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Polarization identity
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<h2 id="polarization-identities">Polarization identities</h2> <p>Any <a href="/facts/Inner_product_space/HoyjElEy">inner product</a> on a vector space induces a norm by the equation ‖ x ‖ = ⟨ x , x ⟩ . {\displaystyle \|x\|={\sqrt {\langle x,x\rangle }}.} The polarization identities reverse this relationship, recovering the inner product from the norm. Every inner product satisfies: ‖ x + y ‖ 2 = ‖ x ‖ 2 + ‖ y ‖ 2 + 2 Re ⟨ x , y ⟩ for all vectors x , y . {\displaystyle \|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}+2\operatorname {Re} \langle x,y\rangle \qquad {\text{ for all vectors }}x,y.} </p><p>Solving for Re ⟨ x , y ⟩ {\displaystyle \operatorname {Re} \langle x,y\rangle } gives the formula Re ⟨ x , y ⟩ = 1 2 ( ‖ x + y ‖ 2 − ‖ x ‖ 2 − ‖ y ‖ 2 ) . {\displaystyle \operatorname {Re} \langle x,y\rangle ={\frac {1}{2}}\left(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}\right).} If the inner product is real then Re ⟨ x , y ⟩ = ⟨ x , y ⟩ {\displaystyle \operatorname {Re} \langle x,y\rangle =\langle x,y\rangle } and this formula becomes a polarization identity for real inner products. </p> <h3>Real vector spaces</h3> <p>If the vector space is over the <a href="/facts/Real_number/R02gw5Pb">real numbers</a> then the polarization identities are:<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> ⟨ x , y ⟩ = 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 ) = 1 2 ( ‖ x + y ‖ 2 − ‖ x ‖ 2 − ‖ y ‖ 2 ) = 1 2 ( ‖ x ‖ 2 + ‖ y ‖ 2 − ‖ x − y ‖ 2 ) . {\displaystyle {\begin{alignedat}{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{alignedat}}} </p><p>These various forms are all equivalent by the <a href="/facts/Parallelogram_law/JI1LwPxd">parallelogram law</a>:<a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> 2 ‖ x ‖ 2 + 2 ‖ y ‖ 2 = ‖ x + y ‖ 2 + ‖ x − y ‖ 2 . {\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}.} </p><p>This further implies that L p {\displaystyle L^{p}} class is not a <a href="/facts/Hilbert_space/aDFMh4lV">Hilbert space</a> whenever p ≠ 2 {\displaystyle p\neq 2} , as the parallelogram law is not satisfied. For the sake of counterexample, consider x = 1 A {\displaystyle x=1_{A}} and y = 1 B {\displaystyle y=1_{B}} for any two disjoint subsets A , B {\displaystyle A,B} of general domain Ω ⊂ R n {\displaystyle \Omega \subset \mathbb {R} ^{n}} and compute the measure of both sets under parallelogram law. </p> <h3>Complex vector spaces</h3> <p>For vector spaces over the <a href="/facts/Complex_number/2w7mqI7e">complex numbers</a>, the above formulas are not quite correct because they do not describe the <a href="/facts/Imaginary_part/2w7mqI7e">imaginary part</a> 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 <a href="/facts/Antilinear_map/GTXzWMKc">antilinear</a> in the first or the second argument. The notation ⟨ x | y ⟩ , {\displaystyle \langle x|y\rangle ,} which is commonly used in physics will be assumed to be <a href="/facts/Antilinear_map/GTXzWMKc">antilinear</a> in the first argument while ⟨ x , y ⟩ , {\displaystyle \langle x,\,y\rangle ,} which is commonly used in mathematics, will be assumed to be antilinear in its second argument. They are related by the formula: ⟨ x , y ⟩ = ⟨ y | x ⟩ for all x , y ∈ H . {\displaystyle \langle x,\,y\rangle =\langle y\,|\,x\rangle \quad {\text{ for all }}x,y\in H.} </p><p>The <a href="/facts/Real_part/2w7mqI7e">real part</a> 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 x , y ∈ H {\displaystyle x,y\in H} is always equal to:<a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a><a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a> R ( x , y ) : = Re ⟨ x ∣ y ⟩ = Re ⟨ x , y ⟩ = 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 ) = 1 2 ( ‖ x + y ‖ 2 − ‖ x ‖ 2 − ‖ y ‖ 2 ) = 1 2 ( ‖ x ‖ 2 + ‖ y ‖ 2 − ‖ x − y ‖ 2 ) . {\displaystyle {\begin{alignedat}{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{alignedat}}} </p><p>It is always a <a href="/facts/Symmetric_map/6XbIgHGx">symmetric map</a>, meaning that<a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a> R ( x , y ) = R ( y , x ) for all x , y ∈ H , {\displaystyle R(x,y)=R(y,x)\quad {\text{ for all }}x,y\in H,} and it also satisfies:<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a> R ( y , i x ) = − R ( x , i y ) for all x , y ∈ H . {\displaystyle R(y,ix)=-R(x,iy)\quad {\text{ for all }}x,y\in H.