View source for Polarization identity

{{short description|Formula relating the norm and the inner product in a inner product space}}

[[File:Parallelogram law.svg|thumb|Vectors involved in the polarization identity <math>2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2.</math> ]]
In [[Linear algebra|linear algebra]], a branch of [[Mathematics|mathematics]], the '''polarization identity''' is any one of a family of formulas that express the inner product of two vectors in terms of the [[Norm (mathematics)|norm]] of a [[Normed vector space|normed vector space]]. 
If a [[Norm (mathematics)|norm]] arises from an inner product then the polarization identity can be used to express this inner product entirely in terms of the norm. The polarization identity shows that a norm can arise from at most one inner product; however, there exist norms that do not arise from any inner product. 

The norm associated with any [[Inner product space|inner product space]] satisfies the [[Parallelogram law|parallelogram law]]: <math>\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2.</math> 
In fact, as observed by [[Biography:John von Neumann|John von Neumann]],{{sfn|Lax|2002|p=53}} the parallelogram law characterizes those norms that arise from inner products. 
Given a [[Normed vector space|normed space]] <math>(H, \|\cdot\|)</math>, the parallelogram law holds for <math>\|\cdot\|</math> if and only if there exists an inner product <math>\langle \cdot, \cdot \rangle</math> on <math>H</math> such that <math>\|x\|^2 = \langle x,\ x\rangle</math> for all <math>x \in H,</math> in which case this inner product is uniquely determined by the norm via the polarization identity.<ref name=Blanchard>{{cite book|author=[[Biography:Philippe Blanchard|Philippe Blanchard]], Erwin Brüning|chapter=Proposition 14.1.2 (Fréchet–von Neumann–Jordan)|chapter-url=https://books.google.com/books?id=1g2rikccHcgC&pg=PA192|page=192|title=Mathematical methods in physics: distributions, Hilbert space operators, and variational methods|year=2003|publisher=Birkhäuser|isbn=0817642285}}</ref><ref name=Teschl>{{cite book|author=[[Biography:Gerald Teschl|Gerald Teschl]]|title=Mathematical methods in quantum mechanics: with applications to Schrödinger operators|chapter=Theorem 0.19 (Jordan–von Neumann)|page=19|url=https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/|isbn=978-0-8218-4660-5|year=2009|publisher=American Mathematical Society Bookstore}}</ref>

==Polarization identities==

Any [[Inner product space|inner product]] on a vector space induces a norm by the equation
<math display=block>\|x\| = \sqrt{\langle x, x \rangle}.</math>
The polarization identities reverse this relationship, recovering the inner product from the norm.
Every inner product satisfies:
<math display=block>\|x + y\|^2 = \|x\|^2 + \|y\|^2 + 2\operatorname{Re}\langle x, y \rangle \qquad \text{ for all vectors } x, y.</math>

Solving for <math>\operatorname{Re}\langle x, y \rangle</math> gives the formula <math>\operatorname{Re}\langle x, y \rangle = \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right).</math> If the inner product is real then <math>\operatorname{Re}\langle x, y \rangle = \langle x, y \rangle</math> and this formula becomes a polarization identity for real inner products.

===Real vector spaces===

If the vector space is over the [[Real number|real number]]s then the polarization identities are:{{sfn|Schechter|1996|pp=601-603}}

<math display="block">\begin{alignat}{4}
\langle x, y \rangle
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x\|^2 + \|y\|^2 - \|x-y\|^2\right). \\[3pt]
\end{alignat}</math>


These various forms are all equivalent by the [[Parallelogram law|parallelogram law]]:<ref group="proof" name="RealPartEquivalentFormsAndFormulasProof" />

<math display="block">2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2.</math>


This further implies that <math>L^p</math> class is not a [[Hilbert space]] whenever <math>p\neq 2</math>, as the parallelogram law is not satisfied. For the sake of counterexample, consider <math>x=1_A</math> and <math>y=1_B</math> for any two disjoint subsets <math>A,B</math> of general domain <math>\Omega\subset\mathbb{R}^n</math> and compute the measure of both sets under parallelogram law.
===Complex vector spaces===

