Bounded Linear Operators from X to X

# Bounded Linear Operators from X to X

Recall that if $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ are normed linear spaces then a linear operator $T : X \to Y$ is said to be bounded if there exists an $M > 0$ such that for every $x \in X$:

(1)
\begin{align} \quad \| T(x) \|_Y \leq M \| x \|_X \end{align}

The "smallest" such $M$ is denoted $\| T \|$ and is defined by:

(2)
\begin{align} \quad \| T \| = \inf \{ M : \|T(x) \|_Y \leq M \| x \|_X, \: \forall x \in X \} \end{align}

We will now shortly discuss bounded linear operators from $X$ to $X$.

 Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space. If $S, T : X \to X$ are bounded linear operators then $ST = S \circ T : X \to X$ is a bounded linear operator. Moreover, $\| ST \| \leq \| S \| \| T \|$.
• Proof: Since $S$ and $T$ are bounded linear operators there exists $M_1, M_2 > 0$ such that $\| S(x) \|_X \leq M_1 \| x \|_X$ for all $x \in X$ and $\| T(x) \|_X \leq M_2 \| x \|_X$ for all $x \in X$. Hence:
(3)
\begin{align} \quad \| S(T(x)) \|_X \leq M_1 \| T(x) \|_X \leq M_1M_2 \| x \|_X \end{align}
• So $ST$ is a bounded linear operator, and by setting $M_1 = \| S \|$ and $M_2 = \| T \|$ we have that:
(4)
\begin{align} \quad \| ST \| \leq \| S \| \| T \| \quad \blacksquare \end{align}
 Proposition 2: Let $(X, \| \cdot \|_X)$ be a normed linear space. If $X$ is a Banach space and if $T : X \to X$ is a bounded linear operator and $\| T \| < 1$ then $I - T$ is invertible and $(I - T)^{-1} = \sum_{n=0}^{\infty} T^n$.
• Proof: Since $X$ is a Banach space, so is $\mathcal B(X, X)$. From the Absolute Summability Criterion for Completeness we have that every absolutely summable series in $\mathcal B(X, X)$ converges in $\mathcal B(X, X)$. Consider the series $\sum_{n=0}^{\infty} T^n$. We have that:
(5)
\begin{align} \quad \sum_{n=0}^{\infty} \| T^n \| \leq \sum_{n=0}^{\infty} \| T \|^n \end{align}
• And the righthand numerical series converges since $\| T \| < 1$. Therefore $\sum_{n=0}^{\infty} T^n$ converges to some $S \in \mathcal B(X, X)$. We have that:
(6)
\begin{align} \quad (I - T)S = (I - T) \sum_{n=0}^{\infty} T^n = \sum_{n=0}^{\infty} (I - T)T^n = \sum_{n=0}^{\infty} (T^n - T^{n+1}) = T^0 = I \end{align}
• Therefore $(I - T)$ is invertible and $(I - T)^{-1} = \sum_{n=0}^{\infty} T^n$. $\blacksquare$
 Corollary 3: Let $(X, \| \cdot \|_X)$ be a normed linear space. If $X$ is a Banach space then the set of all invertible bounded linear operators in $\mathcal B(X, X)$ is open in $\mathcal B(X, X)$.
• Proof: Let $O$ be the set of all invertible bounded linear operators in $\mathcal B(X, X)$. Let $T \in O$. We will show that there exists an $\epsilon > 0$ such that for all bounded linear operators $S$ with $\| T - S \| < \epsilon$ then $S \in O$, which shows that the open ball centered at $T$ with radius $\epsilon$ is fully contained in $O$.
• Let $\epsilon = \frac{1}{\| T^{-1} \|}$. Note that $\| T^{-1} \| \neq 0$ since $T$ is invertible and so $T \neq 0$. Let $S \in \mathcal B(X, X)$ such that $\| S - T \| < \frac{1}{\| T^{-1} \|}$. Then:
(7)
\begin{align} \quad \| T^{-1} (S - T) \| \leq \| T^{-1} \| \| S - T \| < \| T^{-1} \| \cdot \frac{1}{ \| T^{-1} \|} = 1 \end{align}
• So by Lemma 2 we have that $I - (T^{-1} (S - T))$ is invertible. But note that:
(8)
\begin{align} \quad I - (T^{-1} (S - T)) = I - (T^{-1}S - I) = T^{-1} S \end{align}
• Since $T$ is invertible and $T^{-1}S$ is invertible, so is the composition $S = TT^{-1} S$. So $S$ is invertible, and hence the open ball centered at $T$ with radius $\epsilon = \frac{1}{\| T^{-1} \|}$ is fully contained in $O$. So $O$ is open in $\mathcal B(X, X)$. $\blacksquare$