Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law

Assume first that $V$ is real and satisfies the parallelogram law, and define $b(x,y)=\frac{1}{4}(q(x+y)-q(x-y))$; then $b(x,x)=q(x)$, $b(y,x)=b(x,y)$, $b(-x,y)=b(x,-y)=-b(x,y)$, $b(0,y)=0$, and $q(\lambda z)={\lambda }^{2}q(z)$ for real $\lambda$.

Applying the parallelogram law to the four pairs $(p+q,r)$, $(p-q,r)$, $(p+r,q)$ and $(p-r,q)$ and subtracting the fourth identity from the third gives $8b(p,r)=4b(p+q,r)+4b(p-q,r)$, that is $b(p+q,r)+b(p-q,r)=2b(p,r)$ for all $p,q,r$.

Adding that identity at $(p,q)=(u,v)$ and at $(p,q)=(v,u)$ gives $b(u+v,w)+b(v-u,w)=2b(v,w)$ and $b(u+v,w)+b(u-v,w)=2b(u,w)$; since $b(v-u,w)=-b(u-v,w)$ by step 1.1, the two relations add to $2b(u+v,w)=2b(u,w)+2b(v,w)$, so $b(u+v,w)=b(u,w)+b(v,w)$, and symmetry gives additivity in the second argument as well.

Induction on the natural number $n\ge 0$ using step 2.1 gives $b(nx,y)=nb(x,y)$, and $b(-x,y)=-b(x,y)$ is step 1.1, so $b(mx,y)=mb(x,y)$ for every integer $m$.

For $n\ge 1$ the additivity of step 2.1 gives $nb(x\mathrm{/}n,y)=b(x,y)$, so $b(x\mathrm{/}n,y)=b(x,y)\mathrm{/}n$, and together with step 3.1 this yields $b(rx,y)=rb(x,y)$ for every rational $r$.

Consequently for every rational $t$ the form $b$ is a symmetric rational-bilinear pairing with $b(w+ty,w+ty)=b(w,w)+2tb(w,y)+t^{2}b(y,y)$, that is $q(w+ty)=q(w)+2tb(w,y)+t^{2}q(y)$ by steps 1.1 and 4.1.

If $q(y)>0$, put $C=q(w)-b(w,y{)}^2\mathrm{/}q(y)$ and $t^{\ast }=-b(w,y)\mathrm{/}q(y)$, so that step 5.1 reads $q(y)(t-t^{\ast }{)}^2+C\ge 0$ for every rational $t$; rationals approach $t^{\ast }$ within any $\delta >0$ by [A5], whence $C>-q(y){\delta }^2$ for every $\delta >0$, and $C\ge 0$ because a negative $C$ would give $q(y){\delta }^2<-C$ for some $\delta >0$ by [A6]; if instead $q(y)=0$ then $q(w)+2tb(w,y)\ge 0$ for all rational $t$ forces $b(w,y)=0$, since otherwise [A6] supplies a rational $t$ with $2tb(w,y)<-q(w)$; in both cases $b(w,y{)}^2\le q(w)q(y)$, so $\mathrm{\mid }b(w,y)\mathrm{\mid }\le \mathrm{\parallel }w\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by [A8].

For fixed $x,y$ the map $\phi (\lambda )=b(\lambda x,y)$ is additive in $\lambda$ by step 2.1 and satisfies $\mathrm{\mid }\phi (\lambda )\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by step 6.1, hence $\mathrm{\mid }\phi (h)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ for $\mathrm{\mid }h\mathrm{\mid }\le 1$; given $\epsilon >0$ choose $n\ge 1$ with $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}n<\epsilon$ by [A6], then $\mathrm{\mid }h\mathrm{\mid }\le 1\mathrm{/}n$ gives $\mathrm{\mid }\phi (h)\mathrm{\mid }=\mathrm{\mid }\phi (nh)\mathrm{\mid }\mathrm{/}n\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}n<\epsilon$, so $\phi$ is continuous at $0$.

For real $\lambda$ and rational $r$ one has $\mathrm{\mid }\phi (\lambda )-\lambda \phi (1)\mathrm{\mid }\le \mathrm{\mid }\phi (\lambda -r)\mathrm{\mid }+\mathrm{\mid }r-\lambda \mathrm{\mid }\mathrm{\mid }\phi (1)\mathrm{\mid }$ with $\phi (r)=r\phi (1)$ by step 4.1, so continuity at $0$ from step 7.1 and the rational approximation of $\lambda$ from [A5] give $\phi (\lambda )=\lambda \phi (1)$, that is $b(\lambda x,y)=\lambda b(x,y)$ for every real $\lambda$.

Therefore, in the real case, $b$ is symmetric, additive in each argument and real-homogeneous in the first argument, with $b(x,x)=q(x)\ge 0$ and $b(x,x)=0$ exactly for $x=0$; so $b$ is a real inner product on $V$ whose induced length is the given norm $\mathrm{\parallel }x\mathrm{\parallel }=\sqrt{b(x,x)}$.

Now let $V$ be complex with a norm satisfying the parallelogram law; viewing $V$ as a real vector space with the same norm, to which step 9.1 applies, gives a real inner product $b$ with $b(x,x)=q(x)$, and $q(iz)=q(z)$ together with the definition of $b$ gives $b(iu,iv)=b(u,v)$, hence $b(iu,v)=-b(u,iv)$ by the argument $b(iu,v)=b(i(iu),iv)=b(-u,iv)=-b(u,iv)$.

Define $⟨x,y⟩=b(x,y)-ib(ix,y)$; then additivity in both arguments and $⟨ix,y⟩=i⟨x,y⟩$ follow from the real bilinearity of $b$, and conjugate symmetry $⟨y,x⟩=\overline{⟨x,y⟩}$ follows from $b(y,x)=b(x,y)$ and from $b(iy,x)=-b(y,ix)=-b(ix,y)$ in step 10.1.

Positivity: $⟨x,x⟩=b(x,x)-ib(ix,x)=q(x)$ because $b(ix,x)=-b(x,ix)$ and $b(x,ix)=b(ix,x)$ force $b(ix,x)=0$; hence $⟨\cdot ,\cdot ⟩$ is a complex inner product whose induced length is $\mathrm{\parallel }x\mathrm{\parallel }$, and expanding $b$ in terms of $q$ gives the complex polarisation formula $⟨x,y⟩=\frac{1}{4}(q(x+y)-q(x-y)+iq(x+iy)-iq(x-iy))$.

Conversely, if the given norm is induced by an inner product on $V$, then it satisfies the parallelogram law by [A4] and expanding the pairing in terms of $q$ recovers, in the real case, $⟨x,y⟩=\frac{1}{4}(q(x+y)-q(x-y))=b(x,y)$ and, in the complex case, the four-term formula of step 12.1; with steps 9.1 and 12.1 this proves that a real or complex norm is induced by an inner product exactly when it satisfies the parallelogram law, and that the polarisation formulas display that inner product.