For vector spaces over the [[Complex number|complex number]]s, the above formulas are not quite correct because they do not describe the imaginary part of the (complex) inner product. 
However, an analogous expression does ensure that both real and imaginary parts are retained. 
The complex part of the inner product depends on whether it is [[Antilinear map|antilinear]] in the first or the second argument. 
The notation <math>\langle x | y \rangle,</math> which is commonly used in physics will be assumed to be [[Antilinear map|antilinear]] in the {{em|first}} argument while <math>\langle x,\, y \rangle,</math> which is commonly used in mathematics, will be assumed to be antilinear its the {{em|second}} argument. 
They are related by the formula:
<math display=block>\langle x,\, y \rangle = \langle y \,|\, x \rangle \quad \text{ for all } x, y \in H.</math>

The real part of any inner product (no matter which argument is antilinear and no matter if it is real or complex) is a symmetric bilinear map that for any <math>x, y \in H</math> is always equal to:{{sfn|Schechter|1996|pp=601-603}}<ref group=proof name=RealPartEquivalentFormsAndFormulasProof />
<math display=block>\begin{alignat}{4}
R(x, y) 
:&= \operatorname{Re} \langle x \mid y \rangle = \operatorname{Re} \langle x, y \rangle \\
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \\
&= \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right) \\[3pt]
&= \frac{1}{2} \left(\|x\|^2 + \|y\|^2 - \|x-y\|^2\right). \\[3pt]
\end{alignat}</math>

It is always a symmetric map, meaning that<ref group=proof name=RealPartEquivalentFormsAndFormulasProof />
<math display=block>R(x, y) = R(y, x) \quad \text{ for all } x, y \in H,</math>
and it also satisfies:<ref group=proof name=RealPartEquivalentFormsAndFormulasProof /> 
<math display=block>R(y, ix) = - R(x, iy) \quad \text{ for all } x, y \in H.</math>
Thus <math>R(ix, y) = - R(x, iy),</math> which in plain English says that to move a factor of <math>i</math> to the other argument, introduce a negative sign. 

{{collapse top|title={{anchor|Proof of formulas and equivalent forms}}Proof of properties of <math>R</math>|left=true}}
Let 
<math display=block>R(x, y) := \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right).</math> 
Then <math>2\|x\|^2 + 2\|y\|^2 = \|x+y\|^2 + \|x-y\|^2</math> implies 
<math display=block>R(x, y) = \frac{1}{4} \left(\left(2\|x\|^2 + 2\|y\|^2 - \|x-y\|^2\right) - \|x-y\|^2\right) = \frac{1}{2} \left(\|x\|^2 + \|y\|^2 - \|x-y\|^2\right)</math> 
and 
<math display=block>R(x, y) = \frac{1}{4} \left(\|x+y\|^2 - \left(2\|x\|^2 + 2\|y\|^2 - \|x+y\|^2\right)\right) = \frac{1}{2} \left(\|x+y\|^2 - \|x\|^2 - \|y\|^2\right).</math> 

Moreover, 
<math display=block>4R(x, y) = \|x+y\|^2 - \|x-y\|^2 = \|y+x\|^2 - \|y-x\|^2 = 4R(y, x),</math> 
which proves that <math>R(x, y) = R(y, x).</math> 

From <math>1 = i (-i)</math> it follows that <math>y-ix = i(-iy-x) = -i(x+iy)</math> and <math>y+ix = i(-iy+x) = i(x-iy)</math> so that 
<math display=block>-4R(y, ix) = \|y-ix\|^2 - \|y+ix\|^2 = \|(-i)(x+iy)\|^2 - \|i(x-iy)\|^2 = \|x+iy\|^2 - \|x-iy\|^2 = 4R(x, iy),</math> 
which proves that <math>R(y, ix) = - R(x, iy).</math> 
<math>\blacksquare</math>
<!--Proof:
<math display=block>\begin{alignat}{4}
R(y,ix)&=\frac{1}{2}\left(\langle y,ix \rangle+\langle ix,y \rangle\right) \\
&=\frac{1}{2i}\left(\langle iy,ix \rangle-\langle ix,iy \rangle\right) \\
&=\frac{1}{2i}\left(-i\langle iy,x \rangle-i\langle x,iy \rangle\right) \\
&=-\frac{1}{2}\left(\langle x,iy \rangle+\langle iy,x \rangle\right) \\
&=-R(x,iy)
\end{alignat}</math>-->

{{collapse bottom}}

Unlike its real part, the imaginary part of a complex inner product depends on which argument is antilinear. 

