Past exam of the analysis course of the University of Oxford / 2010 (B4, DMAB 2643, DMCN 2643, DMPB 2643, DMST 2643)

For $\alpha ,\beta \in {\ell }_{1,2}$ and real scalars $c,d$, the scalar triangle inequality givesThe zero sequence belongs to the space, so it is a real vector space. The weighted sum is nonnegative and homogeneous, and the same inequality with $c=d=1$ proves its triangle inequality. Since every weight $j^2$ is strictly positive, zero weighted sum forces every coordinate to vanish. Therefore $\Vert \cdot {\Vert }_{1,2}$ is a norm.

Define $U:{\ell }_{1,2}\to {\ell }_1$ by $(U\alpha {)}_j=j^2{\alpha }_j$. This is a linear bijection, with inverse $(U^{-1}\gamma {)}_j=j^{-2}{\gamma }_j$, andThus $U$ is an isometric isomorphism of normed spaces onto the absolutely summable sequence space, which is a Banach space. Explicitly, a Cauchy sequence ${\alpha }^{(n)}$ maps to a Cauchy sequence $U{\alpha }^{(n)}\to \gamma \in {\ell }_1$; setting $\alpha =U^{-1}\gamma$ gives convergence in the weighted norm. Hence ${\ell }_{1,2}$ is a Banach space, an instance of weighted summable sequence space isometry.

Let $D$ be the set of finitely supported sequences with rational coordinates. It is countable: it is the union over $N$ of copies of ${\mathbb{Q}}^N$. Given $\alpha \in {\ell }_{1,2}$ and $\epsilon >0$, choose $N$ with ${\sum }_{j>N}{j}^2|{\alpha }_j|<\epsilon /2$. Choose rational numbers $q_j$ for $1\le j\le N$ so that ${\sum }_{j=1}^N{j}^2|q_j-{\alpha }_j|<\epsilon /2$, and put $q_j=0$ thereafter. Then $q\in D$ and $\Vert q-\alpha {\Vert }_{1,2}<\epsilon$. Thus $D$ is countable and dense, so the space is separable.

For the first linear map, take $(\alpha ,\beta )=(0,e_n)$. Its input norm is one, while $T_1(0,e_n)=(e_n,0)$ has norm $n^2$. Thus $T_1$ is unbounded.

For the second map, since $j^2\ge 1$,Equality holds at $(0,e_1)$, soBoth formulas map $c_{00}\times c_{00}$ into itself because adding two finitely supported sequences preserves finite support.

Equip the continuous dual space $X^{\prime }$ with the operator normContinuity of a linear functional makes this finite; homogeneity, definiteness and the triangle inequality follow directly from the supremum formula.

For completeness of the dual space, let $(f_n)$ be Cauchy in this norm. For each $x\in X$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $(f_n(x))$ is Cauchy in $\mathbb{R}$. Define $f(x)={\lim }_n{f}_n(x)$. Passing to the limit in the linearity identities makes $f$ linear. Since a Cauchy sequence is bounded, ${\sup }_n\Vert f_n\Vert \le C$ and $|f(x)|\le C\Vert x\Vert$, so $f\in X^{\prime }$.

Given $\epsilon >0$, choose $N$ with $\Vert f_n-f_m\Vert \le \epsilon$ for $n,m\ge N$. Let $m\to \infty$ pointwise; then $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $X^{\prime }$ is a Banach space. This proof in fact works even if $X$ itself is not complete.

If a bilinear map is continuous, continuity at $(0,0)$ gives $\delta >0$ such that $\Vert F(x,y)\Vert \le 1$ whenever $\Vert x\Vert ,\Vert y\Vert <\delta$. For nonzero $x,y$, apply this to $\delta x/(2\Vert x\Vert )$ and $\delta y/(2\Vert y\Vert )$. Bilinearity givesIf either input is zero, the inequality follows from bilinearity directly.

Conversely, suppose the product bound holds with $C\ge 0$. At a fixed point $(x,y)$,Its norm is at most $C(\Vert h\Vert \Vert y\Vert +\Vert x\Vert \Vert k\Vert +\Vert h\Vert \Vert k\Vert )$, which tends to zero as $(h,k)\to (0,0)$. Thus the product bound is equivalent to joint continuity. This is boundedness criterion for a bilinear map.

