Past exam of the analysis course of the University of Oxford / 2007 / B / 5

Expand both squared norms using sesquilinearity and conjugate symmetry. Their mixed terms $⟨u,v⟩+⟨v,u⟩$ appear with opposite signs and cancel. Hencethe parallelogram identity.

Put $Q(u)=\Vert u{\Vert }^2$ and $B(u,w)=\frac{1}{4}[Q(u+w)-Q(u-w)]$. Then $B(0,w)=0$ and the parallelogram identity, applied to $u+w,v$ and to $u-w,v$ and subtracted, givesTaking $v=u$ shows $B(2u,w)=2B(u,w)$. Replacing $u,v$ in the identity by $(u+v)/2,(u-v)/2$ now givesThe proposed complex pairing is $P(u,w)=B(u,w)+iB(u,iw)$. Applying these same identities with $w$ and $iw$ proves both asserted midpoint and additivity equalities for $P$.

Repeated additivity gives $P(mu,v)=mP(u,v)$ for every positive integer $m$. Since $u$ is the sum of $n$ copies of $u/n$, $P(u/n,v)=P(u,v)/n$. Combining these provesThe case $m=0$ also holds by additivity at zero.

Additivity applied to $u+(-u)=0$ gives $P(-u,v)=-P(u,v)$. Multiplication by $i$ preserves the norm, soFor example the first follows by multiplying $iu\pm v$ by $-i$ inside each norm. ConsequentlyThis proves the two required identities, with the convention of linearity in the first argument.

For a fixed second argument the norm formula makes $P$ continuous in its first argument. Rational homogeneity therefore extends to real homogeneity by rational approximation; additivity and $P(iu,v)=iP(u,v)$ then give full complex linearity in that argument.

The real pairing $B$ is symmetric, and $B(v,iu)=B(iu,v)=-B(u,iv)$. Hence $P(v,u)=\overline{P(u,v)}$, giving conjugate linearity in the second argument. Finally $B(u,u)=Q(u)$ and $B(u,iu)=0$, because $\Vert (1+i)u\Vert =\Vert (1-i)u\Vert$. Thereforewhich is positive for $u\ne 0$. This proves $P$ is a complex inner product inducing exactly the original norm. The assumed Banach completeness makes the resulting inner product space Hilbert. This is the Jordan-von Neumann theorem.