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A Pythagorean inequality

Author: Russell M. Reid
Journal: Proc. Amer. Math. Soc. 123 (1995), 831-839
MSC: Primary 42C30; Secondary 15A99
MathSciNet review: 1231041
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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\Vert {\sum {{a_i}{v_i}} } \right\Vert}^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\Vert {\sum {{a_i}{v_i}} } \right\Vert}^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 $,

$\displaystyle d_0^2 = {d^2} + \frac{1}{2}\left( {{r^2} - {d^2} - \sqrt {{{({r^2} + {d^2})}^2} - 4{d^2}{\delta ^2}} } \right)$


$\displaystyle D_0^2 = {D^2} + \frac{1}{2}\left( {{r^2} - {D^2} + \sqrt {{{({r^2} + {D^2})}^2} - 4{D^2}{\delta ^2}} } \right).$

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\Vert {a{v_0} + b{v_1}} \right\Vert^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\Vert {\sum\nolimits_{i = 0}^n {{a_i}{v_i}} } \right\Vert \geq {V_n}/{2^{n/2}}$.

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Keywords: Norm inequality, Gram matrix, nonharmonic Fourier series
Article copyright: © Copyright 1995 American Mathematical Society

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