📂Banach Space

Operator Norm

Definition

Let $X$ and $Y$ be normed spaces. Let $T : X \to Y$ be a bounded linear operator between the two spaces. The quantity $\| \cdot \|$ below is called the operator norm.

$$ \| T \| := \inf\limits_{x \in X} \left\{ C : \| T x \| \le C \| x \| \right\} $$

Explanation

To denote each norm precisely, one can write as follows. For the normed spaces $(X, \| \cdot \|_{X})$ and $(Y, \| \cdot \|_{Y})$,

$$ \| T \|_{\text{op}} := \inf\limits_{x \in X} \left\{ C : \| T x \|_{Y} \le C \| x \|_{X} \right\} $$

This notation expresses the objects unambiguously, but it is relatively less readable. Since the elements inside $\| \cdot \|$ can be distinguished, it is often written more simply.

By the definition, the following holds.

$$ \| T x \| \le \| T \| \| x \|, \quad \forall x \in X $$

If $x \ne 0$, then the above equality becomes $\dfrac{\| T x \|}{\| x \|} \le \| T \|$, so the operator norm of $T$ represents an upper bound on how much $T$ stretches vectors. A slightly more intuitive definition is as follows. For $x$ in $\| x \| = 1$, since $\| Tx \| \le \| T \|$, the definition of the operator norm can be written as

$$ \| T \| := \sup\limits_{\substack{x \in X \\ \| x \| = 1}} \| T x \| $$

Properties

(a) Let $B(X, Y)$ denote the set of all bounded linear operators from $X$ to $Y$. $$ B(X, Y) = \left\{ T : X \to Y \mid T \text{ is a bounded linear operator} \right\} $$ Then $(B(X, Y), \| \cdot \|_{\text{op}})$ is a normed space. If $Y$ is a Banach space, then $(B(X, Y), \| \cdot \|_{\text{op}})$ is also a Banach space.

(b) For $T_{1} : X \to Y$ and $T_{2} : T_{1}(Y) \to Z$ the following holds. $$ \| T_{2}T_{1} \| \le \| T_{2} \| \| T_{1} \| $$ As a corollary of this, the following holds. For $T \in B(X, X)$,

$$ \| T^{n} \| \le \| T \|^{n}, \quad n \in \mathbb{N} $$

Proof

(a)

See here.

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(b)

First, by the definition of the operator norm of $T_{2}$, the following holds.

$$ \| T_{2}T_{1} x \| = \| T_{2}(T_{1} x) \| \le \| T_{2} \| \| T_{1} x \| $$

But since $\| T_{1} x \| \le \| T_{1} \| \| x \|$, we obtain

$$ \| T_{2}T_{1} x \| \le \| T_{2} \| \| T_{1} x \| \le \| T_{2} \| \| T_{1} \| \| x \| $$

$\| T_{2}T_{1} \|$ is the smallest $C$ satisfying $\| T_{2}T_{1} x \| \le C \| x \|$, therefore

$$ \| T_{2}T_{1} \| \le \| T_{2} \| \| T_{1} \| $$

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