ideas?

You can write $f(x)=(x,e)$ where $e=(1,1)$ and $(\cdot,\cdot)$ is the usual inner product on $\mathbb{R}^{2}$. So,
$$
                 |f(x)| \le \|x\|\|e\| \implies \|f\| \le \|e\|.
$$
On the other hand, by the definition of $\|f\|$,
$$
           \|e\|^{2}=f(e) \le \|f\|\|e\| \implies \|e\|\le \|f\|.
$$
So $\|f\|=\|e\|=\sqrt{2}$. The geometric interpretation is that you are trying to find the maximum of the projection onto $e$ on the unit circle.
