Past exam of the analysis course of the University of Oxford / 2011 / 1

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