The metric properties of vectors have certain important implications for the metric properties of linear transformations, which we now begin to study.
Definition 1. A linear transformation A on an inner product space \mathcal{V} is bounded if there exists a constant K such that \|A x\| \leq K\|x\| for every vector x in \mathcal{V} . The greatest lower bound of all constants K with this property is called the norm (or bound ) of A and is denoted by \|A\| .
Clearly if A is bounded, then \|A x\| \leq \|A\| \cdot\|x\| for all x . For examples we may consider the cases where A is a (non-zero) perpendicular projection or an isometry; Section: Perpendicular projections , Theorem 1, and the theorem of Section: Isometries , respectively, imply that in both cases \|A\|=1 . Considerations of the vectors defined by x_{n}(t)=t^{n} in \mathcal{P} shows that the differentiation transformation is not bounded.
Because in the sequel we shall have occasion to ccnsider quite a few upper and lower bounds similar to \|A\| , we introduce a convenient notation. If P is any possible property of real numbers t , we shall denote the set of all real numbers t possessing the property P by the symbol \{t: P\} , and we shall denote greatest lower bound and least upper bound by inf (for infimum) and sup (for supremum) respectively. In this notation we have, for example, \|A\|=\inf \{K : \|A x\| \leq K\|x\| \text { for all } x\}.
The notion of boundedness is closely connected with the notion of continuity. If A is bounded and if \epsilon is any positive number, by writing \delta = \frac{\epsilon}{\|A\|} we make sure that \|x - y\| < \delta implies that \begin{align} \|Ax - Ay\| &= \|A(x - y)\|\\ &\leq \|A\| \cdot \|x - y\|\\ &< \epsilon; \end{align}in other words boundedness implies (uniform) continuity. (In this proof we tacitly assumed that \|A\| \neq 0 ; the other case is trivial.) In view of this fact the following result is a welcome one.
Theorem 1. Every linear transformation on a finite-dimensional inner product space is bounded.
Proof. Suppose that A is a linear transformation on \mathcal{V} ; let \{x_{1}, \ldots, x_{N}\} be an orthonormal basis in \mathcal{V} and write K_{0}=\max \big\{\|A x_{1}\|, \ldots,\|A x_{N}\|\big\}. Since an arbitrary vector x may be written in the form x=\sum_{i}(x, x_{i}) x_{i} , we obtain, applying the Schwarz inequality and remembering that \|x_{i}\|=1 , \begin{align} \|A x\| &= \Big\|A\Big(\sum_{i}(x, x_{i}) x_{i}\Big)\Big\| \\ &= \Big\|\sum_{i}(x, x_{i}) A x_{i}\Big\|\\ &\leq \sum_{i}|(x, x_{i})| \cdot\|A x_{i}\| \\ &\leq \sum_{i}\|x\| \cdot\|x_{i}\| \cdot\|A x_{i}\|\\ &\leq K_{0} \sum_{i}\|x\| \\ &= N K_{0}\|x\|. \end{align}In other words, K=N K_{0} is a bound of A , and the proof is complete. ◻
It is no accident that the dimension N of \mathcal{V} enters into our evaluation; we have already seen that the theorem is not true in infinite-dimensional spaces.
EXERCISES
Exercise 1.
- Prove that the inner product is a continuous function (and therefore so also is the norm); that is, if x_{n} \to x and y_{n} \to y , then (x_{n}, y_{n}) \to(x, y) .
- Is every linear functional continuous? How about multilinear forms?
Exercise 2. A linear transformation A on an inner product space is said to be bounded from below if there exists a (strictly) positive constant K such that \|A x\| \geq K\|x\| for every x . Prove that (on a finite-dimensional space) A is bounded from below if and only if it is invertible.
Exercise 3. If a linear transformation on an inner product space (not necessarily finite-dimensional) is continuous at one point, then it is bounded (and consequently continuous over the whole space).
Exercise 4. For each positive integer n construct a projection E_{n} (not a perpendicular projection) such that \|E_{n}\| \geq n .
Exercise 5.
- If U is a partial isometry other than 0 , then \|U\|=1 .
- If U is an isometry, then \|U A\|=\|A U\|=\|A\| for every linear transformation A .
Exercise 6. If E and F are perpendicular projections, with ranges \mathcal{M} and \mathcal{N} respectively, and if \|E-F\|<1 , then \operatorname{dim} \mathcal{M}=\operatorname{dim} \mathcal{N} .
Exercise 7.
- If A is normal, then \|A^{n}\|=\|A\|^{n} for every positive integer n .
- If A is a linear transformation on a 2 -dimensional unitary space and if \|A^{2}\|=\|A\|^{2} , then A is normal.
- Is the conclusion of (b) true for transformations on a 3 -dimensional space?