} Thus R ( i x , y ) = − R ( x , i y ) {\displaystyle R(ix,y)=-R(x,iy)} , which in plain English says that to move a factor of i {\displaystyle i} to the other argument, introduce a negative sign. </p> <table><tbody><tr><th>Proof of properties of R {\displaystyle R} </th></tr><tr><td><p>Let R ( x , y ) := 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 ) . {\displaystyle R(x,y):={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right).} Then 2 ‖ x ‖ 2 + 2 ‖ y ‖ 2 = ‖ x + y ‖ 2 + ‖ x − y ‖ 2 {\displaystyle 2\|x\|^{2}+2\|y\|^{2}=\|x+y\|^{2}+\|x-y\|^{2}} implies R ( x , y ) = 1 4 ( ( 2 ‖ x ‖ 2 + 2 ‖ y ‖ 2 − ‖ x − y ‖ 2 ) − ‖ x − y ‖ 2 ) = 1 2 ( ‖ x ‖ 2 + ‖ y ‖ 2 − ‖ x − y ‖ 2 ) {\displaystyle 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)} and R ( x , y ) = 1 4 ( ‖ x + y ‖ 2 − ( 2 ‖ x ‖ 2 + 2 ‖ y ‖ 2 − ‖ x + y ‖ 2 ) ) = 1 2 ( ‖ x + y ‖ 2 − ‖ x ‖ 2 − ‖ y ‖ 2 ) . {\displaystyle 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).} </p><p>Moreover, 4 R ( x , y ) = ‖ x + y ‖ 2 − ‖ x − y ‖ 2 = ‖ y + x ‖ 2 − ‖ y − x ‖ 2 = 4 R ( y , x ) , {\displaystyle 4R(x,y)=\|x+y\|^{2}-\|x-y\|^{2}=\|y+x\|^{2}-\|y-x\|^{2}=4R(y,x),} which proves that R ( x , y ) = R ( y , x ) {\displaystyle R(x,y)=R(y,x)} . </p><p>From 1 = i ( − i ) {\displaystyle 1=i(-i)} it follows that y − i x = i ( − i y − x ) = − i ( x + i y ) {\displaystyle y-ix=i(-iy-x)=-i(x+iy)} and y + i x = i ( − i y + x ) = i ( x − i y ) {\displaystyle y+ix=i(-iy+x)=i(x-iy)} so that − 4 R ( y , i x ) = ‖ y − i x ‖ 2 − ‖ y + i x ‖ 2 = ‖ ( − i ) ( x + i y ) ‖ 2 − ‖ i ( x − i y ) ‖ 2 = ‖ x + i y ‖ 2 − ‖ x − i y ‖ 2 = 4 R ( x , i y ) , {\displaystyle -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),} which proves that R ( y , i x ) = − R ( x , i y ) . {\displaystyle R(y,ix)=-R(x,iy).} ◼ {\displaystyle \blacksquare } </p></td></tr></tbody></table> <p>Unlike its real part, the <a href="/facts/Imaginary_part/2w7mqI7e">imaginary part</a> of a complex inner product depends on which argument is antilinear. </p><p>Antilinear in first argument </p><p>The polarization identities for the inner product ⟨ x | y ⟩ , {\displaystyle \langle x\,|\,y\rangle ,} which is <a href="/facts/Antilinear_map/GTXzWMKc">antilinear</a> in the first argument, are </p> ⟨ x | y ⟩ = 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 − i ‖ x + i y ‖ 2 + i ‖ x − i y ‖ 2 ) = 1 4 ∑ k = 0 3 i k ‖ x + ( − i ) k y ‖ 2 = R ( x , y ) − i R ( x , i y ) = R ( x , y ) + i R ( i x , y ) {\displaystyle {\begin{alignedat}{4}\langle x\,|\,y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}-i\|x+iy\|^{2}+i\|x-iy\|^{2}\right)\\&={\frac {1}{4}}\sum _{k=0}^{3}i^{k}\|x+(-i)^{k}y\|^{2}\\&=R(x,y)-iR(x,iy)\\&=R(x,y)+iR(ix,y)\\\end{alignedat}}} <p>where x , y ∈ H . {\displaystyle x,y\in H.} The second to last equality is similar to the formula <a href="/facts/Real_and_imaginary_parts_of_a_linear_functional/oGh7Asrz">expressing a linear functional</a> φ {\displaystyle \varphi } in terms of its real part: φ ( y ) = Re φ ( y ) − i ( Re φ ) ( i y ) . {\displaystyle \varphi (y)=\operatorname {Re} \varphi (y)-i(\operatorname {Re} \varphi )(iy).} </p><p>Antilinear in second argument </p><p>The polarization identities for the inner product ⟨ x , y ⟩ , {\displaystyle \langle x,\ y\rangle ,} which is <a href="/facts/Antilinear_map/GTXzWMKc">antilinear</a> in the second argument, follows from that of ⟨ x | y ⟩ {\displaystyle \langle x\,|\,y\rangle } by the relationship: ⟨ x , y ⟩ := ⟨ y | x ⟩ = ⟨ x | y ⟩ ¯ for all x , y ∈ H . {\displaystyle \langle x,\ y\rangle :=\langle y\,|\,x\rangle ={\overline {\langle x\,|\,y\rangle }}\quad {\text{ for all }}x,y\in H.} So for any x , y ∈ H , {\displaystyle x,y\in H,} <a class="footnote-ref" id="fnref:10" href="#fn:10"><sup>10</sup></a> </p> ⟨ x , y ⟩ = 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 + i ‖ x + i y ‖ 2 − i ‖ x − i y ‖ 2 ) = R ( x , y ) + i R ( x , i y ) = R ( x , y ) − i R ( i x , y ) . {\displaystyle {\begin{alignedat}{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)+iR(x,iy)\\&=R(x,y)-iR(ix,y).