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

If $|\lambda |>\Vert T\Vert$, write $T-\lambda I=-\lambda (I-T/\lambda )$. The Neumann series applies to $T/\lambda$, so $\lambda$ is outside the spectrum of a bounded operator. Thus $\sigma (T)\subseteq \{\lambda :|\lambda |\le \Vert T\Vert \}$ and $r(T)\le \Vert T\Vert$.

The polynomial spectral mapping theorem gives $\sigma (T^m)=\{{\lambda }^m:\lambda \in \sigma (T)\}$, and henceTaking $m$th roots and then the limit, whose existence is allowed, provesOnly this inequality is required; the reverse inequality is the additional content of the full spectral radius formula.

For the multiplication operator, $\Vert Tf{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }\Vert f{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. Taking $f=1$ gives equality, so $\Vert T\Vert =1$.

An eigenfunction obeys $(g(t)-\lambda )f(t)=0$ at every $t$. If $\lambda \in [0,1)$, its level set is the single point $t=\lambda$, and a continuous function supported at a single point is zero. If $\lambda \notin [0,1]$, that level set is empty. For $\lambda =1$, every continuous function vanishing on $[0,1]$ is an eigenfunction, and nonzero examples exist, such as $f(t)=\max \{t-1,0\}$. Thus ${\sigma }_p(T)=\{1\}$.

If $\lambda \notin [0,1]$, multiplication by $(g-\lambda {)}^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda \in [0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore

The unilateral shift operator $R$ is not onto, since its outputs have first coordinate zero. The left shift operator $L$ is not injective, since $L{e}_1=0$. Both have operator norm one, and $LR=I$. ConsequentlyIf $\Vert A\Vert <1$, then $\Vert LA\Vert <1$, so the Neumann series makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would givea product of invertible operators, contradicting the nontrivial kernel of $L$. ThusThis is norm stability of noninvertibility for the unilateral right shift.

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets. The Uniform boundedness principle says that if $V$ is a Banach space, $W$ a normed vector space, and $\mathcal{T}\subseteq B(V,W)$ is pointwise bounded, meaning ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in V$, then ${\sup }_{T\in \mathcal{T}}\Vert T\Vert <\infty$.

To prove it, let $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Each set is closed by continuity of the individual operators, and their union is $V$. The Baire category theorem gives a ball $B(x_0,r)$ inside some $E_N$. For $\Vert h\Vert <r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\Vert Th\Vert \le \Vert T(x_0+h)\Vert +\Vert T{x}_0\Vert \le 2N$ for every $T$. Scaling gives $\Vert T\Vert \le 2N/r$ uniformly. The zero-space and empty-family cases are immediate.

For the first family on a Hilbert space, fix $x$. The functionals $y\mapsto ⟨y,Tx⟩$ have operator norm $\Vert Tx\Vert$ by the Cauchy-Schwarz inequality and equality on the direction of $Tx$. Their values are pointwise bounded by the hypothesis, after taking complex conjugates if necessary. Apply the Uniform boundedness principle to obtain ${\sup }_{T\in \mathcal{E}}\Vert Tx\Vert <\infty$. Apply it a second time to the operator family itself to conclude ${\sup }_{T\in \mathcal{E}}\Vert T\Vert <\infty$.

For the diagonal hypothesis, the answer depends on the scalar field. Over $\mathbb{C}$, let $q_T(x)=⟨Tx,x⟩$, using the convention that the inner product is linear in its first variable. The complex polarization identity givesThe assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus the complex-space answer is yes, even without assuming self-adjointness.

Over $\mathbb{R}$, take $X={\mathbb{R}}^2$ andEvery diagonal value $⟨nJx,x⟩$ is zero, whereas $\Vert nJ\Vert =n$. Thus the real-space answer is no in general. Real diagonal values do not detect the skew-symmetric part of an operator. This is diagonal boundedness of operator families over real and complex Hilbert spaces.

A bounded linear operator on a complex Hilbert space is self-adjoint if $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y$.

If $Tx=0$, then $⟨x,Ty⟩=⟨Tx,y⟩=0$ for all $y$, so $x\in (\mathrm{Ran}T{)}^{\perp }$. Conversely, membership in that orthogonal complement gives $⟨Tx,y⟩=0$ for every $y$, and choosing $y=Tx$ yields $Tx=0$. Therefore

Self-adjointness and conjugate symmetry of the inner product giveA complex number equal to its conjugate is real, so $⟨Tx,x⟩\in \mathbb{R}$ for every $x$.

