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

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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.

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