\\\end{alignedat}}} <p>This expression can be phrased symmetrically as:<a class="footnote-ref" id="fnref:11" href="#fn:11"><sup>11</sup></a> ⟨ x , y ⟩ = 1 4 ∑ k = 0 3 i k ‖ x + i k y ‖ 2 . {\displaystyle \langle x,y\rangle ={\frac {1}{4}}\sum _{k=0}^{3}i^{k}\left\|x+i^{k}y\right\|^{2}.} </p><p>Summary of both cases </p><p>Thus if R ( x , y ) + i I ( x , y ) {\displaystyle R(x,y)+iI(x,y)} denotes the real and imaginary parts of some inner product's value at the point ( x , y ) ∈ H × H {\displaystyle (x,y)\in H\times H} of its domain, then its imaginary part will be: I ( x , y ) = { R ( i x , y ) if antilinear in the 1 st argument R ( x , i y ) if antilinear in the 2 nd argument {\displaystyle 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}}} where the scalar i {\displaystyle i} is always located in the same argument that the inner product is antilinear in. </p><p>Using R ( i x , y ) = − R ( x , i y ) {\displaystyle R(ix,y)=-R(x,iy)} , the above formula for the imaginary part becomes: I ( x , y ) = { − R ( x , i y ) if antilinear in the 1 st argument − R ( i x , y ) if antilinear in the 2 nd argument {\displaystyle 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}}} </p> <h2 id="reconstructing-the-inner-product">Reconstructing the inner product</h2> <p>In a normed space ( H , ‖ ⋅ ‖ ) , {\displaystyle (H,\|\cdot \|),} if the <a href="/facts/Parallelogram_law/JI1LwPxd">parallelogram law</a> ‖ x + y ‖ 2 + ‖ x − y ‖ 2 = 2 ‖ x ‖ 2 + 2 ‖ y ‖ 2 {\displaystyle \|x+y\|^{2}~+~\|x-y\|^{2}~=~2\|x\|^{2}+2\|y\|^{2}} holds, then there exists a unique <a href="/facts/Inner_product/HoyjElEy">inner product</a> ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\ \cdot \rangle } on H {\displaystyle H} such that ‖ x ‖ 2 = ⟨ x , x ⟩ {\displaystyle \|x\|^{2}=\langle x,\ x\rangle } for all x ∈ H . {\displaystyle x\in H.} <a class="footnote-ref" id="fnref:12" href="#fn:12"><sup>12</sup></a><a class="footnote-ref" id="fnref:13" href="#fn:13"><sup>13</sup></a> </p> Proof <p>We will only give the real case here; the proof for complex vector spaces is analogous. </p><p>By the above formulas, if the norm is described by an inner product (as we hope), then it must satisfy ⟨ x , y ⟩ = 1 4 ( ‖ x + y ‖ 2 − ‖ x − y ‖ 2 ) for all x , y ∈ H , {\displaystyle \langle x,\ y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)\quad {\text{ for all }}x,y\in H,} which may serve as a definition of the unique candidate ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } for the role of a suitable inner product. Thus, the uniqueness is guaranteed. </p><p>It remains to prove that this formula indeed defines an inner product and that this inner product induces the norm ‖ ⋅ ‖ . {\displaystyle \|\cdot \|.} Explicitly, the following will be shown: </p> <ol><li> ⟨ x , x ⟩ = ‖ x ‖ 2 , x ∈ H {\displaystyle \langle x,x\rangle =\|x\|^{2},\quad x\in H} </li> <li> ⟨ x , y ⟩ = ⟨ y , x ⟩ , x , y ∈ H {\displaystyle \langle x,y\rangle =\langle y,x\rangle ,\quad x,y\in H} </li> <li> ⟨ x + z , y ⟩ = ⟨ x , y ⟩ + ⟨ z , y ⟩ for all x , y , z ∈ H , {\displaystyle \langle x+z,y\rangle =\langle x,y\rangle +\langle z,y\rangle \quad {\text{ for all }}x,y,z\in H,} </li> <li> ⟨ α x , y ⟩ = α ⟨ x , y ⟩ for all x , y ∈ H and all α ∈ R {\displaystyle \langle \alpha x,y\rangle =\alpha \langle x,y\rangle \quad {\text{ for all }}x,y\in H{\text{ and all }}\alpha \in \mathbb {R} } </li></ol> <p>(This axiomatization omits <a href="/facts/Positive-definite_bilinear_form/utlBuVpN">positivity</a>, which is implied by (1) and the fact that ‖ ⋅ ‖ {\displaystyle \|\cdot \|} is a norm.) </p><p>For properties (1) and (2), substitute: ⟨ x , x ⟩ = 1 4 ( ‖ x + x ‖ 2 − ‖ x − x ‖ 2 ) = ‖ x ‖ 2 , {\textstyle \langle x,x\rangle ={\frac {1}{4}}\left(\|x+x\|^{2}-\|x-x\|^{2}\right)=\|x\|^{2},} and ‖ x − y ‖ 2 = ‖ y − x ‖ 2 . {\displaystyle \|x-y\|^{2}=\|y-x\|^{2}.} </p><p>For property (3), it is convenient to work in reverse. It remains to show that ‖ x + z + y ‖ 2 − ‖ x + z − y ‖ 2 = ? ‖ x + y ‖ 2 − ‖ x − y ‖ 2 + ‖ z + y ‖ 2 − ‖ z − y ‖ 2 {\displaystyle \|x+z+y\|^{2}-\|x+z-y\|^{2}{\overset {?}{=}}\|x+y\|^{2}-\|x-y\|^{2}+\|z+y\|^{2}-\|z-y\|^{2}} or equivalently, 2 ( ‖ x + z + y ‖ 2 + ‖ x − y ‖ 2 ) − 2 ( ‖ x + z − y ‖ 2 + ‖ x + y ‖ 2 ) = ? 2 ‖ z + y ‖ 2 − 2 ‖ z − y ‖ 2 . {\displaystyle 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}.