I'm trying to solve this problem:
The unit sphere is $S = \{(b,a) \in \mathbb R^2 \mid \sup_{t \in [0,1]} |b+ta| = 1\}$. Since the function $f(t)=|b+ta|$ is continuous on the compact set $[0,1]$, $\sup_{t \in [0,1]} |b+ta| = \max_{t \in [0,1]} |b+ta|$. As such, $$S = \left \{(b,a) \in \mathbb R^2 \,\middle\vert\, \max_{x \in [0,1]} |ax+b| = 1 \right \}$$
For a given $(b,a)\in \mathbb R^2$, $y=ax+b$ is the equation of a line. It seems very close to me, but I'm unable to proceed.
Could you please shed me some light? Thank you so much!
I suggest to look what is happening in each quadrant. For example, if $a \geq 0$ and $b \geq 0$, then $b \leq b+ta \leq b+a$, and therefore $\sup_{t\in[0,1]} |b+ta| = b+a$. Hence, $a+b = 1$. If $a \geq 0$ and $b \leq 0$, then again $b \leq b+ta \leq b+a$. But then we need to inspect more carefully. Either $a+b \leq 0$, then $\sup_{t\in[0,1]} |b+ta| = |b|.$ Hence $b = -1.$ Or $a+b \geq 0,$ then $\sup_{t\in[0,1]} |b+ta| = \max\{a+b,-b\}.$ If $a+b \geq -b$ or $a+2b \geq 0$, then $\max \{a+b,-b\} = a+b$, and we have again $a+b = 1$. Otherwise, $\max\{a+b,-b\} = -b$, and again we have $b = -1$.
In summary, for $a \geq 0$, the unit sphere is defined as $b = -1$, if $a+2b \leq 0$, and $a+b =1$ otherwise.