We return now to the consideration of convergence problems. There are three obvious senses in which we may try to define the convergence of a sequence \((A_{n})\) of linear transformations to a fixed linear transformation \(A\) . \begin{align} &\|A_n \to A\| \text{ as } n \to \infty. \tag{i}\\ &\|A_nx - Ax\| \to 0 \text{ as } n \to \infty \text{ for each fixed } x. \tag{ii}\\ &|(A_nx, y) - (Ax, y)| \to 0 \text{ as } n \to \infty \text{ for each fixed $x$ and $y$}. \tag{iii} \end{align} 

If (i) is true, then, for every \(x\) , \begin{align} \|A_{n} x-A x\| &= \|(A_{n}-A) x\|\\ &\leq \|A_{n}-A\| \cdot \|x\|\\ &\to 0 \end{align} so that (i) \(\implies\) (ii). We have already seen ( Section: Convergence of vectors ) that (ii) \(\implies\) (iii) and that in finite-dimensional spaces (iii) \(\implies\) (ii). It is even true that in finite-dimensional spaces (ii) \(\implies\) (i), so that all three conditions are equivalent. To prove this, let \(\{x_{1}, \ldots, x_{N}\}\) be an orthonormal basis in \(\mathcal{V}\) . If we suppose that (ii) holds, then, for each \(\epsilon>0\) , we may find an \(n_{0}=n_{0}(\epsilon)\) such that \(\|A_{n} x_{i}-A x_{i}\|<\epsilon\) for \(n \geq n_{0}\) and for \(i=1, \ldots, N\) . It follows that for an arbitrary \(x=\sum_{i}(x, x_{i}) x_{i}\) we have \begin{align} \|(A_{n}-A) x\| &= \Big\|\sum_{i}(x, x_{i})(A_{n}-A) x_{i}\Big\| \\ &\leq \sum_{i}\|x\| \cdot\|(A_{n}-A) x_{i}\|\\ &\leq \epsilon N\|x\|, \end{align} and this implies (i).

It is also easy to prove that if the norm is used to define a distance for transformations, then the resulting metric space is complete, that is, if \(\|A_{n}-A_{m}\| \to 0\) as \(n, m \to \infty\) , then there is an \(A\) such that \(\|A_{n}-A\| \to 0\) . The proof of this fact is reduced to the corresponding fact for vectors. If \(\|A_{n}-A_{m}\| \to 0\) , then \(\|A_{n} x-A_{m} x\| \to 0\) for each \(x\) , so that we may find a vector corresponding to \(x\) , which we may denote by \(A x\) , say, such that \(\|A_{n} x-A x\| \to 0\) . It is clear that the correspondence from \(x\) to \(A x\) is given by a linear transformation \(A\) ; the implication relation (ii) \(\implies\) (i) proved above completes the proof.

Now that we know what convergence means for linear transformations, it behooves us to examine some simple functions of these transformations in order to verify their continuity. We assert that \(\|A\|\) , \(\|A x\|\) , \((A x, y)\) , \(A x\) , \(A+B\) , \(\alpha A\) , \(A B\) , and \(A^{*}\) all define continuous functions of all their arguments simultaneously. (Observe that the first three are numerical-valued functions, the next is vector-valued, and the last four are transformation-valued.) The proofs of these statements are all quite easy, and similar to each other; to illustrate the ideas we discuss \(\|A\|\) , \(A x\) , and \(A^{*}\) .

1. If \(A_{n} \to A\) , that is, \(\|A_{n}-A\| \to 0\) , then, since the relations \[\|A_{n}\| \leq\|A_{n}-A\|+\|A\|,\] and \[\|A\| \leq\|A-A_{n}\|+\|A_{n}\|\] imply that \[\big|\|A_{n}\|-\|A\|\big| \leq \|A_{n}-A\|,\] we see that \(\|A_{n}\| \to\|{A}\|\) .
2. If \(A_{n} \to A\) and \(x_{n} \to x\) , then \[\|A_{n} x_{n}-A x\| \leq\|A_{n} x_{n}-A x_{n}\|+\|A x_{n}-A x\| \to 0,\] so that \(A_{n} x_{n} \to A x\) .
3. If \(A_{n} \to A\) , then, for each \(x\) and \(y\) , \begin{align} (A_{n}^{*} x, y) &= (x, A_{n}y)\\ &= \overline{(A_{n} y, x)}\\ &\to \overline{(A y, x)} \\ &= \overline{(y, A^{*} x)}\\ &= (A^{*} x, y) \end{align} whence \(A_{n}^{*} \to A^{*}\) .

EXERCISES