'''Antilinear in first argument'''

The polarization identities for the inner product <math>\langle x \,|\, y \rangle,</math> which is [[Antilinear map|antilinear]] in the {{em|first}} argument, are
:<math>\begin{alignat}{4}
\langle x \,|\, y \rangle 
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2 - i\|x + iy\|^2 + i\|x - iy\|^2\right) \\
&= R(x, y) - i R(x, iy) \\
&= R(x, y) + i R(ix, y) \\
\end{alignat}</math>
where <math>x, y \in H.</math> 
The second to last equality is similar to the formula expressing a linear functional <math>\varphi</math> in terms of its real part: <math>\varphi(y) = \operatorname{Re} \varphi(y) - i (\operatorname{Re} \varphi)(i y).</math> 

'''Antilinear in second argument'''

The polarization identities for the inner product <math>\langle x, \ y \rangle,</math> which is [[Antilinear map|antilinear]] in the {{em|second}} argument, follows from that of <math>\langle x \,|\, y \rangle</math> by the relationship: 
<math>\langle x, \ y \rangle := \langle y \,|\, x \rangle = \overline{\langle x \,|\, y \rangle} \quad \text{ for all } x, y \in H.</math> 
So for any <math>x, y \in H,</math>{{sfn|Schechter|1996|pp=601-603}}
:<math>\begin{alignat}{4}
\langle x,\, y \rangle 
&= \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2\right) \\
&= R(x, y) + i R(x, iy) \\
&= R(x, y) - i R(ix, y). \\
\end{alignat}</math>

This expression can be phrased symmetrically as:<ref>{{Cite web|last=Butler|first=Jon|date=20 June 2013|title=norm - Derivation of the polarization identities?|url=https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities|url-status=live|archive-url=https://archive.today/raxIv|archive-date=14 October 2020|access-date=2020-10-14|website=Mathematics Stack Exchange}} See Harald Hanche-Olson's answer.</ref>
<math display=block>\langle x, y \rangle = \frac{1}{4} \sum_{k=0}^3 i^k \left\|x + i^k y\right\|^2.</math>

'''Summary of both cases'''

Thus if <math>R(x, y) + i I(x, y)</math> denotes the real and imaginary parts of some inner product's value at the point <math>(x, y) \in H \times H</math> of its domain, then its imaginary part will be:
<math display=block>I(x, y) ~=~ 
\begin{cases}
~R({\color{red}i} x, y)  & \qquad \text{ if antilinear in the } {\color{red}1} \text{st argument} \\
~R(x, {\color{blue}i} y) & \qquad \text{ if antilinear in the } {\color{blue}2} \text{nd argument} \\
\end{cases}</math>
where the scalar <math>i</math> is always located in the same argument that the inner product is antilinear in.

Using <math>R(ix, y) = - R(x, iy),</math> the above formula for the imaginary part becomes:
<math display=block>I(x, y) ~=~ 
\begin{cases}
-R(x, {\color{black}i} y) & \qquad \text{ if antilinear in the } {\color{black}1} \text{st argument} \\
-R({\color{black}i} x, y) & \qquad \text{ if antilinear in the } {\color{black}2} \text{nd argument} \\
\end{cases}</math>

==Reconstructing the inner product==

In a normed space <math>(H, \|\cdot\|),</math> if the [[Parallelogram law|parallelogram law]]
<math display=block>\|x+y\|^2 ~+~ \|x-y\|^2 ~=~ 2\|x\|^2+2\|y\|^2</math>
holds, then there exists a unique inner product <math>\langle \cdot,\ \cdot\rangle</math> on <math>H</math> such that <math>\|x\|^2 = \langle x,\ x\rangle</math> for all <math>x \in H.</math>{{sfn|Schechter|1996|pp=601-603}}{{sfn|Lax|2002|p=53}}

{{math proof|proof=
We will only give the real case here; the proof for complex vector spaces is analogous.

