The Space of Bounded Linear Operators

Definition: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. The Space of Bounded Linear Operators is the set $\mathcal B(X, Y)$ of bounded linear operators with the operator norm $\| \cdot \|_{\mathrm{op}} : \mathcal B(X, Y) \to [0, \infty)$ defined for all $T \in \mathcal B(X, Y)$ by $\| T \|_{\mathrm{op}} = \inf \{ M : \| T(x) \|_Y \leq M \| x \|_X, \: \forall x \in X \}$.

Again recall that if $Y = \mathbb{R}$ with the usual Euclidean norm then the space of bounded linear operators from $X$ to $Y$ is precisely the space of all bounded linear operators from $X$ to $Y$, i.e., $X^*$.

Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. Then $(\mathcal B(X, Y), \| \cdot \|_{\mathrm{op}})$ is a normed linear space.

• Proof: $\mathcal B(X, Y)$ is a subspace of the linear space of all functions from $X$ to $Y$, so to show that $\mathcal B(X, Y)$ is a linear space we only need to show that it is closed under addition, closed under scalar multiplication, and contains the zero operator.

• Let $S, T \in \mathcal B(X, Y)$. Then there exists $M_S, M_T > 0$ such that $\| S(x) \|_Y \leq M_S \| x \|_X$ and $\| T(x) \|_Y \leq M_T \| x \|_X$ for all $x \in X$.

• We have that $S + T$ is linear since for all $\alpha, \beta \in \mathbb{R}$ and for all $x_1, x_2 \in X$ we have that:

(1)

\begin{align} \quad (S + T)(\alpha x_1 + \beta x_1) = S(\alpha x_1 + \beta x_2) + T(\alpha x_1 + \beta x_2) = \alpha (S + T)(x+1) + \beta(S + T)(x_2) \end{align}

• It is also bounded, since for all $x \in X$:

(2)

\begin{align} \quad \| (S + T)(x) \|_Y = \| S(x) + T(x) \|_Y \leq \| S(x) \|_Y + \| T(x) \|_Y \leq M_S \| x \|_X + M_T \| x \|_X = (M_S + M_T) \| x \|_X \end{align}

• Therefore $(S + T) \in \mathcal B(X, Y)$.

• Also, if $t \in \mathbb{R}$ then $t T$ is linear since:

(3)

\begin{align} \quad (\alpha T)(\alpha x_1 + \beta x_2) = tT(\alpha x_1 + \beta x_2) = \alpha (tT)(x_1) + \beta (tT)(x_2) \end{align}

• And it is bounded since for all $x \in X$:

(4)

\begin{align} \quad \| (tT)(x) \|_Y = \| tT(x) \|_Y = |t| \|T(x) \|_Y \leq [|t|M_T] \| x \|_X \end{align}

• So $tT \in \mathcal B(X, Y)$.

• It is trivially clear that $0 \in \mathcal B(X, Y)$ too. Hence $\mathcal B(X, Y)$ is a linear space.

• All that remains to show is that $\| \cdot \|_{\mathrm{op}}$ is a norm on $\mathcal B(X, Y)$.
• Showing that $\| T \|_{\mathrm{op}} = 0$ if and only if $T = 0$: Suppose that $\| T \|_{\mathrm{op}} = 0$. Then $\inf \{ M : \|T(x) \|_Y \leq M\| x \|_X, \: \forall x \in X \} = 0$. So $\|T(x)\|_Y \leq 0$ for all $x \in X$ implying that $T = 0$. Conversely, suppose that $T = 0$. Then $\inf \{ M : |0| \leq M \| x \|_X, \: \forall x \in X \} = 0$. Hence $\| T \|_{\mathrm{op}} = \| 0 \|_{\mathrm{op}} = 0$.

• Showing that $\| \alpha T \|_{\mathrm{op}} = |\alpha| \| T \|_{\mathrm{op}}$: Let $\alpha \in \mathbb{R}$ and let $T \in X^*$. Then:

(5)

\begin{align} \quad \| \alpha T \|_{\mathrm{op}} = \inf \{ M : \|\alpha T(x)\|_Y \leq M \| x \|_X, \: \forall x \in X \} = |\alpha| \inf \{ M^* : \|T(x)\|_Y \leq M^* \| x \|_X, \: \forall x \in X \} = |\alpha| \| T \|_{\mathrm{op}} \end{align}

• Showing that $\| S + T \|_{\mathrm{op}} \leq \| S \|_{\mathrm{op}} + \| T \|_{\mathrm{op}}$: Let $S, T \in X^*$. Then:

(6)

\begin{align} \quad \| S + T \|_{\mathrm{op}} = \inf \{ M : \|S(x) + T(x)\|_Y \leq M\| x \|_X, \: \forall x \in X \} &\leq \inf \{ M : \|S(x)\|_Y + \|T(x)\|_Y \leq M\| x \|_X, \: \forall x \in X \} \\ &\leq \inf \{ M : \|S(x)\|_Y \leq M_1\| x \|_X, \: \forall x \in X \} + \inf \{ M_2 : \|T(x)\|_Y \leq M_2\| x \|_X, \: \forall x \in X \} \\ & \leq \| S \|_{\mathrm{op}} + \| T \|_{\mathrm{op}} \end{align}

• Hence $(\mathcal B(X, Y), \| \cdot \|_{\mathrm{op}})$ is a normed linear space. $\blacksquare$