A Pythagorean inequality
Authors: Reid Russell M. 
Published: June 2010

Let

{

v
1

,

v
2

,

v
3

,
…
}

\{ {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
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="d squared sigma-summation a Subscript i Superscript 2 Baseline less-than-or-equal-to double-vertical-bar sigma-summation a Subscript i Baseline v Subscript i Baseline double-vertical-bar squared less-than-or-equal-to upper D squared sigma-summation a Subscript i Superscript 2">

d
2

∑

a
i
2

≤

‖

∑

a
i

v
i

‖

2

≤

D
2

∑

a
i
2

{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

}

\{ {a_i}\}

. If an element

v
0

{v_0}

is adjoined to

{

v
i

}

\{ {v_i}\}

, then the resulting set satisfies (one or both of)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="d 0 squared sigma-summation a Subscript i Superscript 2 Baseline less-than-or-equal-to double-vertical-bar sigma-summation a Subscript i Baseline v Subscript i Baseline double-vertical-bar squared less-than-or-equal-to upper D 0 squared sigma-summation a Subscript i Superscript 2">

d
0
2

∑

a
i
2

≤

‖

∑

a
i

v
i

‖

2

≤

D
0
2

∑

a
i
2

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

{v_0}

by r and its distance from the closed linear span of the

v
i

{v_i}

by

δ
\delta

,
\[

d
0
2

=

d
2

+

1
2

(

r
2

−

d
2

−

(

r
2

+

d
2

)

2

−
4

d
2

δ
2

)

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
\[
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="upper D 0 squared equals upper D squared plus one half left-parenthesis r squared minus upper D squared plus StartRoot left-parenthesis r squared plus upper D squared right-parenthesis squared minus 4 upper D squared delta squared EndRoot right-parenthesis period">

D
0
2

=

D
2

+

1
2

(

r
2

−

D
2

+

(

r
2

+

D
2

)

2

−
4

D
2

δ
2

)

.

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

{v_0}

is in the span of the original set, the expressions above simplify to

d
0

=
0

{d_0} = 0

and

D
0
2

=

D
2

+

r
2

D_0^2 = {D^2} + {r^2}

. If the original set is a single unit vector

v
1

{v_1}

, so

d
=
D
=
1

d = D = 1

, and if

v
0

⊥

v
1

{v_0} \bot {v_1}

is a unit vector so

δ
=
1

\delta = 1

, then the above is

(

a
2

+

b
2

)
≤

‖

a

v
0

+
b

v
1

‖

2

≤
(

a
2

+

b
2

)

({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

{v_i}

are unit vectors,

∑

a
i
2

=
1

\sum {a_i^2 = 1}

, and

δ
i

{\delta _i}

is the distance from

v
i

{v_i}

to the span of its predecessors (so that the volume of the parallelotope spanned by the

v
i

{v_i}

is

V
n

=

δ
1

δ
2

⋯

δ
n

{V_n} = {\delta _1}{\delta _2} \cdots {\delta _n}

), the above result is used to show that

‖

∑

i
=
0

n

a
i

v
i

‖

≥

V
n

/

2

n

/

2

\left \| {\sum \nolimits _{i = 0}^n {{a_i}{v_i}} } \right \| \geq {V_n}/{2^{n/2}}

.