By the above formulas, if the norm is described by an inner product (as we hope), then it must satisfy
<math display=block>\langle x, \ y \rangle = \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \quad \text{ for all } x, y \in H,</math>

which may serve as a definition of the unique candidate <math>\langle \cdot, \cdot \rangle  </math> for the role of a suitable inner product. Thus, the uniqueness is guaranteed.

It remains to prove that this formula indeed defines an inner product and that this inner product induces the norm <math>\|\cdot\|.</math> 
Explicitly, the following will be shown:

#<math>\langle x, x \rangle = \|x\|^2, \quad x \in H</math>
#<math>\langle x, y \rangle = \langle y, x \rangle, \quad x, y \in H</math>
#<math>\langle x+z, y\rangle = \langle x, y\rangle + \langle z, y\rangle \quad \text{ for all } x, y, z \in H,</math> 
#<math>\langle \alpha x, y \rangle = \alpha\langle x, y \rangle \quad \text{ for all } x, y \in H \text{ and all } \alpha \in \R</math>
(This axiomatization omits positivity, which is implied by (1) and the fact that <math>\|\cdot\|</math> is a norm.)

For properties (1) and (2), substitute: <math display=inline>\langle x, x \rangle = \frac{1}{4} \left(\|x+x\|^2 - \|x-x\|^2\right) = \|x\|^2,</math> and <math>\|x-y\|^2 = \|y-x\|^2.</math>

For property (3), it is convenient to work in reverse. 
It remains to show that 
<math display=block>\|x+z+y\|^2 - \|x+z-y\|^2 \overset{?}{=} \|x+y\|^2 - \|x-y\|^2 + \|z+y\|^2 - \|z-y\|^2</math>
or equivalently,
<math display=block>2\left(\|x+z+y\|^2 + \|x-y\|^2\right) - 2\left(\|x+z-y\|^2 + \|x+y\|^2\right) \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2.</math>

Now apply the parallelogram identity:
<math display=block>2\|x+z+y\|^2 + 2\|x-y\|^2 = \|2x+z\|^2 + \|2y+z\|^2</math>
<math display=block>2\|x+z-y\|^2 + 2\|x+y\|^2 = \|2x+z\|^2 + \|z-2y\|^2</math>
Thus it remains to verify:
<math display=block>\cancel{\|2x+z\|^2} + \|2y+z\|^2 - (\cancel{\|2x+z\|^2} + \|z-2y\|^2) \overset{?}{{}={}} 2\|z+y\|^2 - 2\|z-y\|^2</math>
<math display=block>\|2y+z\|^2 - \|z-2y\|^2 \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2</math>

But the latter claim can be verified by subtracting the following two further applications of the parallelogram identity:
<math display=block>\|2y+z\|^2 + \|z\|^2 = 2\|z+y\|^2 + 2\|y\|^2</math>
<math display=block>\|z-2y\|^2 + \|z\|^2 = 2\|z-y\|^2 + 2\|y\|^2</math>

Thus (3) holds.

It can be verified by induction that (3) implies (4), as long as <math>\alpha \in \Z.</math> 
But "(4) when <math>\alpha \in \Z</math>" implies "(4) when <math>\alpha \in \Q</math>". 
And any positive-definite, real-valued, <math>\Q</math>-bilinear form satisfies the [[Cauchy–Schwarz inequality]], so that <math>\langle \sdot,\sdot \rangle</math> is continuous. 
Thus <math>\langle \sdot,\sdot \rangle</math> must be <math>\R</math>-linear as well. 
}}

Another necessary and sufficient condition for there to exist an inner product that induces a given norm <math>\|\cdot\|</math> is for the norm to satisfy [[Ptolemy's inequality]], which is:<ref>{{Cite journal|last=Apostol|first=Tom M.|date=1967|title=Ptolemy's Inequality and the Chordal Metric|url=https://www.tandfonline.com/doi/pdf/10.1080/0025570X.1967.11975804|journal=Mathematics Magazine|volume=40|issue=5|pages=233–235| language=en| doi=10.2307/2688275|jstor=2688275}}</ref>
<math display=block>\|x - y\| \, \|z\| ~+~ \|y - z\| \, \|x\| ~\geq~ \|x - z\| \, \|y\| \qquad \text{ for all vectors } x, y, z.</math>

