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?
 A: (1) No. Your result happens to be correct but from $\|f\|\le K$ you cannot deduce that $\|f\|=K$. You have to prove it, for example like this:
We have $\|x\| = 1$ if and only if $\|x\|^2 = 1$ if and only if $x_1^2 + x_2^2 = 1$.
Also, $|x_1 + x_2| = \sqrt{(x_1 + x_2)^2} $ so that $|x_1 + x_2|^2 = x_1^2 + 2x_1 x_2 + x_2^2$.
Therefore on the unit circle $|x_1 + x_2|^2 = 1 + 2 x_1 x_2$.
To find the maximum value of $x_1 x_2$ we can write it as a function of the angle: $x_1 x_2 = \sin \theta \cos \theta$ then take the derivative 
$$ {\partial \over \partial \theta} \sin \theta \cos \theta = \cos^2 \theta - \sin^2 \theta$$
and set it zero:
$\cos^2 \theta - \sin^2 \theta = 0$ if and only if $\cos^2 \theta = \sin^2 \theta$ if and only if $\theta \in \{\pi/4, {3\pi\over 4}, {5\pi \over 4}, {7\pi \over 4}\}$.
For $\theta = \pi/4$ we have $\sin \theta = \cos \theta = {1\over \sqrt{2}}$ and therefore 
$$ 1+ 2 x_1 x_2 = 1 + 2\cdot{{1\over 2}} = 2$$
hence 
$$ \|f\| = \sqrt{2}$$
(2) Yes.
(3) It's not clear to me how you could do (2) so easily but can't do (3). If $f(x) = a_1 x_1 + a_2 x_2$ assume $a_2 \neq 0$ and then $f(x) = 1 =  a_1 x_1 + a_2 x_2$ is the line 
$$ x_2 = {1\over a_2}( 1 - a_1 x_1)$$
It seems to me that finding the closest point to the origin on this line is no more difficult than it was for the line $a_1 = a_2 = 1$ but maybe I am missing something. 
You express a point on the line in terms of $x_1$ then calculate the distance to $0$ and take the derivative to find the minimum:
$$ x_2 = {1\over a_2}( 1 - a_1 x_1)$$ 
so
$$ \|(x_1, x_2)\|^2 = x_1^2 +  ({1\over a_2}( 1 - a_1 x_1))^2 = x_1^2 +  {1\over a_2^2}( 1 - 2 a_1 x_1 + a_1^2 x_1^2)  = \dots$$
and taking the derivative
$$ 2 x_1 - 2 {a_1 \over a_2^2} + 2 {a_1^2 \over a_2^2} x_1 \stackrel{!}{=} 0$$
we get
$$ (2   + 2 {a_1^2 \over a_2^2} ) x_1 = 2 {a_2^2 + a_1^2 \over a_2^2}x_1= 2 {a_1 \over a_2^2}$$
so that
$$ x_1 =  {a_1 \over a_2^2} {a_2^2 \over a_2^2 + a_1^2 } = {a_1 \over a_2^2 + a_1^2 }$$
and
$$ x_2 = {1\over a_2}( 1 - a_1 x_1)={1\over a_2}( 1 -  {a_1^2 \over a_2^2 + a_1^2 })$$ 
A: 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.
