Bounded Linear Operators

Basics

Let $X$ and $Y$ be Banach spaces and denote by $\|\cdot \|_X, \|\cdot \|_Y$ their corresponding norms.

Definition 1.1: A map $T : X \rightarrow Y$ is called a linear operator if:

• For all $x,\xi \in X$, $T(x + \xi) = Tx + T\xi$;
• For all $x \in X$ and $\alpha \in \mathbb{C}$, $T(\alpha x) = \alpha Tx$.

In particular, if $T$ is a linear operator then $T0 = 0$.

Definition 1.2: A linear map $T: X \rightarrow Y$ is said to be bounded if there exists $C > 0$ such that $\|Tx\|_Y \leq C \|x\|_X$ for all $x \in X$.

Theorem 1.3: A linear map $T: X \rightarrow Y$ is continuous if and only if it is bounded.

Proof: Assume first that $T$ is bounded, that is $\|Tx\|_Y \leq C \|x\|_X$. Let $x_0 \in X$ and $\varepsilon >0$. Take $\delta = C^{-1}\varepsilon$. Then for all $x \in X$ satisfying $\|x-x_0\|_X \leq \delta$, we have:

This implies that $T$ is continuous.

Assume now that $T$ is continuous. Then $T$ is continuous at $0$, and therefore there exists $\delta >0$ such that:

whenever $\| \xi \|_X \leq \delta$.

If $x \in X$, let us denote $c = \delta \|x\|_X^{-1}$ and $\xi = cx$. Then $\|\xi\|_X = \delta$. Since $T$ is a linear map, the equation above implies:

which means that $T$ is bounded. $\blacksquare$

Notation: If $T$ is bounded, then there is a minimal constant $C$ such that $\|Tx\|_Y \leq C\|x\|_X$, which is denoted $\|T\|$ and called the operator norm:

The set of all bounded operators from $X$ to $Y$ is denoted by $\mathcal{B}(X,Y)$. We will mostly be interested in the case $X=Y$; then the notation is $\mathcal{B}(X)= \mathcal{B}(X,X)$.

Theorem 1.4: Given two Banach spaces, the set $\mathcal{B}(X,Y)$ is a Banach space with respect to the operator norm defined above. Proof: Exercise. $\blacksquare$

Theorem 1.5: Given $A \in \mathcal{B}(Y,Z)$ and $B \in \mathcal{B}(X,Y)$, the operator $AB$ is bounded and its norm satisfies:

Proof: Exercise. $\blacksquare$

The previous two theorems together mean that $\mathcal{B}(X)$ forms a Banach algebra, in other words a complete normed vector space along with a composition rule.

Definition 1.6: Let $T \in \mathcal{B}(X,Y)$. The set:

is called the kernel of $T$ and the set:

is called the range of $T$.

Theorem 1.7: For every $T \in \mathcal{B}(X,Y)$, $\ker T$ is a closed set.

Proof: Exercise. $\blacksquare$

Notation 1.8: By $I$ we always denote the identity operator $Ix = x$ in a Banach space $X$. If the Banach space for which it is the identity needs to be made clear, we write $I_X$.

Definition 1.9: Let $X,Y$ be Banach spaces, $E \subset X$ a subspace and $T \in \mathcal{B}(E,Y)$. We say that $T$ extends to $X$ if there exists $\hat{T} \in \mathcal{B}(X,Y)$ such that $Tx = \hat{T}x$ for all $x \in E$. We often abuse notation and write $T \in \mathcal{B}(X,Y)$ for the extension as well.

Theorem 1.10: Let $D \subset X$ be a dense subspace, and let $T \in \mathcal{B}(D,Y)$. Then $T$ extends to a bounded operator from $X$ to $Y$ with the same norm.

Proof: Exercise. $\blacksquare$

Our main interest in this course will be $\mathcal{B}(H)$, the class of bounded operators from a separable Hilbert space $\mathcal{H}$ to itself. In this case, the following objects are useful:

Definition 1.11: A map $\phi : \mathcal{H} \times \mathcal{H} \rightarrow \mathbb{C}$ is called a bounded sesquilinear form if it is linear in the first argument, anti-linear in the second argument and satisfies:

for some $C >0$ and for all $x,y \in \mathcal{H}$.

Proposition 1.12: For each $T \in \mathcal{B}(\mathcal{H})$ there exists a unique bounded sesquilinear form $\phi$ such that:

Conversely, for each bounded sesquilinear form $\phi$ there is a unique $T \in \mathcal{B}(\mathcal{H})$ such that the above identity holds.

Proof: Exercise. $\blacksquare$