Past exam of the analysis course of the University of Oxford / 2009 (B4, DMAB 2643, DMCN 2643, DMPB 2643)

A normed vector space is a real or complex vector space $X$ with a norm satisfying positivity, definiteness, absolute homogeneity and the triangle inequality. It is separable if it has a countable dense subset for the norm distance.

First prove the unlabelled general assertions. If $(x_n)$ is dense in $X$, choose $y_{n,m}\in Y\cap B(x_n,1/m)$ whenever that intersection is nonempty. These choices form a countable subset of $Y$. For any $y\in Y$ and $\epsilon >0$, choose $m$ with $2/m<\epsilon$ and $n$ with $\Vert x_n-y\Vert <1/m$. The corresponding intersection is nonempty and $\Vert y-y_{n,m}\Vert <2/m<\epsilon$. Thus every subspace of a separable normed vector space is separable.

If a countable $S$ has dense linear span, take all finite linear combinations of its elements with rational coefficients in the real case, or coefficients in $\mathbb{Q}+i\mathbb{Q}$ in the complex case. This set is countable and dense in the linear span, by approximating the finitely many coefficients of each combination. Its closure is therefore $X$, proving separability. Conversely, a countable dense set already has dense linear span.

For $c_0$, use all finitely supported sequences with coordinates in $\mathbb{Q}+i\mathbb{Q}$. Truncate a sequence tending to zero to control its supremum tail, then approximate the finite initial coordinates. This produces a countable dense subset, so

Map each bounded sequence $a=(a_n)$ to the continuous function that takes value $a_n$ at the integer $n\ge 1$ and is affine on every interval $[n,n+1]$. Convexity of absolute value gives $|F_a(t)|\le {\sup }_n|a_n|$, and evaluation at integers gives equality of suprema. Thus $a\mapsto F_a$ is an isometric embedding of the l-infinity sequence space into $C^b([1,\infty ))$.

The former is not separable by the allowed fact. If the latter were separable, the subspace result just proved would make its isometric copy of ${\ell }_{\infty }$ separable, a contradiction. Hence

For $1<p<\infty$, truncation makes the ${\ell }^p$ tail arbitrarily small, since ${\sum }_j|a_j{|}^p$ converges. Approximate the finite remaining coordinates by numbers in $\mathbb{Q}+i\mathbb{Q}$. The resulting countable set of finitely supported sequences is dense. ThereforeThe same construction also works for $p=1$, though that endpoint is not needed here.

The norm-preserving form of the Hahn-Banach theorem states: if $M$ is a linear subspace of a real or complex normed vector space $X$ and $f:M\to \mathbb{K}$ is a bounded linear functional, there exists $F\in X^{\ast }$ with $F{|}_M=f$ and $\Vert F\Vert =\Vert f\Vert$.

For a real separable $X$, put $C=\Vert f\Vert$. The case $C=0$ is immediate. To extend over one vector $x_0\notin M$, choose a real number $c$ betweenThe lower endpoint is no greater than the upper: for any $m,n\in M$,The endpoints are finite because testing zero bounds them between finite numbers. Define $F(m+t{x}_0)=f(m)+tc$. The upper inequality ensures $F(z)\le C\Vert z\Vert$ for vectors with positive $x_0$ coefficient, and the lower inequality gives it for negative coefficient; it already holds on $M$. Applying the same bound to $-z$ gives $|F(z)|\le C\Vert z\Vert$. Thus the extension preserves the norm.

Now add a countable dense sequence of vectors one at a time. The consistent extensions define a bounded functional of norm $C$ on the union of the increasing domains. That union is dense in $X$, so extension by continuity gives the desired functional on all of $X$. This proves the real separable case without a maximality argument.

For the density application, suppose the closed linear span $M$ were proper. The Hahn-Banach theorem supplies a nonzero continuous linear functional annihilating $M$. By duality of sequence spaces, it has the form $f(z)={\sum }_{j\ge 1}{a}_j{z}_j$ with $a\in {\ell }_1$, not identically zero. Let $m$ be the first index with $a_m\ne 0$. Writing $t_n=(n+1{)}^{-1}$, annihilation of $x_n$ givesThe tail has absolute value at most $t_n{\sum }_{j>m}|a_j|\to 0$. Letting $n\to \infty$ contradicts $a_m\ne 0$. ThereforeThis is geometric sequences with accumulating ratios span c0.

First give the general definitions and radius facts. The spectrum of a bounded operator is $\sigma (A)=\{\lambda :A-\lambda I\text{ has no bounded inverse}\}$, and the point spectrum is $p\sigma (A)=\{\lambda :\ker (A-\lambda I)\ne 0\}$. If $|\lambda |>\Vert A\Vert$, the Neumann series inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda (I-A/\lambda )$. Thus $\sigma (A)$ is contained in the closed disc of radius $\Vert A\Vert$ and is bounded.

The spectral radius is $\nu (A)={\sup }_{\lambda \in \sigma (A)}|\lambda |$. The spectral radius formula isFor the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j={\lambda }^{j-1}{a}_1$. A nonzero such sequence is in ${\ell }_1$ exactly when $|\lambda |<1$. Since $\Vert R\Vert =1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as point spectrum and therefore its closure by closedness of the spectrum. Hence

For the diagonal operator $S$, the equation $Sa=\lambda a$ gives $(j^{-1}-\lambda )a_j=0$ for every $j$. Thus a nonzero eigenvector exists exactly for $\lambda =1/j$, with eigenvector $e_j$; zero is not an eigenvalue because all diagonal entries are nonzero.

