I've been trying to calculate the norm of $\phi\colon \mathbb R^3\rightarrow \mathbb R$ defined by $\phi(x,y,z)=1.2x-0.6y$.
I really don't know how to this. I did note that $\phi(-2y,y,z)=-3y$, therefore $|\phi(-2y,y,z)|\leq (3/\sqrt5)\|(-2y,y,z)\|$ meaning that $\|f\|\leq 3/\sqrt5$. my assumption that the other inequallity is also true, but I can't prove it.
Thanks
On
Consider the vector $${\bf a}:=\left({6\over5},\ -{3\over5}, \ 0\right)\ .$$ Then $\phi({\bf r})={\bf a}\cdot {\bf r}$ and therefore $$\bigl|\phi({\bf r})\bigr|\leq |{\bf a}|\>|{\bf r}|={3\over\sqrt{5}}\>|{\bf r}|\qquad\forall\> {\bf r}\in{\mathbb R}^3\ .$$ This proves $$\|\phi\|\leq {3\over\sqrt{5}}\ .$$ (Note that your proof of this fact is not watertight since you only considered "special" vectors ${\bf r}$.)
On the other hand, $\phi({\bf a})=|{\bf a}|^2={3\over\sqrt{5}}|{\bf a}|$, which implies $\|\phi\|\geq{3\over\sqrt{5}}$.
All in all we have proven that $$\|\phi\|={3\over\sqrt{5}}\ .$$ When the euclidean norm in ${\mathbb R}^3$ is replaced by the $\|\cdot\|_1$-norm we can argue as follows: The unit sphere with respect to this norm is the octahedron $$O:\quad |x|+|y|+|z|=1\ .$$ The linear functional $\phi$ takes its maximum on $O$ at one of the six vertices (this is a fundamental principle of linear programming), and a quick check shows that this is the vertex $(1,0,0)$. It follows that $$\|\phi\|=\phi(1,0,0)=1.2\ .$$
$$\|\varphi\|=\max_{\|v\|=1}\|\varphi(v)\|=\max_{x^2+y^2+z^2=1}|1.2x-0.6y|={3\over \sqrt{5}}$$
You can get the last equality several ways, one of which is by using Lagrange Multipliers on the constrained optimization problem. This will work when your functional is more complicated.
Here, though, you could note that since $\varphi$ is independent of $z$, then the constraint says $y=\pm\sqrt{1-x^2}$ so that $$|1.2x-0.6y|=|1.2x\mp 0.6\sqrt{1-x^2}\,|,$$ which is maximized when $x=\pm 2/\sqrt{5}\implies y=\mp 1/\sqrt{5}$, and the maximum is $3/\sqrt{5}$.
Looking at this might help solidify the geometry of this concept.