} </p><p>Now apply the parallelogram identity: 2 ‖ x + z + y ‖ 2 + 2 ‖ x − y ‖ 2 = ‖ 2 x + z ‖ 2 + ‖ 2 y + z ‖ 2 {\displaystyle 2\|x+z+y\|^{2}+2\|x-y\|^{2}=\|2x+z\|^{2}+\|2y+z\|^{2}} 2 ‖ x + z − y ‖ 2 + 2 ‖ x + y ‖ 2 = ‖ 2 x + z ‖ 2 + ‖ z − 2 y ‖ 2 {\displaystyle 2\|x+z-y\|^{2}+2\|x+y\|^{2}=\|2x+z\|^{2}+\|z-2y\|^{2}} Thus it remains to verify: ‖ 2 x + z ‖ 2 + ‖ 2 y + z ‖ 2 − ( ‖ 2 x + z ‖ 2 + ‖ z − 2 y ‖ 2 ) = ? 2 ‖ z + y ‖ 2 − 2 ‖ z − y ‖ 2 {\displaystyle {\cancel {\|2x+z\|^{2}}}+\|2y+z\|^{2}-({\cancel {\|2x+z\|^{2}}}+\|z-2y\|^{2}){\overset {?}{{}={}}}2\|z+y\|^{2}-2\|z-y\|^{2}} ‖ 2 y + z ‖ 2 − ‖ z − 2 y ‖ 2 = ? 2 ‖ z + y ‖ 2 − 2 ‖ z − y ‖ 2 {\displaystyle \|2y+z\|^{2}-\|z-2y\|^{2}{\overset {?}{=}}2\|z+y\|^{2}-2\|z-y\|^{2}} </p><p>But the latter claim can be verified by subtracting the following two further applications of the parallelogram identity: ‖ 2 y + z ‖ 2 + ‖ z ‖ 2 = 2 ‖ z + y ‖ 2 + 2 ‖ y ‖ 2 {\displaystyle \|2y+z\|^{2}+\|z\|^{2}=2\|z+y\|^{2}+2\|y\|^{2}} ‖ z − 2 y ‖ 2 + ‖ z ‖ 2 = 2 ‖ z − y ‖ 2 + 2 ‖ y ‖ 2 {\displaystyle \|z-2y\|^{2}+\|z\|^{2}=2\|z-y\|^{2}+2\|y\|^{2}} </p><p>Thus (3) holds. </p><p>It can be verified by induction that (3) implies (4), as long as α ∈ Z . {\displaystyle \alpha \in \mathbb {Z} .} But "(4) when α ∈ Z {\displaystyle \alpha \in \mathbb {Z} } " implies "(4) when α ∈ Q {\displaystyle \alpha \in \mathbb {Q} } ". And any positive-definite, <a href="/facts/Characteristic_zero/D2VzcQaG">real-valued</a>, Q {\displaystyle \mathbb {Q} } -bilinear form satisfies the <a href="/facts/Cauchy%25E2%2580%2593Schwarz_inequality/poMzQUIY">Cauchy–Schwarz inequality</a>, so that ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } is continuous. Thus ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } must be R {\displaystyle \mathbb {R} } -linear as well. </p> <p>Another necessary and sufficient condition for there to exist an inner product that induces a given norm ‖ ⋅ ‖ {\displaystyle \|\cdot \|} is for the norm to satisfy <a href="/facts/Ptolemy%2527s_inequality/rCosKUXL">Ptolemy's inequality</a>, which is:<a class="footnote-ref" id="fnref:14" href="#fn:14"><sup>14</sup></a> ‖ x − y ‖ ‖ z ‖ + ‖ y − z ‖ ‖ x ‖ ≥ ‖ x − z ‖ ‖ y ‖ for all vectors x , y , z . {\displaystyle \|x-y\|\,\|z\|~+~\|y-z\|\,\|x\|~\geq ~\|x-z\|\,\|y\|\qquad {\text{ for all vectors }}x,y,z.} </p> <h2 id="applications-and-consequences">Applications and consequences</h2> <p>If H {\displaystyle H} is a complex Hilbert space then ⟨ x ∣ y ⟩ {\displaystyle \langle x\mid y\rangle } is real if and only if its imaginary part is 0 = R ( x , i y ) = 1 4 ( ‖ x + i y ‖ 2 − ‖ x − i y ‖ 2 ) {\displaystyle 0=R(x,iy)={\frac {1}{4}}\left(\Vert x+iy\Vert ^{2}-\Vert x-iy\Vert ^{2}\right)} , which happens if and only if ‖ x + i y ‖ = ‖ x − i y ‖ {\displaystyle \Vert x+iy\Vert =\Vert x-iy\Vert } . Similarly, ⟨ x ∣ y ⟩ {\displaystyle \langle x\mid y\rangle } is (purely) imaginary if and only if ‖ x + y ‖ = ‖ x − y ‖ {\displaystyle \Vert x+y\Vert =\Vert x-y\Vert } . For example, from ‖ x + i x ‖ = | 1 + i | ‖ x ‖ = 2 ‖ x ‖ = | 1 − i | ‖ x ‖ = ‖ x − i x ‖ {\displaystyle \|x+ix\|=|1+i|\|x\|={\sqrt {2}}\|x\|=|1-i|\|x\|=\|x-ix\|} it can be concluded that ⟨ x | x ⟩ {\displaystyle \langle x|x\rangle } is real and that ⟨ x | i x ⟩ {\displaystyle \langle x|ix\rangle } is purely imaginary. </p> <h3>Isometries</h3> <p>If A : H → Z {\displaystyle A:H\to Z} is a <a href="/facts/Linear_map/Q1PBKNVV">linear</a> <a href="/facts/Isometry/K5wX8ah6">isometry</a> between two Hilbert spaces (so ‖ A h ‖ = ‖ h ‖ {\displaystyle \|Ah\|=\|h\|} for all h ∈ H {\displaystyle h\in H} ) then ⟨ A h , A k ⟩ Z = ⟨ h , k ⟩ H for all h , k ∈ H ; {\displaystyle \langle Ah,Ak\rangle _{Z}=\langle h,k\rangle _{H}\quad {\text{ for all }}h,k\in H;} that is, linear isometries preserve inner products. </p><p>If A : H → Z {\displaystyle A:H\to Z} is instead an <a href="/facts/Antilinear_map/GTXzWMKc">antilinear</a> isometry then ⟨ A h , A k ⟩ Z = ⟨ h , k ⟩ H ¯ = ⟨ k , h ⟩ H for all h , k ∈ H . {\displaystyle \langle Ah,Ak\rangle _{Z}={\overline {\langle h,k\rangle _{H}}}=\langle k,h\rangle _{H}\quad {\text{ for all }}h,k\in H.} </p> <h3>Relation to the law of cosines</h3> <p>The second form of the polarization identity can be written as ‖ u − v ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ( u ⋅ v ) . {\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).