Analysis course of the University of Oxford (B4)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

A normed vector space over $\mathbb{R}$ is a real vector space $X$ with a function $\Vert \cdot \Vert :X\to [0,\infty )$ satisfyingIts distance is $d(x,y)=\Vert x-y\Vert$. It is a Banach space if this metric is complete: every Cauchy sequence converges to an element of $X$.

The series ${\sum }_{n\ge 1}{x}_n$ converges in $X$ when its partial sums $s_N={\sum }_{n=1}^N{x}_n$ converge in norm to an element of $X$. It is absolutely convergent when ${\sum }_n\Vert x_n\Vert <\infty$.

If $X$ is complete, thenfor an absolutely convergent series. Its partial sums are Cauchy and therefore converge.

Conversely suppose every absolutely convergent series converges. Given a Cauchy sequence $(z_n)$, choose a subsequence $(z_{n_k})$ with $\Vert z_{n_{k+1}}-z_{n_k}\Vert \le 2^{-k}$. The seriesis absolutely convergent. Its partial sums are precisely the subsequence, which therefore converges to some $z\in X$. A Cauchy sequence with a convergent subsequence converges to that same limit: estimate $\Vert z_n-z\Vert$ by $\Vert z_n-z_{n_k}\Vert +\Vert z_{n_k}-z\Vert$. Hence

Show that $c$ is a closed linear subspace of ${\ell }^{\infty }$. Suppose $a^{(j)}\in c$ and $\Vert a^{(j)}-a{\Vert }_{\infty }\to 0$. Given $\epsilon >0$, choose $j$ with uniform error below $\epsilon /3$. Since $a^{(j)}$ converges, choose $N$ such that $|a_m^{(j)}-a_n^{(j)}|<\epsilon /3$ for $m,n\ge N$. ThenThus $(a_n)$ is a real Cauchy sequence and converges, so $a\in c$. A closed linear subspace of a Banach space is complete in the inherited norm. Therefore $\boxed{(c,\Vert \cdot {\Vert }_{\infty })\text{ is Banach}}$.

Let $e_j$ be the coordinate vectors and define finitely supported vectorsEach $v_k$ belongs to ${\ell }^1$, and $\Vert v_k{\Vert }_{\infty }=2^{-k}$. Hence ${\sum }_k{v}_k$ is absolutely convergent for the supremum norm. Its partial sums converge uniformly in ${\ell }^{\infty }$ to the sequence $v$ whose entries on the $k$th block all equal $2^{-k}$. But every block contributes one to its ${\ell }^1$ sum:Thus $v\notin {\ell }^1$. Any supremum-norm limit in ${\ell }^1$ would have the same coordinates as $v$, an impossibility. The partial sums are Cauchy and have no limit in the space. Therefore $({\ell }^1,\Vert \cdot {\Vert }_{\infty })$ is not Banach, and the displayed series is the required absolutely convergent but nonconvergent series in that space.

Write $q(x)={\sup }_k|{\sum }_{n=1}^k{x}_n|$. This is indeed a norm: the triangle inequality and homogeneity follow from the scalar absolute value, and vanishing of every partial sum forces every coordinate to vanish. It is finite on ${\ell }^1$ because $q(x)\le \Vert x{\Vert }_1$.

Let $a_n=(-1{)}^{n+1}/n$ and let $a^{(N)}$ be its first $N$ entries followed by zeros. The alternating-series bound gives, for $M>N$,Thus these truncations are Cauchy for $q$. But $|x_n|\le 2q(x)$ for every coordinate, since $x_n$ is the difference of two consecutive partial sums. Any $q$-limit would consequently have coordinates $a_n$. This sequence is not in ${\ell }^1$. Hence

The continuous dual space $X^{\prime }$ is the vector space of bounded scalar-valued linear functionals, with operator normThis makes it a normed space, and in fact a Banach space whether or not $X$ is complete.

The norm-preserving Hahn-Banach theorem states that a bounded linear functional $f_0$ on any linear subspace $M\subseteq X$ extends to a bounded linear functional $f$ on all of $X$ with $f{|}_M=f_0$ and $\Vert f\Vert =\Vert f_0\Vert$. The subspace need not be closed. This holds over both the real and complex scalar fields, using the complex version when necessary.

