Past exam of the analysis course of the University of Oxford / 2010 / A / 1

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.