The Normed Linear Space B(X, Y)

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)
\begin{align} \quad \| T(x) \| \leq M \| x \| \end{align}

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:
(2)
\begin{align} \quad \| T(x) \| \leq r_n \| x \| \end{align}
  • Taking the limit as $n \to \infty$ yields:
(3)
\begin{align} \quad \lim_{n \to \infty} \| T(x) \| &\leq \lim_{n \to \infty} r_n \| x \| \\ \| T(x) \| &\leq \| T \| \| x \| \end{align}
  • Substituting $x$ for $x - y$ and using the additivity property of $T$ gives us that for all $x, y \in X$:
(4)
\begin{align} \quad \| T(x - y) \| \leq \| T \| \| x - y \| \\ \quad \| T(x) - T(y) \| \leq \| T \| \| x - y \| \quad \blacksquare \end{align}
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:
(5)
\begin{align} \quad \| T \| &= \inf \{ M : \| T(x) \| \leq M \| x \|, x \neq 0 \} \\ &= inf \left \{ M : \frac{\| T(x) \|}{\| x \|} \leq M, x \neq 0 \right \} \\ &= \sup \left \{ \frac{\| T(x) \|}{\| x \|} : x \neq 0 \right \} \quad \blacksquare \end{align}
  • Proof of b) If $\| x \| = 1$ then by definition, $\| T(x) \| \leq \| T \| \| x \| = \| T \|$. Therefore :
(6)
\begin{align} \quad \| T \| &= \inf \{ M : \| T(x) \| \leq M\| x \|, x \in X \} \\ &= \inf \{ M : \| T(x) \| \leq M, \| x \| = 1 \| \} \\ &= \sup \{ \| T(x) \| : \| x \| = 1 \} \quad \blacksquare \end{align}

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:
(7)
\begin{align} \quad \| T \| = 0 \end{align}
  • 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:
(8)
\begin{align} \quad \| \lambda T \| &= \sup \{ \| \lambda T(x) \| : \| x \| = 1 \} \\ &= \lambda \sup \{ \| T(x) \| : \| x \| = 1 \} \\ &= | \lambda | \| x \| \quad \end{align}
  • Lastly, let $S, T \in \mathcal B(X, Y)$. Then:
(9)
\begin{align} \quad \| S + T \| &= \sup \{ \| (S + T)(x) \| : \| x \| = 1 \} \\ &= \sup \{ \| S(x) + T(x) \| : \| x \| = 1 \} \\ & \leq \sup \{ \| S(x) \| + \| T(x) \| : \| x \| = 1 \} \\ & \leq \sup \{ \| S(x) \| : \| x \| = 1 \} + \sup \{ \| T(x) \| : \| x \| = 1 \} \\ & \leq \| S \| + \| T \| \end{align}
  • Therefore the operator norm on $\mathcal B(X, Y)$ is indeed a norm. $\blacksquare$
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