# On a certain norm of the identity operator on $\mathbb R^2$

$$\newcommand\R{\mathbb R}\newcommand\Q{\mathcal Q}$$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 $$\R^n$$?

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.