Hilbert Space Geometry and Riesz Representation

If $y=0$, then $⟨x,y⟩=⟨x,0\cdot y⟩=0\cdot ⟨x,y⟩=0$ by conjugate-linearity in the second argument, and $\mathrm{\parallel }y\mathrm{\parallel }=0$; hence $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }=0=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, while $x,y$ are dependent with witness scalars $(a,b)=(0,1)$ because $0\cdot x+1\cdot 0=0$ and $b\mathrm{\ne }0$.

Suppose now $y\mathrm{\ne }0$ and set $c=⟨x,y⟩\mathrm{/}⟨y,y⟩$, a well-defined scalar because $⟨y,y⟩=\mathrm{\parallel }y{\mathrm{\parallel }}^2>0$; expanding with linearity in the first argument, conjugate-linearity in the second and conjugate symmetry gives $0\le \mathrm{\parallel }x-cy{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2-c\overline{⟨x,y⟩}-\bar{c}⟨x,y⟩+\mathrm{\mid }c{\mathrm{\mid }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2-\mathrm{\mid }⟨x,y⟩{\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }y{\mathrm{\parallel }}^2$.

Multiplying step 1.2 by the positive number $\mathrm{\parallel }y{\mathrm{\parallel }}^2$ gives $\mathrm{\mid }⟨x,y⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2$, and since $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }$, $\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }y\mathrm{\parallel }$ are nonnegative, monotonicity of squaring on the nonnegatives gives the inequality $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

If $x,y$ are dependent, then either $y=0$, which is step 1.1, or $x=cy$ for some scalar $c$ with $y\mathrm{\ne }0$; in the second case $⟨x,y⟩=c\mathrm{\parallel }y{\mathrm{\parallel }}^2$ and $\mathrm{\parallel }cy{\mathrm{\parallel }}^2=⟨cy,cy⟩=\mathrm{\mid }c{\mathrm{\mid }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2$ with both norms nonnegative, so $\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\mid }c\mathrm{\mid }\mathrm{\parallel }y\mathrm{\parallel }$ by uniqueness of nonnegative square roots, and $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }=\mathrm{\mid }c\mathrm{\mid }\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

Conversely suppose $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ and $y\mathrm{\ne }0$; then $\mathrm{\parallel }y{\mathrm{\parallel }}^2>0$ and step 1.2 gives $\mathrm{\parallel }x-cy{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2-\mathrm{\parallel }x{\mathrm{\parallel }}^2=0$, so $x-cy=0$ by positive definiteness, that is $x=cy$ and the pair is dependent by [A7]; together with steps 1.1 and 2.2 this proves the inequality and both directions of the equality statement.

Nonnegativity and separation are [A1]: $\mathrm{\parallel }v\mathrm{\parallel }\ge 0$, and $\mathrm{\parallel }v\mathrm{\parallel }=0$ exactly when $⟨v,v⟩=0$, that is exactly when $v=0$.

Homogeneity: $⟨\lambda v,\lambda v⟩=\lambda \bar{\lambda }⟨v,v⟩=\mathrm{\mid }\lambda {\mathrm{\mid }}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by sesquilinearity and [A4], and both $\mathrm{\parallel }\lambda v\mathrm{\parallel }$ and $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ are nonnegative with equal squares, so $\mathrm{\parallel }\lambda v\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ by uniqueness of the nonnegative square root.

Triangle inequality: expanding and using conjugate symmetry gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2+2\mathrm{Re}⟨u,v⟩+\mathrm{\parallel }v{\mathrm{\parallel }}^2$, and $\mathrm{Re}z\le \mathrm{\mid }z\mathrm{\mid }$ for every scalar $z$, since either $\mathrm{Re}z<0\le \mathrm{\mid }z\mathrm{\mid }$ or $\mathrm{Re}z=a\ge 0$ with $a^2\le a^2+b^2=\mathrm{\mid }z{\mathrm{\mid }}^2$ by [A6]; with [A2] this gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2\le \mathrm{\parallel }u{\mathrm{\parallel }}^2+2\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }v{\mathrm{\parallel }}^2=(\mathrm{\parallel }u\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }{)}^2$.

