# The Normed Linear Space B(X, Y)

Recall from the Linear Operators on Linear Spaces page that if $X$ and $Y$ are linear spaces over $\mathbb{R}$ (or $\mathbb{C}$) then a function $T : X \to Y$ is said to be a linear operator from $X$ to $Y$ if it satisfies the following properties:

- $T(x + y) = T(x) + T(y)$ for all $x, y \in X$.

- $T(\lambda x) = \lambda T(x)$ for all $x \in X$ and for all $\lambda \in \mathbb{R}$ (or $\mathbb{C}$)

We denoted the set of all linear operators from $X$ to $Y$ by $\mathcal L(X, Y)$.

Also, if $X$ and $Y$ are normed linear spaces, then a linear operator $T : X \to Y$ is said to be a bounded linear operator if there exists an $M \in \mathbb{R}$, $M \geq 0$ such that for every $x \in X$ we have that:

(1)We denoted the set of all bounded linear operators from $X$ to $Y$ by $\mathcal B(X, Y)$.

Definition: Let $X$ and $Y$ be normed linear spaces. We define the operation of Addition of Bounded Linear Operators for all $S, T \in \mathcal B(X, Y)$ by $(S + T)(x) = S(x) + T(x)$ for all $x \in X$. We define the operator of Scalar Multiplication of Bounded Linear Operators for all $T \in \mathcal B(X, Y)$ and for all $\lambda \in \mathbb{C}$ by $(\lambda T)(x) = \lambda T(x)$. |

It is easy to show that $\mathcal B(X, Y)$ with the operations defined above is a linear space. We are about to show that this space is actually a normed linear space. To do so, we will first need to define an important norm on this space called the operator norm, and then prove a few lemmas regarding it.

Definition: Let $X$ and $Y$ be normed linear spaces. The Operator Norm on $\mathcal B(X, Y)$ is defined for all $T \in \mathcal B(X, Y)$ by $\| T \| = \inf \{ M : \| T(x) \| \leq M\| x \|, x \in X \}$. |

Lemma 1: Let $X$ and $Y$ be normed linear spaces and let $T \in \mathcal B(X, Y)$. Then for all $x, y \in X$ we have that $\| T(x) - T(y) \| \leq \| T \| \| x - y \|$. |

**Proof:**Let $(r_n)_{n=1}^{\infty}$ be any sequence of nonnegative real numbers that are greater than $\| T \|$ and such that $\displaystyle{\lim_{n \to \infty} r_n = \| T \|}$. Since $T$ is a bounded linear operator, for all $n \in \mathbb{N}$ we have that:

- Taking the limit as $n \to \infty$ yields:

- Substituting $x$ for $x - y$ and using the additivity property of $T$ gives us that for all $x, y \in X$:

Lemma 2: Let $X$ and $Y$ be normed linear spaces and let $T \in \mathcal B(X, Y)$. Then:a) $\displaystyle{\| T \| = \sup \left \{ \frac{\| T(x) \|}{\| x \|} : x \neq 0 \right \}}$.b) $\| T \| = \sup \{ \| T(x) \| : \| x \| = 1 \}$ |

**Proof of a)**Observe that if $x = 0$ then $\| T(x) \| \leq M \| x \|$ for every $M \in \mathbb{R}$, $M > 0$ since $\| T(0) \| = 0$ and $\| 0 \| = 0$. So the quantity $\| T \|$ does not depend on $x = 0$. That is:

**Proof of b)**If $\| x \| = 1$ then by definition, $\| T(x) \| \leq \| T \| \| x \| = \| T \|$. Therefore :

We are now ready to prove that $\mathcal B(X, Y)$ with the operator norm $\| \cdot \|$ defined above is a normed linear space.

Theorem 3: Let $X$ and $Y$ be normed linear spaces. Then the operator norm $\| \cdot \| : \mathcal B(X, Y) \to [0, \infty)$ is indeed a norm. |

**Proof:**Suppose that $T = 0$, that is, $T : X \to Y$ is defined for all $x \in X$ by $T(x) = 0$. Then clearly $\| T(x) \| \leq 0 \| x \| = 0$ for all $x \in X$. Therefore:

- Now suppose that $\| T \| = 0$. Then $\| T(x) \| \leq 0\| x \| = 0$ for all $x \in X$. But by the property of the norm on $Y$, $T(x) = 0$ for all $x \in X$. So $T = 0$. Hence $\| T \| = 0$ if and only if $T = 0$.

- Now let $T \in \mathcal B(X, Y)$ and let $\lambda \in \mathbb{C}$. By Lemma 2 we have that:

- Lastly, let $S, T \in \mathcal B(X, Y)$. Then:

- Therefore the operator norm on $\mathcal B(X, Y)$ is indeed a norm. $\blacksquare$