Hello i'm currently trying to find this :
Let $f\in \mathcal{L}(\mathbb{R}^2)$ and $\| \ \|_{\infty}$ be the uniform norm over $\mathbb{R}^2$ defined as, $\forall (x,y) \in \mathbb{R}^2, \|(x,y)\|_{\infty}=\max_{\{ (x,y)\in\mathbb{R}^2, \| (x,y)\|_{\infty} \leq 1 \}}(|x|,|y|). $
Find $\|f\|$.
What i did is :
Since $f\in \mathcal{L}(\mathbb{R}^2)$, then $\exists a,b,c,d\in \mathbb{R}$ such that $\forall v \in \mathbb{R}^2, f(v)=(ax+by,cx+dy)$.
We have to find $\|f\|=\max_{\{ (x,y)\in\mathbb{R}^2, \| (x,y)\|_{\infty} \leq 1 \}} \|f(x,y)\|_{\infty} $
I noted that, \begin{align}\max_{\{ (x,y)\in\mathbb{R}^2, \| (x,y)\|_{\infty} \leq 1 \}} \|f(x,y)\|_{\infty} &=\max_{|x| \leq 1 \text{ and } |y| \leq 1} \|f(x,y)\|_{\infty} \\ &=\max_{|x| \leq 1 \text{ and } |y| \leq 1} \|f(x,y)\|_{\infty} \\&=\max_{|x| \leq 1 \text{ and } |y| \leq 1}( \max_{|x| \leq 1 \text{ and } |y| \leq 1} (|ax+by|,|cx+dy|)) \end{align}
Now $|ax+by| \leq |a||x]+|b||y|\leq |a| + |b|$ is an upper bound and is atteined for $|a|=a* sign(a), |b|=b*sign(b)$ so it is the maximum of $|ax+by|$ over $ \{(x,y)\in \mathbb{R}^2 \text{ such that }|x| \leq 1 \text{ and } |y| \leq 1 \}$.
So we have $ \max_{|x| \leq 1 \text{ and } |y| \leq 1} (|ax+by|,|cx+dy|))=|a|+|b| $ or $|c|+|d|$.
Now suppose $ \max_{|x| \leq 1 \text{ and } |y| \leq 1} (|ax+by|,|cx+dy|))=|a|+|b| $
Then we have, \begin{align} \max_{|x| \leq 1 \text{ and } |y| \leq 1}( \max_{|x| \leq 1 \text{ and } |y| \leq 1} (|ax+by|,|cx+dy|)) = \max_{|x| \leq 1 \text{ and } |y| \leq 1}(|a|+|b|,|cx+dy|) \end{align}
and it is here that i'm not sure the second max gives what i've just written.
If it is it is easy to conclude by setting $x=sign(c)$ and $y=sign(d)$ so we have $|cx+dy|=|(c*sign(c)+d*sign(d))|=|(|c|+|d|)|=|c|+|d| \leq |a|+|b|$.
We do the same for $ \max_{|x| \leq 1 \text{ and } |y| \leq 1} (|ax+by|,|cx+dy|))=|c|+|d| $
And we conclude that $ \| f\|=\max_{|x| \leq 1 \text{ and } |y| \leq 1}(|a|+|b|,|c|+|d|)$