} </p><p>This is essentially a vector form of the <a href="/facts/Law_of_cosines/RRTGcW0V">law of cosines</a> for the <a href="/facts/Triangle/cRXUwsMn">triangle</a> formed by the vectors u {\displaystyle {\textbf {u}}} , v {\displaystyle {\textbf {v}}} , and u − v {\displaystyle {\textbf {u}}-{\textbf {v}}} . In particular, u ⋅ v = ‖ u ‖ ‖ v ‖ cos θ , {\displaystyle {\textbf {u}}\cdot {\textbf {v}}=\|{\textbf {u}}\|\,\|{\textbf {v}}\|\cos \theta ,} where θ {\displaystyle \theta } is the angle between the vectors u {\displaystyle {\textbf {u}}} and v {\displaystyle {\textbf {v}}} . </p><p>The equation is numerically unstable if u and v are similar because of <a href="/facts/Catastrophic_cancellation/2K543P23">catastrophic cancellation</a> and should be avoided for numeric computation. </p> <h3>Derivation</h3> <p>The basic relation between the norm and the <a href="/facts/Dot_product/tNz8MLjT">dot product</a> is given by the equation ‖ v ‖ 2 = v ⋅ v . {\displaystyle \|{\textbf {v}}\|^{2}={\textbf {v}}\cdot {\textbf {v}}.} </p><p>Then ‖ u + v ‖ 2 = ( u + v ) ⋅ ( u + v ) = ( u ⋅ u ) + ( u ⋅ v ) + ( v ⋅ u ) + ( v ⋅ v ) = ‖ u ‖ 2 + ‖ v ‖ 2 + 2 ( u ⋅ v ) , {\displaystyle {\begin{aligned}\|{\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{aligned}}} and similarly ‖ u − v ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ( u ⋅ v ) . {\displaystyle \|{\textbf {u}}-{\textbf {v}}\|^{2}=\|{\textbf {u}}\|^{2}+\|{\textbf {v}}\|^{2}-2({\textbf {u}}\cdot {\textbf {v}}).} </p><p>Forms (1) and (2) of the polarization identity now follow by solving these equations for u ⋅ v {\displaystyle {\textbf {u}}\cdot {\textbf {v}}} , while form (3) follows from subtracting these two equations. (Adding these two equations together gives the parallelogram law.) </p> <h2 id="generalizations">Generalizations</h2> <h3>Jordan–von Neumann theorems</h3> <p>The standard <a href="/facts/Pascual_Jordan/YNLoGJ5d">Jordan</a>–<a href="/facts/John_von_Neumann/aUk6urof">von Neumann</a> theorem, as stated previously, is that the if a norm satisfies the parallelogram law, then it can be induced by an inner product defined by the polarization identity. There are variants of the theorem.<a class="footnote-ref" id="fnref:15" href="#fn:15"><sup>15</sup></a> </p><p>Define various senses of orthogonality: </p> <ul><li>isosceles: ‖ x + y ‖ = ‖ x − y ‖ {\textstyle \|x+y\|=\|x-y\|} </li> <li>Roberts’: ‖ x + t y ‖ = ‖ x − t y ‖ {\textstyle \left\|x+ty\right\|=\left\|x-ty\right\|} for all scalar t {\textstyle t} .</li> <li>Pythagorean: ‖ x + y ‖ 2 = ‖ x ‖ 2 + ‖ y ‖ 2 {\textstyle \left\|x+y\right\|^{2}=\|x\|^{2}+\left\|y\right\|^{2}} </li> <li>Birkhoff–James: ‖ x ‖ ≤ ‖ x + t y ‖ {\textstyle \|x\|\leq \|x+ty\|} for all scalar t {\textstyle t} .</li></ul> <p>Let V {\textstyle V} be a vector space over the real or complex numbers. Let ‖ ⋅ ‖ {\textstyle \|\cdot \|} be a norm over V {\textstyle V} . We consider conditions for which the norm is induced by an inner product. In the following statements, whenever a scalar appears, the scalar may be restricted to be merely real, even when V {\textstyle V} is over the complex numbers. </p> <ul><li>(von Neumann–Jordan condition) The norm satisfies the parallelogram identity.</li> <li>(weakened von Neumann–Jordan condition) ‖ x + y ‖ 2 + ‖ x − y ‖ 2 = 4 {\textstyle \|x+y\|^{2}+\|x-y\|^{2}=4} for all unit vectors x , y {\textstyle x,y} . That is, the norm satisfies the parallelogram identity for unit vectors.</li> <li>For any x , y ∈ V {\textstyle x,y\in V} , the set of points equidistant to x , y {\textstyle x,y} is flat, that is, an affine subspace.</li> <li>Orthogonality in either isosceles or Roberts’ sense is either additive or homogeneous on one variable.</li> <li>For every two-dimensional subspace W ⊂ V {\textstyle W\subset V} , for every x ∈ W {\textstyle x\in W} , there exists y ∈ W {\textstyle y\in W} that is Roberts’ orthogonal to x {\textstyle x} .</li> <li>Isosceles orthogonality implies Pythagorean orthogonality.</li> <li>Pythagorean orthogonality implies isosceles orthogonality.</li> <li>If x , y {\textstyle x,y} are Pythagorean orthogonal, then so are x , − y {\textstyle x,-y} .