Put $d=\mathrm{dist}(x,Z)$. If $d=0$, the zero functional works. If $d>0$, then $x\notin \overline{Z}$, and on $M=Z+\mathrm{span}\{x\}$ defineThe representation is unique, so this is linear. For $a\ne 0$,and the bound also holds for $a=0$. Thus $\Vert f_0\Vert \le 1$. The Hahn-Banach theorem extends it to $f\in X^{\prime }$ with the same bound, provingNo closedness assumption on $Z$ was required; distance is unchanged on taking its closure.

For the centered ball, the Hahn-Banach theorem supplies a norm-one real functional with $f(y)=\Vert y\Vert >r$. Hence $f(z)\le r$ on the ball, and any $t\in (r,\Vert y\Vert )$ separates it strictly from $y$.

For the general center, apply the same construction to $y-x$: choose $\Vert f\Vert =1$ and $f(y-x)=\Vert y-x\Vert$. ThenFor example,gives $f(z)<t<f(y)$ for every $z\in B$. The strict gap uses exactly the assumption that $y$ lies outside the closed ball.

Let $Z=\overline{\mathrm{span}}\{x_n:n\ge 1\}$. If $Z\ne X$, part (a) provides a nonzero bounded functional annihilating $Z$; normalize it to have norm one. Density of the sequence in the dual unit ball gives an index $n$ with $\Vert f_n-f\Vert <1/4$. Since $f(x_n)=0$ and $\Vert x_n\Vert \le 1$,But the hypothesis and reverse triangle inequality givea contradiction. Therefore the span is dense.

If $X^{\prime }$ is separable, its unit ball has a countable dense set. For each nonzero $f_n$, the definition of its norm permits choosing $x_n$ in the unit ball with $|f_n(x_n)|\ge \Vert f_n\Vert /2$; choose zero for a zero functional. The preceding result gives a dense countable span over $\mathbb{Q}$, or over $\mathbb{Q}+i\mathbb{Q}$ in the complex case. Thus $X$ is separable.

The converse fails: ${\ell }^1$ is separable, since finite rational-coordinate sequences are dense, but $({\ell }^1{)}^{\prime }\cong {\ell }^{\infty }$ is not separable. The dual identification sends a bounded sequence $a$ to $x\mapsto {\sum }_n{a}_n{x}_n$, isometrically; conversely a bounded functional is determined by its values on the coordinate vectors and hence has this form by truncation. The uncountable family of zero-one sequences in ${\ell }^{\infty }$ has pairwise distance one, which no countable dense set can approximate within a radius smaller than $1/2$.

For a bounded linear operator on a complex Banach space, its spectrum isThe complement is the resolvent set. For a real space, one may use its complexification when speaking of the complex spectrum; the algebraic arguments below apply over the underlying scalar field as well.

The bounded operators on a Banach space form a Banach algebra in the operator norm. If $\Vert T\Vert <1$, the Neumann series $A={\sum }_{k=0}^{\infty }{T}^k$ converges absolutely in that norm. Its partial sums satisfyTaking limits proves $\boxed{(I-T{)}^{-1}=\sum _{k\ge 0}{T}^k}$.

If $|\lambda |>\Vert T\Vert$, factor $\lambda I-T=\lambda (I-T/\lambda )$ and apply the same argument. Thus such $\lambda$ are in the resolvent set and

If $T$ has a bounded inverse, it is onto and $\Vert x\Vert \le \Vert T^{-1}\Vert \Vert Tx\Vert$, giving the claimed lower bound. Conversely a positive lower bound makes $T$ injective. Together with surjectivity it gives an everywhere-defined linear inverse with norm at most $1/\epsilon$. Hence it is bounded, proving the equivalence.

If $R,S$ are individually invertible, their product has inverse $S^{-1}{R}^{-1}$. Conversely assume $RS=SR$ and put $A=RS$, with $A$ invertible. Both factors commute with $A$ and therefore with $A^{-1}$. Direct multiplication givesThese are bounded two-sided inverses, so an invertible commuting product has both factors invertible. The commutation assumption is used when verifying the second side of each inverse.

The factorizationhas commuting factors because both are polynomials in $T$. If ${\lambda }^n\notin \sigma (T^n)$, their product is invertible. The preceding part would then make $T-\lambda I$ invertible, contradicting $\lambda \in \sigma (T)$. ThereforeThis proves the required inclusion without assuming the full spectral mapping theorem.

