calculating norm of a linear functional I was solving a problem regarding Hanh-Banach extension of a linear functional and I encountered the following problem as a subpart of the original one -
Consider the normed linear space ($ \mathbb C^2$, $ $|| ||$_\infty$). $f$ is a linear functional on the space given by, $f(x,y)=ax+by$, for some constant a,b. Calculate norm of the functional $f$.
Now, it can be easily shown that $\frac {|f(x,y)|} {||(x,y)||_\infty} \leq |a|+|b|$ $\Rightarrow$ $||f|| \leq |a|+|b|$ . I know that the norm of the functional is $|a|+|b|$. However I am not being able to establish the equality $||f|| = |a|+|b|$. Someone kindly help me in proving the opposite inequaity. I have seen many hints that were available online and still have not been able to figure it out. So please be a little elaborate.
Thank you for your time.
 A: We can put $\tilde{x}=\overline{a}/\vert a \vert$ and $\tilde{y}=\overline{b}/\vert b \vert$. Then we have $ \Vert (\tilde{x}, \tilde{y}) \Vert_\infty = 1 $ and $$f(\tilde{x}, \tilde{y})= a \overline{a}/\vert a \vert + b\overline{b}/b = \vert a \vert + \vert b \vert.$$
Hence, we get
$$ \Vert f \Vert = \sup_{\Vert (x,y) \Vert_\infty=1} \vert f(x,y) \vert \geq \vert f(\tilde{x}, \tilde{y}) \vert=\vert a \vert + \vert b \vert.$$
In fact this is the general strategy to get the norm of a functional. When we have an upper bound (and believe that is indeed the norm), then we try to construct vectors, we will yield better approximations from below. Here we are lucky and can directly construct one vector where the norm is attained.
A: Note that the map
$$(\Bbb{C}^2,\|\cdot\|_1) \to (\Bbb{C}^2, \|\cdot\|_\infty)^*$$
where $(a,b) \in \Bbb{C}^2$ maps to the functional $f(x,y) = ax+by$ is an isometric isomorphism. In particular the operator norm of $f$ is equal to the $\|\cdot\|_1$-norm of the pair which induces the functional:
$$\|f\| = \|(a,b)\|_1 = |a|+|b|.$$