</li> <li>Birkhoff–James orthogonality is symmetric.</li> <li>If ‖ x ‖ = ‖ y ‖ {\textstyle \|x\|=\|y\|} and t , s {\textstyle t,s} are real, then ‖ t x + s y ‖ = ‖ s x + t y ‖ {\textstyle \|tx+sy\|=\|sx+ty\|} .</li></ul> <p>For the real vector space, there is also the condition: </p> <ul><li>Any two-dimensional slice of the unit sphere is an ellipse, that is, parameterizable as { x cos θ + y sin θ : θ ∈ [ 0 , 2 π ] } {\textstyle \{x\cos \theta +y\sin \theta :\theta \in [0,2\pi ]\}} , for some unit vectors x , y {\textstyle x,y} .</li></ul> Proof <p>Since the internal <a href="/facts/John_ellipsoid/70N6wDYd">John ellipse</a> E {\textstyle E} is unique, for any bijective linear map T {\textstyle T} that preserves the unit circle, it must have T E = E {\textstyle TE=E} . Since E {\textstyle E} must touch the circle at some point x {\textstyle x} , we may map x {\textstyle x} to any other point y {\textstyle y} on the circle, thus every point y {\textstyle y} touches the ellipse E {\textstyle E} . Thus the circle is the ellipse. </p> <p>The Banach-Mazur rotation problem: Given a <a href="/facts/Separable_space/2bmU73bk">separable</a> <a href="/facts/Banach_space/sDEIk7EC">Banach space</a> V {\textstyle V} such that any two unit vectors x , y {\textstyle x,y} , there exists a linear surjective isometry T {\textstyle T} such that T ( x ) = y {\textstyle T(x)=y} or T ( y ) = x {\textstyle T(y)=x} , is V {\textstyle V} isometrically isomorphic to a Hilbert space? </p><p>The general case of the problem is <a href="/facts/Open_problem/VuR6yLsW">open</a>. When the space is parable finite-dimensional, the answer is yes. In other words, given a finite-dimensional normed vector space over the real or complex numbers, if any point on the unit sphere can be mapped (rotated) to any other point by a linear isometry, then the norm is induced by an inner product.<a class="footnote-ref" id="fnref:16" href="#fn:16"><sup>16</sup></a> </p> <h3>Symmetric bilinear forms</h3> <p>The polarization identities are not restricted to inner products. If B {\displaystyle B} is any <a href="/facts/Symmetric_bilinear_form/UI9b6RK7">symmetric bilinear form</a> on a vector space, and Q {\displaystyle Q} is the <a href="/facts/Quadratic_form/l2n1jXPe">quadratic form</a> defined by Q ( v ) = B ( v , v ) , {\displaystyle Q(v)=B(v,v),} then 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 ) . {\displaystyle {\begin{aligned}2B(u,v)&=Q(u+v)-Q(u)-Q(v),\\2B(u,v)&=Q(u)+Q(v)-Q(u-v),\\4B(u,v)&=Q(u+v)-Q(u-v).\end{aligned}}} </p><p>The so-called <a href="/facts/Homogeneous_polynomial/WUHL7Kct">symmetrization map</a> generalizes the latter formula, replacing Q {\displaystyle Q} by a homogeneous polynomial of degree k {\displaystyle k} defined by Q ( v ) = B ( v , … , v ) , {\displaystyle Q(v)=B(v,\ldots ,v),} where B {\displaystyle B} is a symmetric k {\displaystyle k} -linear map.<a class="footnote-ref" id="fnref:17" href="#fn:17"><sup>17</sup></a> </p><p>The formulas above even apply in the case where the <a href="/facts/Field_(mathematics)/xAjAS4ko">field</a> of <a href="/facts/Scalar_(mathematics)/MkDbA4Eo">scalars</a> has <a href="/facts/Characteristic_(algebra)/D2VzcQaG">characteristic</a> 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 <a href="/facts/L-theory/DjSy9S3p">L-theory</a>; for brevity, in this context "symmetric bilinear forms" are often referred to as "symmetric forms". </p><p>These formulas also apply to bilinear forms on <a href="/facts/Module_(mathematics)/jP0LIhFL">modules</a> over a <a href="/facts/Commutative_ring/QS1r2ovR">commutative ring</a>, though again one can only solve for B ( u , v ) {\displaystyle B(u,v)} if 2 is invertible in the ring, and otherwise these are distinct notions. For example, over the integers, one distinguishes <a href="/facts/Integral_quadratic_form/l2n1jXPe">integral quadratic forms</a> from integral symmetric forms, which are a narrower notion. </p><p>More generally, in the presence of a ring involution or where 2 is not invertible, one distinguishes <a href="/facts/%25CE%2595-quadratic_form/1TsXq6eM"> ε {\displaystyle \varepsilon } -quadratic forms</a> and <a href="/facts/%25CE%2595-symmetric_form/1TsXq6eM"> ε {\displaystyle \varepsilon } -symmetric forms</a>; 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 "<a href="/facts/Symmetrization/6XbIgHGx">symmetrization</a> 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 quadratic form) and "twos in" (integral symmetric form) was understood – see discussion at <a href="/facts/Integral_quadratic_form/l2n1jXPe">integral quadratic form</a>; and in the algebraization of <a href="/facts/Surgery_theory/5Ip2RAGW">surgery theory</a>, Mishchenko originally used symmetric <i>L</i>-groups, rather than the correct quadratic <i>L</i>-groups (as in Wall and Ranicki) – see discussion at <a href="/facts/L-theory/DjSy9S3p">L-theory</a>. </p> <h3>Homogeneous polynomials of higher degree</h3> <p>Finally, in any of these contexts these identities may be extended to <a href="/facts/Homogeneous_polynomial/WUHL7Kct">homogeneous polynomials</a> (that is, <a href="/facts/Algebraic_form/WUHL7Kct">algebraic forms</a>) of arbitrary <a href="/facts/Degree_of_a_polynomial/Bf8vEIhf">degree</a>, where it is known as the <a href="/facts/Polarization_formula/ysDI99Eo">polarization formula</a>, and is reviewed in greater detail in the article on the <a href="/facts/Polarization_of_an_algebraic_form/ysDI99Eo">polarization of an algebraic form</a>. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Inner_product_space/HoyjElEy">Inner product space</a> – Vector space with generalized dot product</li> <li><a href="/facts/Law_of_cosines/RRTGcW0V">Law of cosines</a> – Generalization of Pythagorean theorem</li> <li><a href="/facts/Mazur%25E2%2580%2593Ulam_theorem/JGwqSehd">Mazur–Ulam theorem</a> – Surjective isometries are affine mappings</li> <li><a href="/facts/Minkowski_distance/4KXAMfE2">Minkowski distance</a> – Vector distance using pth powers</li> <li><a href="/facts/Parallelogram_law/JI1LwPxd">Parallelogram law</a> – Sides and diagonals have equal sums of squares</li> <li><a href="/facts/Ptolemy%2527s_inequality/rCosKUXL">Ptolemy's inequality</a> – Relation between distances of four points</li></ul> <h2 id="notes-and-references">Notes and references</h2> <h2 id="bibliography">Bibliography</h2> <ul><li><a href="/facts/Peter_D._Lax/sptzPICM">Lax, Peter D.</a> (2002). <a href="http://www.math.univ-metz.fr/~gnc/bibliographie/Functional%20Analysis/Lax,.Functional.Analysis,.Wiley,.2002,.603s.pdf"><i>Functional Analysis</i></a> (PDF). Pure and Applied Mathematics. New York: Wiley-Interscience. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-471-55604-6. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/47767143">47767143</a>. Retrieved July 22, 2020.</li> <li><a href="/facts/Walter_Rudin/9mxRAECY">Rudin, Walter</a> (1991). <a href="https://archive.org/details/functionalanalys00rudi"><i>Functional Analysis</i></a>. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: <a href="/facts/McGraw-Hill_Science%2fEngineering%2fMath/srSyQSsY">McGraw-Hill Science/Engineering/Math</a>. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-07-054236-5. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/21163277">21163277</a>.</li> <li><a href="/facts/Eric_Schechter/cApyLyxB">Schechter, Eric</a> (1996). <i>Handbook of Analysis and Its Foundations</i>. San Diego, CA: Academic Press. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-12-622760-4. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/175294365">175294365</a>.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Lax 2002, p. 53. - Lax, Peter D. (2002). Functional Analysis (PDF). Pure and Applied Mathematics. New York: Wiley-Interscience. ISBN 978-0-471-55604-6. OCLC 47767143. Retrieved July 22, 2020. <a href="http://www.math.univ-metz.fr/~gnc/bibliographie/Functional%20Analysis/Lax,.Functional.Analysis,.Wiley,.2002,.603s.pdf" target="_blank">http://www.math.univ-metz.fr/~gnc/bibliographie/Functional%20Analysis/Lax,.Functional.Analysis,.Wiley,.2002,.603s.