Problem Sheet 2 Q4

Linearity of \(T\) is obvious. For \(T\) to be a bounded linear operator from \(\ell_1\) to itself, it is necessary and sufficient for \(a_n\) to be pointwise bounded, that is exist \(M\) such that each \(\left\lvert a_n\right\rvert < M\).

Suppose exists \(M\) such that for all \(n\), \(\left\lvert a_n\right\rvert < M\). Let \(x\in \ell_1\) and consider \(Tx = (a_nx_n)\). Now \[\left\lVert Tx\right\rVert = \sum_n \left\lvert a_nx_n\right\rvert \leq \sum_n M\left\lvert x_n\right\rvert = M \left\lVert x\right\rVert.\]

Conversely suppose \(T\) is bounded with \(\left\lVert T\right\rVert\leq N\). For each \(j\in\mathbb{N}\) consider \(e_j = (\delta_{ij})_{i\in\mathbb{N}}\), and we know that \(\left\lVert e_j\right\rVert = 1\). As \(\left\lVert T\right\rVert < N\), \[ \left\lVert Te_j\right\rVert = \left\lvert a_j\right\rvert \leq N \] and as a result \(N\) also bounds each \(\left\lvert a_j\right\rvert\).

Problem Sheet 3 Q1

Choose a basis \(\left\{e_1,\dots,e_n\right\}\) for \(V\). To show that \(T\) is continuous we just need to show that \(T\) is bounded. Consider any \(v\in V\) \[\begin{align*} \left\lVert Tv\right\rVert &= \left\lVert T\left( \sum_i v_i e_i \right) \right\rVert \\ &\leq \sum_i \left\lvert v_i\right\rvert \left\lVert Te_i\right\rVert \end{align*}\] now let \(M = \max_i \left\lVert T e_i\right\rVert\) and as any two finite-dimensional norms are equivalent, there exists \(N\) such that \[ \sum_i \left\lvert v_i\right\rvert \leq N\left\lVert v\right\rVert \] and therefore \(\left\lVert Tv\right\rVert \leq MN\left\lVert v\right\rVert\).