Both sides of the inequality in step 1.3 are nonnegative, so monotonicity of squaring on the nonnegatives turns it into $\mathrm{\parallel }u+v\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }$.

Steps 1.1, 1.2 and 2.1 are exactly the clauses of [A3] over $\mathbb{R}$, and over $\mathbb{C}$ they are the same clauses with the complex modulus, so $\mathrm{\parallel }\cdot \mathrm{\parallel }$ is a norm on $V$ in either scalar field.

Expanding with linearity in the first argument, conjugate-linearity in the second and conjugate symmetry gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2=⟨x+y,x+y⟩=\mathrm{\parallel }x{\mathrm{\parallel }}^2+⟨x,y⟩+\overline{⟨x,y⟩}+\mathrm{\parallel }y{\mathrm{\parallel }}^2$ and, replacing $y$ by $-y$, $\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2-⟨x,y⟩-\overline{⟨x,y⟩}+\mathrm{\parallel }y{\mathrm{\parallel }}^2$.

The cross terms in step 1.1 cancel when the two expansions are added, so $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2+\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=2\mathrm{\parallel }x{\mathrm{\parallel }}^2+2\mathrm{\parallel }y{\mathrm{\parallel }}^2$, 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.

Inserting and subtracting the mixed pairing gives $⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩=⟨x-x^{\mathrm{\prime }},y⟩+⟨x^{\mathrm{\prime }},y-y^{\mathrm{\prime }}⟩$, so Cauchy–Schwarz applied to the two summands and the triangle inequality for scalars give $\mathrm{\mid }⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩\mathrm{\mid }\le \mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }$.

Given $\epsilon >0$, put $\delta =\min \{1,\epsilon \mathrm{/}(1+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel })\}>0$; if $\mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }<\delta$ and $\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }<\delta$, then $\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel }+\delta <1+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel }$ by the triangle inequality, so step 1.1 gives $\mathrm{\mid }⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩\mathrm{\mid }<\delta (1+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel })\le \epsilon$.

The product $B(x^{\mathrm{\prime }},\delta )\times B(y^{\mathrm{\prime }},\delta )$ is a basic product-open set containing $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, and step 2.1 shows that on it the pairing stays within every ball about $⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩$, so the pairing is continuous at every point of $V\times V$; if moreover $x_n\to x$ and $y_n\to y$, then for every $\epsilon >0$ the pair $(x_n,y_n)$ eventually lies in the corresponding $\delta$-box, whence $\mathrm{\mid }⟨x_n,y_n⟩-⟨x,y⟩\mathrm{\mid }<\epsilon$ and $⟨x_n,y_n⟩\to ⟨x,y⟩$.

Let $\xi ,\eta \in \widehat{X}$. For every $n$, density makes the sets $\{x\in X:\mathrm{\parallel }\kappa x-\xi \mathrm{\parallel }<1\mathrm{/}(n+1)\}$ and $\{y\in X:\mathrm{\parallel }\kappa y-\eta \mathrm{\parallel }<1\mathrm{/}(n+1)\}$ nonempty. Countable Choice selects $x_n,y_n$ from these sets. Thus $\kappa (x_n)\to \xi$ and $\kappa (y_n)\to \eta$, the selected sequences are Cauchy, and continuity of the vector operations gives $\kappa (x_n\pm y_n)=\kappa (x_n)\pm \kappa (y_n)\to \xi \pm \eta$.

The parallelogram identities $\mathrm{\parallel }{x}_n+y_n{\mathrm{\parallel }}^2+\mathrm{\parallel }{x}_n-y_n{\mathrm{\parallel }}^2=2\mathrm{\parallel }{x}_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }{y}_n{\mathrm{\parallel }}^2$ hold in $X$ by [A6]; passing to the limit using continuity of the norm and of sums, and the isometry $\mathrm{\parallel }\kappa z\mathrm{\parallel }=\mathrm{\parallel }z\mathrm{\parallel }$, gives $\mathrm{\parallel }\xi +\eta {\mathrm{\parallel }}^2+\mathrm{\parallel }\xi -\eta {\mathrm{\parallel }}^2=2\mathrm{\parallel }\xi {\mathrm{\parallel }}^2+2\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$, so the completion norm satisfies the parallelogram law.

