Norm

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 $𝒱$ is bounded if there exists a constant $K$ such that $‖ A x ‖ \leq K ‖ x ‖$ for every vector $x$ in $𝒱$ . 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 $𝒫$ 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, $‖ A ‖ = inf {K : ‖ A x ‖ \leq K ‖ x ‖ for all x} .$

The notion of boundedness is closely connected with the notion of continuity. If $A$ is bounded and if $ϵ$ is any positive number, by writing $δ = \frac{ϵ}{‖ A ‖}$ we make sure that $‖ x - y ‖ < δ$ 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 $𝒱$ ; let ${x_{1}, \dots, x_{N}}$ be an orthonormal basis in $𝒱$ and write $K_{0} = max {‖ A x_{1} ‖, \dots, ‖ A x_{N} ‖} .$ 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 $𝒱$ 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 $ℳ$ and $𝒩$ respectively, and if $‖ E - F ‖ < 1$ , then $\dim ℳ = \dim 𝒩$ .

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?

EXERCISES

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