Isomorphism Linear Operators on Normed Linear Spaces

Isomorphism Linear Operators on Normed Linear Spaces

Let $X$ and $Y$ be normed linear spaces and let $T \in \mathcal B(X, Y)$. Suppose that $T : X \to Y$ is bijective. Then $T^{-1} : Y \to X$ is well-defined. However, it $T^{-1}$ a linear operator? The answer is yes, as we prove in the following theorem.

Theorem 1: Let $X$ and $Y$ be normed linear spaces and let $T : X \to Y$ be a bijective linear operator from $X$ to $Y$. Then $T^{-1} : Y \to X$ is a linear operator from $Y$ to $X$.
  • Proof: Let $s, t \in Y$. Since $T$ is bijective there exists unique $x, y \in X$ such that $T(x) = s$ and $T(y) = t$, and if $u = s + t$ then there exists a $z \in X$ such that $T(z) = u = s + t = T(x) + T(y)$. Since $T$ is bijective, this shows that $z = x + y$. So:
(1)
\begin{align} \quad T^{-1} (s + t) = T^{-1} (u) = z = x + y = T^{-1}(s) + T^{-1}(t) \end{align}
  • Now let $s \in Y$ and let $\lambda \in \mathbb{C}$. Then there exists a unique $x \in X$ such that $T(x) = s$ and if $u = \lambda s$ then there exists a unique $y \in X$ such that $T(y) = \lambda s$. But then $T(y) = \lambda s = \lambda T(x) = T(\lambda x)$. So $y = \lambda x$. So:
(2)
\begin{align} \quad T^{-1} (\lambda s) = y = \lambda x = \lambda T^{-1} (s) \end{align}
  • So $T^{-1} : Y \to X$ is a linear operator from $Y$ to $X$. $\blacksquare$

Then next question we might ask whether or not $T^{-1}$ is a bounded linear operator if $T$ is a bounded linear operator. The answer is no in general. However, the bijective bounded linear operators with bounded inverses are given a special name which we define below.

Definition: Let $X$ and $Y$ be normed linear spaces. A bijective bounded linear operator $T \in \mathcal B(X, Y)$ is said to be an Isomorphism if $T^{-1} : Y \to X$ is a bounded linear operator, and the normed linear spaces $X$ and $Y$ are said to be Isomorphic.

We make a couple of alternative characterization of isomorphism below.

Theorem 2: Let $X$ and $Y$ be normed linear spaces. A bijective bounded linear operator $T : X \to Y$ is an isomorphism if and only if $T^{-1}$ is continuous.
Theorem 3: Let $X$ and $Y$ be normed linear spaces. A bijective bounded linear operator $T : X \to Y$ is an isomorphism if and only if there exists $C, D \in \mathbb{R}$ with $C, D > 0$ such that $C \| x \| \leq \| T(x) \| \leq D \| x \|$ for all $x \in X$.
  • Proof: $\Rightarrow$ Suppose that $T : X \to Y$ is a bijective bounded linear operator. Then $T : X \to Y$ is a bounded linear operator and $T^{-1} : Y \to X$ is a bounded linear operator.
  • Since $T$ is bounded there exists an $M_1 \in \mathbb{R}$, $M_1 > 0$ such that:
(3)
\begin{align} \quad \| T(x) \| \leq M_1 \| x \| \quad \forall x \in X \quad (*) \end{align}
  • And since $T^{-1}$ is bounded there exists an $M_2 \in \mathbb{R}$, $M_2 > 0$ such that:
(4)
\begin{align} \quad \| T^{-1} (y) \| \leq M_2 \| y \| \quad \forall y \in Y \end{align}
  • Since $T$ is bijective, for each $y \in Y$ there exists an $x \in X$ such that $T(x) = y$. So:
(5)
\begin{align} \quad \| T^{-1}(y) \| &= \| T^{-1}(T(x)) \| \leq M_2 \| T(x) \| \\ \| x \| & \leq M_2 \| T(x) \| \\ \frac{1}{M_2} \| x \| & \leq \| T(x) \| \quad \forall x \in X \quad (**) \end{align}
  • By letting $\displaystyle{C = \frac{1}{M_2}}$ and $D = M_1$ we see that:
(6)
\begin{align} \quad C \| x \| \leq \| T(x) \| \leq D \| x \| \quad \forall x \in X \end{align}
  • $\Leftarrow$ Suppose there exists $C, D \in \mathbb{R}$ with $C, D > 0$ such that:
(7)
\begin{align} \quad C \| x \| \overset{(*)} \leq \| T(x) \| \leq D \| x \| \quad \forall x \in X \end{align}
  • Consider the inequality at $(*)$. Since $T$ is bijective, there exists a unique element $y \in Y$ such that $T(x) = y$. So $x = T^{-1}(y)$. Therefore:
(8)
\begin{align} \quad C \| T^{-1}(y) \| & \leq \| T(T^{-1}(y)) \| \\ C \| T^{-1}(y) \| & \leq \| y \| \\ \| T^{-1}(y) \| & \leq \frac{1}{C} \| y \| \quad \forall y \in Y \end{align}
  • So $T^{-1}$ is a bounded linear operator from $Y$ to $X$. That is, $T$ is an isomorphism. $\blacksquare$
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