==Applications and consequences==

If <math>H</math> is a complex Hilbert space then <math>\langle x \mid y \rangle</math> is real if and only if its imaginary part is <math>0 = R(x, iy) = \frac{1}{4} \left(\|x+iy\|^2 - \|x-iy\|^2\right),</math> which happens if and only if <math>\|x+iy\| = \|x-iy\|.</math> 
Similarly, <math>\langle x \mid y \rangle</math> is (purely) imaginary if and only if <math>\|x+y\| = \|x-y\|.</math> 
For example, from <math>\|x+ix\| = |1+i| \|x\| = \sqrt{2} \|x\| = |1-i| \|x\| = \|x-ix\|</math> it can be concluded that <math>\langle x | x \rangle</math> is real and that <math>\langle x | ix \rangle</math> is purely imaginary. 

===Isometries===

If <math>A : H \to Z</math> is a [[Linear map|linear]] [[Isometry|isometry]] between two Hilbert spaces (so <math>\|A h\| = \|h\|</math> for all <math>h \in H</math>) then 
<math display=block>\langle A h, A k \rangle_Z = \langle h, k \rangle_H \quad \text{ for all } h, k \in H;</math>
that is, linear isometries preserve inner products. 

If <math>A : H \to Z</math> is instead an [[Antilinear map|antilinear]] isometry then 
<math display=block>\langle A h, A k \rangle_Z = \overline{\langle h, k \rangle_H} = \langle k, h \rangle_H \quad \text{ for all } h, k \in H.</math> 

===Relation to the law of cosines===

The second form of the polarization identity can be written as
<math display=block>\|\textbf{u}-\textbf{v}\|^2 = \|\textbf{u}\|^2 + \|\textbf{v}\|^2 - 2(\textbf{u} \cdot \textbf{v}).</math>

This is essentially a vector form of the [[Law of cosines|law of cosines]] for the [[Triangle|triangle]] formed by the vectors <math>\textbf{u}, \textbf{v},</math> and <math>\textbf{u}-\textbf{v}.</math> 
In particular,
<math display=block>\textbf{u}\cdot\textbf{v} = \|\textbf{u}\|\,\|\textbf{v}\| \cos\theta,</math>
where <math>\theta</math> is the angle between the vectors <math>\textbf{u}</math> and <math>\textbf{v}.</math>

The equation is numerically unstable if u and v are similar because of [[Catastrophic cancellation|catastrophic cancellation]] and should be avoided for numeric computation.

===Derivation===

The basic relation between the norm and the [[Dot product|dot product]] is given by the equation
<math display=block>\|\textbf{v}\|^2 = \textbf{v} \cdot \textbf{v}.</math>

Then
<math display=block>\begin{align}
\|\textbf{u} + \textbf{v}\|^2
&= (\textbf{u} + \textbf{v}) \cdot (\textbf{u} + \textbf{v}) \\[3pt]
&= (\textbf{u} \cdot \textbf{u}) + (\textbf{u} \cdot \textbf{v}) + (\textbf{v} \cdot \textbf{u}) + (\textbf{v} \cdot \textbf{v}) \\[3pt]
&= \|\textbf{u}\|^2 + \|\textbf{v}\|^2 + 2(\textbf{u} \cdot \textbf{v}),
\end{align}</math>
and similarly
<math display=block>\|\textbf{u} - \textbf{v}\|^2 = \|\textbf{u}\|^2 + \|\textbf{v}\|^2 - 2(\textbf{u} \cdot \textbf{v}).</math>

Forms (1) and (2) of the polarization identity now follow by solving these equations for <math>\textbf{u} \cdot \textbf{v},</math> while form (3) follows from subtracting these two equations. 
(Adding these two equations together gives the parallelogram law.)