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Philippe Blanchard, Erwin Brüning (2003). "Proposition 14.1.2 (Fréchet–von Neumann–Jordan)". Mathematical methods in physics: distributions, Hilbert space operators, and variational methods. Birkhäuser. p. 192. ISBN 0817642285. <a href="0817642285" target="_blank">0817642285</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Gerald Teschl (2009). "Theorem 0.19 (Jordan–von Neumann)". Mathematical methods in quantum mechanics: with applications to Schrödinger operators. American Mathematical Society Bookstore. p. 19. ISBN 978-0-8218-4660-5. <a href="978-0-8218-4660-5" target="_blank">978-0-8218-4660-5</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>Schechter 1996, pp. 601–603. - Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365. <a href="https://search.worldcat.org/oclc/175294365" target="_blank">https://search.worldcat.org/oclc/175294365</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>A proof can be found here. <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>Schechter 1996, pp. 601–603. - Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365. <a href="https://search.worldcat.org/oclc/175294365" target="_blank">https://search.worldcat.org/oclc/175294365</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> <li id="fn:7"><p>A proof can be found here. <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li> <li id="fn:8"><p>A proof can be found here. <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li> <li id="fn:9"><p>A proof can be found here. <a href="#fnref:9" class="footnote-back-ref">↩</a></p></li> <li id="fn:10"><p>Schechter 1996, pp. 601–603. - Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365. <a href="https://search.worldcat.org/oclc/175294365" target="_blank">https://search.worldcat.org/oclc/175294365</a> <a href="#fnref:10" class="footnote-back-ref">↩</a></p></li> <li id="fn:11"><p>Butler, Jon (20 June 2013). "norm - Derivation of the polarization identities?". Mathematics Stack Exchange. Archived from the original on 14 October 2020. Retrieved 2020-10-14. See Harald Hanche-Olson's answer. <a href="https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities" target="_blank">https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities</a> <a href="#fnref:11" class="footnote-back-ref">↩</a></p></li> <li id="fn:12"><p>Schechter 1996, pp. 601–603. - Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365. <a href="https://search.worldcat.org/oclc/175294365" target="_blank">https://search.worldcat.org/oclc/175294365</a> <a href="#fnref:12" class="footnote-back-ref">↩</a></p></li> <li id="fn:13"><p>Lax 2002, p. 53. - Lax, Peter D. (2002). Functional Analysis (PDF). Pure and Applied Mathematics. New York: Wiley-Interscience. ISBN 978-0-471-55604-6. OCLC 47767143. Retrieved July 22, 2020. <a href="http://www.math.univ-metz.fr/~gnc/bibliographie/Functional%20Analysis/Lax,.Functional.Analysis,.Wiley,.2002,.603s.pdf" target="_blank">http://www.math.univ-metz.fr/~gnc/bibliographie/Functional%20Analysis/Lax,.Functional.Analysis,.Wiley,.2002,.603s.pdf</a> <a href="#fnref:13" class="footnote-back-ref">↩</a></p></li> <li id="fn:14"><p>Apostol, Tom M. (1967). "Ptolemy's Inequality and the Chordal Metric". Mathematics Magazine. 40 (5): 233–235. doi:10.2307/2688275. JSTOR 2688275. <a href="https://www.tandfonline.com/doi/pdf/10.1080/0025570X.1967.11975804" target="_blank">https://www.tandfonline.com/doi/pdf/10.1080/0025570X.1967.11975804</a> <a href="#fnref:14" class="footnote-back-ref">↩</a></p></li> <li id="fn:15"><p>Day, Mahlon M. (1947). "Some Characterizations of Inner-Product Spaces". Transactions of the American Mathematical Society. 62 (2): 320–337. doi:10.2307/1990458. ISSN 0002-9947. <a href="https://www.jstor.org/stable/1990458" target="_blank">https://www.jstor.org/stable/1990458</a> <a href="#fnref:15" class="footnote-back-ref">↩</a></p></li> <li id="fn:16"><p>Becerra Guerrero, Julio; Rodríguez-Palacios, A. (2002). "Transitivity of the norm on Banach spaces". Extracta Mathematicae. 17 (1): 1–58. hdl:10662/18957. MR 1914238. <a href="/wiki/Hdl_(identifier)" target="_blank">/wiki/Hdl_(identifier)</a> <a href="#fnref:16" class="footnote-back-ref">↩</a></p></li> <li id="fn:17"><p>Butler 2013. See Keith Conrad (KCd)'s answer. - Butler, Jon (20 June 2013). "norm - Derivation of the polarization identities?". Mathematics Stack Exchange. Archived from the original on 14 October 2020. Retrieved 2020-10-14. <a href="https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities" target="_blank">https://math.stackexchange.com/questions/425173/derivation-of-the-polarization-identities</a> <a href="#fnref:17" class="footnote-back-ref">↩</a></p></li> </ol>