Linearity and $\Vert Tf{\Vert }_{\infty }\le 2\pi \Vert f{\Vert }_{\infty }$ show that the integral defines a bounded linear operator. Inductively,For $k=0$ this is immediate. If it holds for $k$, then $|\sin s|\le 1$ gives $|T^{k+1}f(t)|\le {\int }_0^t{s}^k\Vert f{\Vert }_{\infty }/k!ds$, which proves the next case. Thus $\Vert T^k\Vert \le (2\pi {)}^k/k!$.

For every $\lambda \ne 0$, the series $A_{\lambda }={\sum }_{k\ge 0}{\lambda }^{-k-1}{T}^k$ converges in operator norm, dominated by $|\lambda {|}^{-1}{e}^{2\pi /|\lambda |}$. Multiplying its partial sums by $\lambda I-T$ on either side gives the identity minus a remainder ${\lambda }^{-N-1}{T}^{N+1}$, which tends to zero by the factorial estimate. Therefore $A_{\lambda }$ is the bounded inverse of $\lambda I-T$.

Finally every function in the image of $T$ vanishes at zero, so $T$ is not onto the continuous functions. Hence $0$ belongs to its spectrum and

The graph $G(T)=\{(x,Tx):x\in X\}$ is closed if it is a closed subset of $X\times Y$. Equivalently, $x_n\to x$ and $T{x}_n\to y$ imply $Tx=y$.

The closed graph theorem states that an everywhere-defined linear map between Banach spaces is bounded if and only if its graph is closed. Completeness of both spaces is part of the theorem. A bounded map has a closed graph by continuity.

Conversely, if the graph is closed, it is a Banach space in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. The coordinate projection $P:G(T)\to X$, $P(x,Tx)=x$, is a bounded linear bijection. The open mapping theorem makes $P^{-1}$ bounded. The other coordinate projection $Q:G(T)\to Y$ is bounded, and $T=Q{P}^{-1}$ is therefore bounded. This proves the theorem from the permitted result.

For scalars $a,b$ and $f,h\in X$, linearity of the Fourier coefficients givesThe assumed uniqueness of the function with these coefficients implies $T(af+bh)=aTf+bTh$. Thus $T$ is linear.

For each fixed $n$, the coefficient functional is bounded sinceSuppose $f_j\to f$ and $T{f}_j\to g$ uniformly. Taking coefficients and passing to the limit givesfor every fixed $n$. No uniform bound on ${\alpha }_n$ is needed for this passage. Uniqueness gives $g=Tf$, so $T$ has a closed graph. The space $X$ is Banach, being the closed subspace defined by $f(-\pi )=f(\pi )$ inside the continuous functions with the uniform norm. The closed graph theorem therefore proves $T$ is bounded.

The normalization givesThus ${\alpha }_n{e}_n$ has exactly the coefficients prescribed for $T{e}_n$. Uniqueness yieldsThese are eigenvectors, including the zero-eigenvalue cases.

Since every $e_n$ is nonzero, boundedness givesCancel its norm to obtain $\boxed{\underset{n\in \mathbb{Z}}{\sup }|{\alpha }_n|\le \Vert T\Vert }$. Thus the multiplier sequence is bounded.

Evaluation at zero is a bounded linear functional for the uniform norm. DefineIt is linear and satisfies $|\varphi (f)|\le \sqrt{2\pi }\Vert T\Vert \Vert f{\Vert }_{\infty }$. Since $e_n(0)=(2\pi {)}^{-1/2}$,The factor $\sqrt{2\pi }$ is required by the source's basis normalization.

The orthogonal complement isIt is a linear subspace and is closed, since it is an intersection of kernels of the continuous linear functionals $w\mapsto ⟨w,y⟩$. Continuity follows from the Cauchy-Schwarz inequality.

The minimizing point and orthogonality proved below write every $x$ as $x=y_0+(x-y_0)\in Y+Y^{\perp }$. If $w\in Y\cap Y^{\perp }$, then $\Vert w{\Vert }^2=⟨w,w⟩=0$, so the intersection is zero. ConsequentlyIn particular the minimizing point is unique: an alternative orthogonal decomposition would have a difference in this zero intersection.