For a positive operator, $⟨Tx,x⟩\ge 0$. We show directly that every nonreal or negative real $\lambda$ lies in the resolvent set.

If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality and reality of $⟨Tx,x⟩$ giveThus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar{\lambda }I$, which has trivial kernel by the same estimate. The identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.

If $\lambda <0$ is real, positivity gives $⟨(T-\lambda I)x,x⟩\ge (-\lambda )\Vert x{\Vert }^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is self-adjoint, so its range is dense because its own kernel is zero. Hence

Set $B(x,y)=⟨Tx,y⟩$. It is a positive semidefinite Hermitian form. Write $A=B(x,x)$, $C=B(y,y)$ and $c=B(x,y)$. For every complex $t$,If $C>0$, choosing $t=-c/C$ gives $A-|c{|}^2/C\ge 0$. If $C=0$ and $c\ne 0$, choosing $t=-sc$ for large positive $s$ makes the expression negative, a contradiction. Thus $c=0$ in that case. In all cases,This is the Cauchy-Schwarz inequality for a positive operator.

For $x\in \ker (S+T)$,Both terms are nonnegative because $S,T$ are positive operators. Their sum is zero, so each vanishes. In particular,

By the Cauchy-Schwarz inequality for a positive operator, for every $y$,Taking $y=Tx$ gives $Tx=0$, and $(S+T)x=0$ then gives $Sx=0$. ThusThe reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.

Commutativity and equality of squares giveThus $S-T$ vanishes on $\mathrm{Ran}(S+T)$, and by continuity on its closure. The preceding kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Since $S+T$ is self-adjoint, part (i) and the Hilbert projection theorem give the orthogonal direct sumTherefore $S-T$ vanishes on all of $X$, provingThis proves commuting positive operators with equal squares are equal without assuming a square-root uniqueness theorem.

For finite subsets $F\subset \mathbb{Z}$, orthonormality givesFor $S_N={\sum }_{|n|\le N}{\alpha }_n{e}_n$, the squared norm of $S_N-S_M$ is the corresponding finite tail sum. Since ${\sum }_{n\in \mathbb{Z}}|{\alpha }_n{|}^2<\infty$, these tails tend to zero. Thus $(S_N)$ is a Cauchy sequence in the complete Hilbert space and converges.

The same estimate for arbitrary finite tails proves that the sum is unconditional, so the bilateral orthogonal series converges in $X$ independently of its ordering. This is the Riesz-Fischer theorem for an orthonormal system.

For $a_n=⟨x,e_n⟩$, the finite orthogonal projection $P_{N}x={\sum }_{|n|\le N}{a}_n{e}_n$ satisfiesThis is Bessel's inequality, so the coefficients are square summable. The preceding result gives a limit $s={\sum }_n{a}_n{e}_n$. For each fixed $k$, continuity of the inner product gives $⟨x-s,e_k⟩=0$. The completeness assumption on the orthonormal system forces $x-s=0$. HenceTaking squared norms also gives Parseval's identity, $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$.

Apply continuity of the inner product to the convergent expansion of $x$:The series is absolutely convergent because the Cauchy-Schwarz inequality for scalar sequences and Bessel's inequality giveThis is the polarized Parseval identity.

Use the inner product convention $⟨f,h⟩={\int }_{-\pi }^{\pi }f\bar{h}$. For $h=1_{[-\pi /2,\pi /2]}$, integration givesThe nonzero even coefficients vanish, while $\sin ((2n+1)\pi /2)=(-1{)}^n$. The preceding polarized Parseval identity therefore yieldsAbsolute convergence follows from square summability of the Fourier coefficients and of $((2n+1{)}^{-1}{)}_{n\in \mathbb{Z}}$.

For the convolution identity, fix any real $t$ and define $h_t(s)=\overline{g(t-s)}$. Periodicity makes $h_t\in L^2(-\pi ,\pi )$ with $\Vert h_t{\Vert }_2=\Vert g{\Vert }_2$. A change of variable and integration over one full period giveApplying polarized Parseval identity to $f,h_t$ gives, for every fixed $t$,The integral exists by the Cauchy-Schwarz inequality. The same inequality gives ${\sum }_n|⟨f,e_n⟩⟨g,e_n⟩|\le \Vert f{\Vert }_2\Vert g{\Vert }_2$, so the displayed series converges absolutely and uniformly in $t$. In particular the equality holds at every $t$, without requiring pointwise convergence of either original Fourier series. This is Fourier convolution of two periodic L2 functions.