Two vectors closed to a third vectors. What can we know about the inner product of these two vector?

Question

Consider in a space of unknown dimension, all we know is that vectors $\vec{v}$ and $\vec{u}$ are both very "close" to $\vec{w}$. To be more specific, they have inner-products $|\langle\vec{v},\vec{w}\rangle|\geq\delta$ and $|\langle\vec{u},\vec{w}\rangle|\geq\delta$. The three vectors are normalized. Can we infer a lower bound on $|\langle\vec{v},\vec{u}\rangle|$ based on this? Thanks!

Answer 1

With the additional assumption that the inner product space $H$ is complete (i.e. a Hilbert space) over $K=\Bbb R$ or $\Bbb C$ we can use the existence of an orthogonal projection $P$ onto the linear subspace generated by $w$. Then $$ u = Pu + (I-P)u = u_1 w + a \quad \text{with } u_1 \in K, \, a \in H, \,\langle a, w \rangle = 0 \, ,\\ v = Pv + (I-P)v = v_1 w + b \quad \text{with } v_1 \in K, \,b \in H, \,\langle b, w \rangle = 0 \, .\\ $$ It follows that $$ |u_1| = |\langle u, w \rangle| \ge \delta \, , \\ |v_1| = |\langle v, w \rangle| \ge \delta \, , \\ $$ and $$ \langle u, v \rangle = u_1 \overline{v_1} + \langle a, b \rangle $$ because the mixed terms are zero. Using the Cauchy-Schwarz inequality and the Pythagorean theorem we get $$ |\langle u, v \rangle| \ge |u_1 v_1| - \Vert a \Vert \cdot \Vert b \Vert = |u_1 v_1| - \sqrt{1 - |u_1|^2} \cdot \sqrt{1 - |v_1|^2} \\ \ge \delta^2 - (1 - \delta^2) = 2 \delta^2 - 1 \, . $$ This a “nontrivial” estimate if $\delta > 1/\sqrt 2$, and in particular $|\langle u, v \rangle|$ is close to one if $\delta $ is close to one.