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


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.