The Jordan–von Neumann theorem applied to the Banach space $\widehat{X}$ therefore produces an inner product $B$ on $\widehat{X}$ whose induced length is exactly the completion norm and which is the unique such inner product.

$B$ extends the original pairing: for $x,y\in X$, the polarisation formula of [A2] together with $\kappa (x\pm y)=\kappa x\pm \kappa y$ and the isometry property gives $B(\kappa x,\kappa y)=\frac{1}{4}(\mathrm{\parallel }\kappa (x+y){\mathrm{\parallel }}^2-\mathrm{\parallel }\kappa (x-y){\mathrm{\parallel }}^2)=\frac{1}{4}(\mathrm{\parallel }x+y{\mathrm{\parallel }}^2-\mathrm{\parallel }x-y{\mathrm{\parallel }}^2)=⟨x,y⟩$ in the real case, and the four-term complex formula likewise in the complex case.

Uniqueness: if $B^{\mathrm{\prime }}$ is any inner product on $\widehat{X}$ that extends the pairing along $\kappa$ and induces the completion norm, then for $\xi ,\eta$ and approximating sequences as in step 1.1 the Cauchy–Schwarz inequality for both forms gives $\mathrm{\mid }{B}^{\mathrm{\prime }}(\xi ,\eta )-B^{\mathrm{\prime }}(\kappa x_n,\kappa y_n)\mathrm{\mid }\le \mathrm{\parallel }\xi -\kappa x_n\mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }+\mathrm{\parallel }\kappa x_n\mathrm{\parallel }\mathrm{\parallel }\eta -\kappa y_n\mathrm{\parallel }$ and the same bound for $B$, so both pairings are the limits of the common values $⟨x_n,y_n⟩$ and $B^{\mathrm{\prime }}=B$.

Hence $\widehat{X}$ carries the inner product $B$ extending the original one with the completion norm as induced length, so it is complete and therefore a Hilbert space by [A7]. Countable Choice is used in the published completion interface and explicitly in step 1.1 to select the two approximating sequences; no further choice enters the limit arguments.

At $n=0$ the sum is $0$ and both sides vanish, and at $n=1$ the identity is $\mathrm{\parallel }{x}_1{\mathrm{\parallel }}^2=\mathrm{\parallel }{x}_1{\mathrm{\parallel }}^2$.

For two orthogonal vectors $x,y$, expansion gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+⟨x,y⟩+⟨y,x⟩+\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2$, and the same computation with $y$ replaced by $-y$ shows that the sum of a finite orthogonal family may be split off one term at a time.

Suppose the identity holds for orthogonal families of $n-1$ terms, $n\ge 2$, and let $x_1,\dots ,x_n$ be pairwise orthogonal; the partial sum $s={\sum }_{j=1}^{n-1}{x}_j$ satisfies $⟨s,x_n⟩={\sum }_{j=1}^{n-1}⟨x_j,x_n⟩=0$ by linearity in the first argument, and $\mathrm{\parallel }s{\mathrm{\parallel }}^2={\sum }_{j=1}^{n-1}\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$ by the induction hypothesis, so $\mathrm{\parallel }{\sum }_{j=1}^n{x}_j{\mathrm{\parallel }}^2=\mathrm{\parallel }s+x_n{\mathrm{\parallel }}^2=\mathrm{\parallel }s{\mathrm{\parallel }}^2+\mathrm{\parallel }{x}_n{\mathrm{\parallel }}^2={\sum }_{j=1}^n\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$.

Induction on $n$ from the cases $n=0,1$ of step 1.1 and the induction step of step 2.1 proves the identity for every $n$, so pairwise orthogonal vectors satisfy $\mathrm{\parallel }{\sum }_{j=1}^n{x}_j{\mathrm{\parallel }}^2={\sum }_{j=1}^n\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$.