Past exam of the analysis course of the University of Oxford / 2009 / A

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.