Let $d={\inf }_{y\in Y}\Vert x-y\Vert$ and choose $y_n\in Y$ with $\Vert x-y_n\Vert \to d$. The parallelogram law givesThe right side tends to zero. Thus $(y_n)$ is Cauchy. The closed subspace $Y$ of a Hilbert space is complete, so $y_n\to y_0\in Y$. Continuity of the norm gives $\Vert x-y_0\Vert =d$, proving the distance is attained.

Put $w=x-y_0$. For any $y\in Y$ and any real $t$, minimality givesThe derivative at zero must vanish, so $\mathrm{Re}⟨w,y⟩=0$. In a complex Hilbert space, apply the same argument to $iy$ to get the imaginary part zero as well. In the real case the first conclusion is already enough. Hence

For $h,g\in Y$, their product $hg$ is holomorphic and integrable on the disk by the Cauchy-Schwarz inequality. Its angular mean on every centered circle is $h(0)g(0)$ by the mean value property. Polar integration therefore givesIf $g(0)=0$, this vanishes for every $h\in Y$, proving $\boxed{Z\subseteq Y^{\perp }}$. Integrability of the product justifies integration over the full disk, even when the individual holomorphic functions grow near its boundary.

The inclusion is strict. Set $w(z)=|z{|}^2-1/2$, a nonzero bounded function in $L^2(D)$. For every $h\in Y$,Thus $w\in Y^{\perp }$. It cannot equal $\overline{g}$ for holomorphic $g$: equality almost everywhere between these continuous functions would give equality everywhere, making $g=w$ a real-valued nonconstant holomorphic function, which is impossible. Therefore

Use the convention that the inner product is linear in its first argument. The precise polarization identity isIf (1) holds, apply this formula before and after $U$ to get preservation of all inner products, proving (2). If (2) holds, the identity $⟨Ux,Uy⟩=⟨x,U^{\ast }Uy⟩$ gives $U^{\ast }U=I$. For $z=Ux$, surjectivity then gives $U{U}^{\ast }z=U{U}^{\ast }Ux=Ux=z$, proving (3).

From (3), $\Vert Ux{\Vert }^2=⟨x,U^{\ast }Ux⟩=\Vert x{\Vert }^2$ and likewise $\Vert U^{\ast }x\Vert =\Vert x\Vert$, proving (4). Finally, if (4) holds, apply the same polarization identity to $U^{\ast }$ to obtain $U{U}^{\ast }=I$. This makes $U$ onto; the other norm identity supplies the remaining requirement of (1). ThusThese four statements are conditions in a single equivalence proof, rather than four independent source questions.

The four claims below show that $I+T$ has a bounded inverse and that the Cayley transform of a bounded accretive operator is contractive. For an arbitrary $z\in X$, write $z=(I+T)x$. ThenTaking the supremum over the unit ball gives

Expanding both squared norms givesBoth norms are nonnegative, so taking square roots proves $\boxed{\Vert (I+T)x\Vert \ge \Vert (I-T)x\Vert }$.

A separate expansion givesHence $\boxed{\Vert (I+T)x\Vert \ge \Vert x\Vert }$. In particular $I+T$ is injective and bounded below by one.

By the definition of the adjoint operator and conjugate symmetry,Taking real parts proves

The lower bound in (ii) makes the image of $I+T$ closed: if $(I+T)x_n$ converges, the bound makes $(x_n)$ Cauchy; completeness gives its limit and hence a image preimage. The same bound applies to $I+T^{\ast }$ by (iii), so its kernel is zero. The orthogonal complement identityshows that the image is dense, using the orthogonal decomposition by a closed subspace for its closure. Being both closed and dense, it equals $X$. Thus $I+T$ is onto and bounded below, and by part 3(a)(iii) it has a bounded inverse. Moreover

If the real part vanishes identically, the norm-difference identity becomes equality, soThus the same substitution as above gives $\Vert Sz\Vert =\Vert z\Vert$ for every $z$. To prove surjectivity as well, observe that both $T$ and $-T$ satisfy the nonnegative-real-part hypothesis. The preceding part makes both $I+T$ and $I-T$ invertible. Their product $S=(I-T)(I+T{)}^{-1}$ is therefore invertible and onto. It is a surjective norm-preserving linear operator, so the equivalences in (a) prove

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.