Chapter 3

The Operator Norm

Turning $B(H)$ into a normed — in fact complete — space, with a norm that interacts well with composition.

Definition

Definition For $T \in B(H)$, the operator norm of $T$ is $$ \|T\| = \sup_{\substack{x \in H \\ x \neq 0}} \frac{\|Tx\|}{\|x\|}. $$

Three equivalent formulas are used interchangeably, and it is worth knowing why:

Equivalent formulations $$ \|T\| \;=\; \sup_{\|x\| \le 1} \|Tx\| \;=\; \sup_{\|x\| = 1} \|Tx\| \;=\; \inf\{ C \ge 0 : \|Tx\| \le C\|x\| \text{ for all } x \}. $$

The equivalence follows from homogeneity of the norm: for $x \neq 0$, write $x = \|x\| \cdot \frac{x}{\|x\|}$ and use $\|T x\| = \|x\| \cdot \left\| T \frac{x}{\|x\|}\right\|$ to pass between a supremum over all nonzero $x$ and one over unit vectors.

Basic properties

Completeness of $B(H)$

Theorem $B(H)$, equipped with the operator norm, is a Banach space: every Cauchy sequence $(T_n)$ in $B(H)$ converges in operator norm to some $T \in B(H)$.

Sketch. For each fixed $x$, $(T_n x)$ is Cauchy in $H$ (since $\|T_n x - T_m x\| \le \|T_n - T_m\| \|x\|$), so it converges to some limit, which we call $Tx$; linearity of $T$ follows from linearity of each $T_n$, and one checks $T$ is bounded and that $\|T_n - T\| \to 0$.

Three notions of convergence Convergence $T_n \to T$ in operator norm ($\|T_n - T\| \to 0$) is the strongest of three standard topologies on $B(H)$. Weaker ones — strong operator convergence ($T_n x \to Tx$ for each $x$) and weak operator convergence ($\langle T_n x, y \rangle \to \langle Tx, y \rangle$ for all $x,y$) — appear throughout operator theory but are not needed for the results on this site.

A worked bound

As an illustration, consider a diagonal operator on $\ell^2$ defined by $T(a_1, a_2, \dots) = (\lambda_1 a_1, \lambda_2 a_2, \dots)$ for a bounded sequence of scalars $(\lambda_n)$. Then $$ \|Tx\|^2 = \sum_n |\lambda_n|^2 |a_n|^2 \le \left(\sup_n |\lambda_n|\right)^2 \sum_n |a_n|^2 = \left(\sup_n |\lambda_n|\right)^2 \|x\|^2, $$ so $\|T\| \le \sup_n |\lambda_n|$; taking $x = e_n$ shows $\|T\| \ge |\lambda_n|$ for every $n$, hence $\|T\| = \sup_n |\lambda_n|$ exactly. This example is revisited in Chapter 8.