A Pythagorean inequality

Abstract:

Let $\{ {v_1},{v_2},{v_3}, \ldots \}$ be a sequence of elements of a Hilbert space, and suppose that (one or both of) the inequalities ${d^2}\sum {a_i^2 \leq {{\left \| {\sum {{a_i}{v_i}} } \right \|}^2} \leq {D^2}} \sum {a_i^2}$ hold for every finite sequence of scalars $\{ {a_i}\}$. If an element ${v_0}$ is adjoined to $\{ {v_i}\}$, then the resulting set satisfies (one or both of) $d_0^2\sum {a_i^2 \leq {{\left \| {\sum {{a_i}{v_i}} } \right \|}^2} \leq D_0^2} \sum {a_i^2}$, where, denoting the norm of ${v_0}$ by r and its distance from the closed linear span of the ${v_i}$ by $\delta$, \[ d_0^2 = {d^2} + \frac {1}{2}\left ( {{r^2} - {d^2} - \sqrt {{{({r^2} + {d^2})}^2} - 4{d^2}{\delta ^2}} } \right )\] and \[ D_0^2 = {D^2} + \frac {1}{2}\left ( {{r^2} - {D^2} + \sqrt {{{({r^2} + {D^2})}^2} - 4{D^2}{\delta ^2}} } \right ).\] Both bounds are best possible. If ${v_0}$ is in the span of the original set, the expressions above simplify to ${d_0} = 0$ and $D_0^2 = {D^2} + {r^2}$. If the original set is a single unit vector ${v_1}$, so $d = D = 1$, and if ${v_0} \bot {v_1}$ is a unit vector so $\delta = 1$, then the above is $({a^2} + {b^2}) \leq {\left \| {a{v_0} + b{v_1}} \right \|^2} \leq ({a^2} + {b^2})$, the Pythagorean Theorem. Several consequences are deduced. If ${v_i}$ are unit vectors, $\sum {a_i^2 = 1}$, and ${\delta _i}$ is the distance from ${v_i}$ to the span of its predecessors (so that the volume of the parallelotope spanned by the ${v_i}$ is ${V_n} = {\delta _1}{\delta _2} \cdots {\delta _n}$), the above result is used to show that $\left \| {\sum \nolimits _{i = 0}^n {{a_i}{v_i}} } \right \| \geq {V_n}/{2^{n/2}}$.