==Generalizations==

===Symmetric bilinear forms===

The polarization identities are not restricted to inner products. 
If <math>B</math> is any symmetric bilinear form on a vector space, and <math>Q</math> is the [[Quadratic form|quadratic form]] defined by
<math display=block>Q(v) = B(v, v),</math>
then
<math display=block>\begin{align}
2 B(u, v) &= Q(u + v) - Q(u) - Q(v), \\
2 B(u, v) &= Q(u)     + Q(v) - Q(u - v), \\
4 B(u, v) &= Q(u + v)        - Q(u - v).
\end{align}</math>

The so-called [[Homogeneous polynomial|symmetrization map]] generalizes the latter formula, replacing <math>Q</math> by a homogeneous polynomial of degree <math>k</math> defined by <math>Q(v) = B(v, \ldots, v),</math> where <math>B</math> is a symmetric <math>k</math>-linear map.<ref>{{harvnb|Butler|2013}}. See Keith Conrad (KCd)'s answer.</ref>

The formulas above even apply in the case where the [[Field (mathematics)|field]] of [[Scalar (mathematics)|scalars]] has [[Characteristic (algebra)|characteristic]] two, though the left-hand sides are all zero in this case. 
Consequently, in characteristic two there is no formula for a symmetric bilinear form in terms of a quadratic form, and they are in fact distinct notions, a fact which has important consequences in [[L-theory]]; for brevity, in this context "symmetric bilinear forms" are often referred to as "symmetric forms".

These formulas also apply to bilinear forms on [[Module (mathematics)|modules]] over a [[Commutative ring|commutative ring]], though again one can only solve for <math>B(u, v)</math> if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishes integral quadratic forms from integral {{em|symmetric}} forms, which are a narrower notion.

More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes [[Ε-quadratic form|<math>\varepsilon</math>-quadratic form]]s and <math>\varepsilon</math>-symmetric forms; a symmetric form defines a quadratic form, and the polarization identity (without a factor of 2) from a quadratic form to a symmetric form is called the "[[Symmetrization|symmetrization]] map", and is not in general an isomorphism. This has historically been a subtle distinction: over the integers it was not until the 1950s that relation between "twos out" (integral {{em|quadratic}} form) and "twos in" (integral {{em|symmetric}} form) was understood – see discussion at integral quadratic form; and in the algebraization of [[Surgery theory|surgery theory]], Mishchenko originally used {{em|symmetric}} ''L''-groups, rather than the correct {{em|quadratic}} ''L''-groups (as in Wall and Ranicki) – see discussion at [[L-theory]].

===Homogeneous polynomials of higher degree===

Finally, in any of these contexts these identities may be extended to [[Homogeneous polynomial|homogeneous polynomial]]s (that is, algebraic forms) of arbitrary [[Degree of a polynomial|degree]], where it is known as the polarization formula, and is reviewed in greater detail in the article on the [[Polarization of an algebraic form|polarization of an algebraic form]].

==See also==

* {{annotated link|Inner product space}}
* {{annotated link|Law of cosines}}
* {{annotated link|Mazur–Ulam theorem}}
* {{annotated link|Minkowski distance}}
* {{annotated link|Parallelogram law}}
* {{annotated link|Ptolemy's inequality}}

==Notes and references==

{{reflist}}
{{reflist|group=note}}
{{reflist|group=proof|refs=
<ref group=proof name=RealPartEquivalentFormsAndFormulasProof>A proof can be found here.</ref>}}

==Bibliography==

* {{Lax Functional Analysis}} <!-- {{sfn|Lax|2002|p=}} -->
* {{Rudin Walter Functional Analysis|edition=2}} <!-- {{sfn|Rudin|1991|p=}} -->
* {{Schechter Handbook of Analysis and Its Foundations}} <!-- {{sfn|Schechter|1996|p=}} -->

{{Functional Analysis}}

[[Category:Abstract algebra]]
[[Category:Linear algebra]]
[[Category:Functional analysis]]
[[Category:Vectors (mathematics and physics)]]
[[Category:Norms (mathematics)]]
[[Category:Mathematical identities]]

{{Sourceattribution|Polarization identity}}