On the approximate Birkhoff orthogonality

Question

In an inner product space $(X,\langle \cdot \vert \cdot \rangle)$, an approximate orthogonality with $\varepsilon \in [0,1)$ is defined as follows: $$x\perp^{\varepsilon}y \iff |\langle\,x\,|\,y\,\rangle| \leq \varepsilon \| x\|\,\|y\|.$$ On the other hand, in a normed space $(X,\Vert \cdot \Vert )$, an approxmiate Birkhoff orthogonality with $\varepsilon \in [0,1)$ is defined as follows: $x\perp_B^{\varepsilon} y$ iff $$\forall \lambda \in \mathbb{K}\; : \; \Vert x+\lambda y\Vert^2 \geq \Vert x\Vert^2 -2\varepsilon \Vert x\Vert \Vert \lambda y\Vert .$$ How can I show that $\perp^{\varepsilon}_B =\perp^{\varepsilon}$ in an inner product space?

Answer 1

Suppose that $x\perp^\varepsilon y$. Then \begin{align*} \|x+\lambda y\|^2 &=\|x\|^2+\|\lambda y\|^2+2\operatorname{Re}\langle x,\lambda y\rangle\\[0.3cm] &\geq\|x\|^2+\|\lambda y\|^2-2|\langle x,\lambda y\rangle|\\[0.3cm] &\geq\|x\|^2-2\varepsilon\,\| x\|\,\|\lambda y\|. \end{align*}

Conversely, suppose that $x\perp_B^\varepsilon y$. We have $$ \|x\|^2-2\varepsilon\|x\|\,\|\lambda y\|\leq\|x+\lambda y\|^2=\|x\|^2+\|\lambda y\|^2+2\operatorname{Re}\langle x,\lambda y\rangle. $$ The left-hand-side does not change if we replace $x$ with $-x$. Thus $$ \|x\|^2-2\varepsilon\|x\|\,\|\lambda y\|\leq\|-x+\lambda y\|^2=\|x\|^2+\|\lambda y\|^2-2\operatorname{Re}\langle x,\lambda y\rangle. $$ This gives us $$\tag1 2\operatorname{Re}\langle x,\lambda y\rangle\leq2\varepsilon\|x\|\,\|\lambda y\|+|\lambda|^2\,\|y\|. $$ Write $\langle x,y\rangle=\beta\,|\langle x,y\rangle|$, with $|\beta|=1$. Putting $\lambda=\beta\,t$, with $t>0$, we have $$\tag2 2\operatorname{Re}\langle x,\lambda y\rangle=2t\,|\langle x,y\rangle|, $$ while $$\tag3 2\varepsilon\|x\|\,\|\lambda y\|+|\lambda|^2\,\|y\| =2t\,\varepsilon\|x\|\,\| y\|+t^2\,\|y\| $$ Using $(2)$ and $(3)$, now $(1)$ reduces to $$ |\langle x,y\rangle|\leq\varepsilon\|x\|\,\|y\|+\frac t2\,\|y\|. $$ As this works for all $t>0$, $$\tag4 |\langle x,y\rangle|\leq\varepsilon\|x\|\,\|y\|. $$