If $\lambda$ lies outside $\{0\}\cup \{1/j:j\ge 1\}$, its distance from this closed set is positive. Dividing each coordinate by $j^{-1}-\lambda$ gives a bounded inverse of $S-\lambda I$. Zero lies in the spectrum: $\Vert S{e}_j{\Vert }_1=1/j\to 0$, excluding a bounded inverse. Therefore

The weighted right shift $T$ is injective, so zero is not an eigenvalue. If $Ta=\lambda a$ with $\lambda \ne 0$, the first coordinate forces $a_1=0$ and subsequent coordinates force every $a_j=0$. Thus its point spectrum is empty.

For $m\ge 1$,since the largest column sum is attained at $j=1$. Therefore the spectral radius formula gives $\nu (T)={\lim }_m(m!{)}^{-1/m}=0$. The limit follows, for example, from $m!\ge (m/2{)}^{m/2}$ for large $m$. Zero is in the spectrum because the output first coordinate is always zero, so $T$ is not onto. HenceThis is a nonzero quasinilpotent element, specifically the factorial weighted right shift.

A complete orthonormal sequence $(e_n)$ has $⟨e_n,e_m⟩={\delta }_{nm}$ and dense linear span; equivalently, the only vector orthogonal to every $e_n$ is zero. The Parseval identity says $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$, with expansion $x={\sum }_n⟨x,e_n⟩e_n$ in the Hilbert norm.

Put $G(t,z)=(1-2zt+z^2{)}^{-1/2}$ and $D=1-2zt+z^2$. Direct differentiation givesSubstitution yields $(1-t^2)G_{tt}-2t{G}_t+z^2{G}_{zz}+2z{G}_z=0$. Extracting the coefficient of $z^n$ in the generating function givesthe Legendre differential equation.

The generating function gives $P_n(-t)=(-1{)}^n{P}_n(t)$, so the odd function $f$ is orthogonal to every even-degree Legendre polynomial. For $n\ge 1$, integrate the self-adjoint form of the Legendre differential equation from zero to one:Thus ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. Differentiating the generating function in $t$ and then setting $t=0$ givesCombining these identities with the normalization of $p_n$ yieldsFor the Legendre expansion of the sign function, $\Vert f{\Vert }_2^2=2$, so the Parseval identity givesTherefore the requested sum equals

Expand both generating functions and use orthogonality of the Legendre polynomials, with ${\int }_{-1}^1{P}_n{P}_m=2{\delta }_{nm}/(2n+1)$. The allowed exchanges giveHence it depends only on $r=wz$. Explicitly,The series form alone already proves the required dependence.

With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.

The Hilbert projection theorem states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed linear subspace $M$, with $m\in M$ and $n\in M^{\perp }$. The vector $m$ is the unique nearest point in $M$.

The Riesz representation theorem states that every bounded complex-linear functional $F$ on a Hilbert space has a unique representation $F(x)=⟨x,y⟩$, with $\Vert F\Vert =\Vert y\Vert$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^{\perp }$ with $F(z)\ne 0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, soThis gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The Cauchy-Schwarz inequality gives $\Vert F\Vert \le \Vert y\Vert$, and evaluation at $y/\Vert y\Vert$ gives equality when $y\ne 0$.

For the adjoint, fix $v$. The functional $u\mapsto ⟨Tu,v⟩$ is bounded with norm at most $\Vert T\Vert \Vert v\Vert$. By the Riesz representation theorem, there is a unique vector $T^{\ast }v$ satisfyingFor scalars $a,b$, conjugate-linearity in the second argument shows that $T^{\ast }(av+bw)=a{T}^{\ast }v+b{T}^{\ast }w$, by uniqueness of the representing vector. The same norm estimate gives $\Vert T^{\ast }v\Vert \le \Vert T\Vert \Vert v\Vert$. Thus $T^{\ast }$ is a unique bounded linear operator, the adjoint operator.

For the adjoint of an integration operator anchored at zero, use the inner product $⟨f,g⟩=\int f\bar{g}$. On the positive half-interval, reversing the triangular integration order givesOn the negative half-interval the corresponding calculation givesThese exchanges are justified by the Fubini theorem, since on a finite interval $L^2$ functions belong to $L^1$ and the absolute double integral is at most $\Vert f{\Vert }_1\Vert g{\Vert }_1$. ThereforeAdding $Tg$ gives ${\int }_0^{\pi }g$ on the positive half and $-{\int }_{-\pi }^{0}g$ on the negative half, exactly the prescribed $Ug$. Values at zero do not matter in $L^2$. Consequently

Since $f(0)=0$, the fundamental theorem of calculus gives $f=T{f}^{\prime }$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued functionChoose any value at zero. This function is bounded and hence belongs to $L^2(-\pi ,\pi ]$. Since it is real, the adjoint operator identity yieldsOne can also check this by integration by parts separately on each half-interval: $g^{\prime }=-t^2$, $g(\pm \pi )=0$, and the two possible jump terms at zero vanish because $f(0)=0$.