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

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.

Differentiating the Rodrigues identity gives $H_0=1$ and $H_{n+1}=2t{H}_n-H_n^{\prime }$. Induct with the proposed finite sum. The coefficient at $t^{n+1-2j}$ in this recurrence is the contribution from $2t{H}_n$ minus the derivative contribution with previous index $j-1$:Using a common denominator, this equals $(n+1)!(-1{)}^j{2}^{n+1-2j}/[j!(n+1-2j)!]$. At an endpoint omit terms whose factorial indices would be negative; the same formula then covers the sole remaining term. ThusThe initial value and recurrence identify these polynomials with the given Rodrigues definition.

Substitute the finite formula and write $n=m+2j$. Under the allowed rearrangements,Hence $G(t,z)=e^{2tz-z^2}$, with the normalization of the physicists' Hermite polynomials used in the source.

Complete the square to obtain, initially for real $y,z$,Compare coefficients of $y^m{z}^n$ on the two sides. The left coefficient is $(m!n!{)}^{-1}\int e^{-t^2}{H}_m(t)H_n(t)dt$, and the right is zero unless $m=n$, in which case it is $\sqrt{\pi }{2}^n/n!$. The polynomials are real on the real line, so these are the complex $L^2$ inner products of the ${\varphi }_n$. ThusPolynomial times Gaussian decay also directly shows each ${\varphi }_n$ lies in $L^2$.

The classical Fourier partial sums areHere $a_0={\pi }^{-1}\int f$, and functions are extended periodically. One version of the Dini test for Fourier series states that if one-sided limits $L_{+},L_{-}$ at $t$ exist andthen $s_n^f(t)\to (L_{+}+L_{-})/2$. A step function on a finite interval is constant on each side of any given point for sufficiently small distance, so its displayed numerator is zero there. Dini test for Fourier series gives exactly the asserted midpoint of the two one-sided values, including at jumps.

Let $P_n=s_n$ on $L^2$. It is the orthogonal projection onto the span of the displayed trigonometric functions through frequency $n$, hence $\Vert P_n\Vert \le 1$. For a step-function approximation $\psi$,The convergence on a dense set under a uniform operator bound, here applied using the allowed convergence for $\psi$ and density of step functions, proves $P_{n}f\to f$ in $L^2$. Consequently the union of these finite trigonometric spans is dense, which is precisely completeness of the orthonormal system. It also proves Parseval's equality by passing to the limit in the Pythagorean projection identity.

For $f(t)=t^2$, oddness gives $b_j=0$, direct integration gives $a_0=2{\pi }^2/3$, and two integrations by parts give $a_j=4(-1{)}^j/j^2$. ThusDini test for Fourier series justifies this pointwise on the interval, or absolute coefficient summability gives its continuous extension after the $L^2$ identification. At zero it yields ${\sum }_{j\ge 1}(-1{)}^{j-1}/j^2={\pi }^2/12$. Parseval givesTherefore ${\sum }_{j\ge 1}{j}^{-4}={\pi }^4/90$.

Anti-self-adjoint means $T^{\ast }=-T$. Unitary means $T^{\ast }T=T{T}^{\ast }=I$, so $T^{-1}=T^{\ast }$. For a unitary operator $\Vert Tu{\Vert }^2=⟨u,T^{\ast }Tu⟩=\Vert u{\Vert }^2$, and the space is nonzero. Thus $\Vert T\Vert =1$.

The series for the exponential converges absolutely in the Banach algebra $B(V)$, since ${\sum }_n\Vert T^n/n!\Vert \le e^{\Vert T\Vert }$. The adjoint is norm continuous and conjugate linear, and its action on these real scalar coefficients gives $(e^T{)}^{\ast }=e^{T^{\ast }}$. The absolutely convergent Cauchy product givesbecause its degree-$n$ coefficient for $n>0$ is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]=0$. If $T^{\ast }=-T$, these identities prove $e^T$ is unitary.

If $T$ is invertible, zero is outside the spectrum of $T^{-1}$. For $\mu \ne 0$ the factorizationshows that the left side is invertible exactly when the final factor is. HenceFor unitary $T$, both $T$ and $T^{-1}=T^{\ast }$ have norm one. The norm bound on each spectrum gives $|\lambda |\le 1$ and $|{\lambda }^{-1}|\le 1$, forcing $\sigma (T)$ to lie on the unit circle.

The converse fails: on ${\mathbb{C}}^2$ with its usual inner product takeIts spectrum is $\{1\}$, but $T^{\ast }T=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\ne I$, so it is not unitary.