$\newcommand\R{\mathbb R}\newcommand\Q{\mathcal Q}$For mutually orthogonal vectors unit vectors $a=[a\_1,\dots,a\_n]^T$ and $b=[b\_1,\dots,b\_n]^T$ in $\R^n=\R^{n\times1}$ (so that $n\ge2$) and for all $x=[x\_1,x\_2]^T\in\R^2$, let
$$\|x\|\_{a,b}:=\sum\_{i=1}^n|a\_i x\_1+b\_i x\_2|\quad\text{and}\quad \|x\|\_1:=|x\_1|+|x\_2|.$$
Let
$$N\_{a,b}:=\max\{\|x\|\_1\colon x\in\R^2,\,\|x\|\_{a,b}\le1\},$$
the norm of the identity operator on $\R^2$ with respect to the norms $\|\cdot\|\_{a,b}$ and $\|\cdot\|\_1$.


Note that $N\_{a,b}=\sqrt2$ if $a=[1,1,0,\dots,0]^T/\sqrt2$ and $b=[-1,1,0,\dots,0]^T/\sqrt2$.



> 
> **Question:** Is it always true that $N\_{a,b}\le\sqrt2$, for all mutually orthogonal unit vectors $a$ and $b$ in $\R^n$?
> 
> 
> 



On a certain norm of the identity operator on $\mathbb R^2$
Simply observe that
$$\|x\|\_{a,b}=\|x\_1a+x\_2b\|\_1\,.$$
Thus, by orthogonality of $a,b$ and the easily-derived inequality $\|y\|\_2\le\|y\|\_1\le\sqrt{n}\|y\|\_2$ for any $y\in\mathbb{R}^n$, we have
\begin{eqnarray\*}
\|x\|\_{a,b} & = & \|x\_1a+x\_2b\|\_1 \\
 & \ge & \|x\_1a+x\_2b\|\_2=\sqrt{x\_1^2+x\_2^2}=\|x\|\_2 \\
& \ge & \frac{1}{\sqrt{2}}\|x\|\_1
\end{eqnarray\*}
for all $x\in\mathbb R^2$, and therefore $N\_{a,b}\le\sqrt{2}$, as desired.


