Isomorphisms Between Normed Linear Spaces

Definition: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. A bounded linear operator $T \in \mathcal B(X, Y)$ is said to be an Isomorphism between $X$ and $Y$ if $T$ is a bijection and there exists constants $m > 0$ and $M > 0$ such that $m \| x \|_X \leq \| T(x) \|_Y \leq M \| x \|_X$ for every $x \in X$. If such an isomorphism exists, then the spaces $X$ and $Y$ are said to be Isomorphic.

Proposition 1: Let $(X, \| \cdot \|_X)$, $(Y, \| \cdot \|_Y)$ and $(Z, \| \cdot \|_Z)$ be normed linear spaces. Then: a) $X$ is isomorphic to $X$ (Reflexive Property). b) If $X$ is isomorphic to $Y$ then $Y$ is isomorphic to $X$ (Symmetry Property). c) If $X$ is isomorphic to $Y$ and $Y$ is isomorphic to $Z$ then $X$ is isomorphic to $Z$ (Transitivity Property).

The proposition above tells us that "isomorphism" between normed linear spaces is an equivalence relation.

• Proof of a) Let $I$ be the identity operator. Then $I$ is clearly bijective. If $m = 1$ and $M = 1$ we have that:

(1)

\begin{align} \quad m\| x \|_X = 1 \| x \|_x \leq \| I(x) \|_X \leq 1 \| x \|_X = M \| x \|_X \end{align}

• So $X$ is isomorphic to $X$.

• **Proof of b) Suppose that $X$ is isomorphic to $Y$. Let $T$ be an isomorphism from $X$ to $Y$. Then $T$ is bijective and there exists $m > 0$, $M > 0$ such that for every $x \in X$:

(2)

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

• Since $T$ is bijective, $T^{-1}$ is bijective. For each $y \in Y$ there is an $x \in X$ such that $T(x) = y$. So the inequality above translates for every $y \in Y$ by:

(3)

\begin{align} \quad m \| T^{-1}(y) \|_X \leq \| y \|_Y \leq M \| T^{-1}(y) \|_X \end{align}

• Hence, for every $y \in Y$ we have that:

(4)

\begin{align} \quad \frac{1}{M} \| y \|_Y \leq \| T^{-1}(y) \|_X \leq \frac{1}{m} \| y \|_Y \end{align}

• So $T^{-1}$ is an isomorphism from $Y$ to $X$. Thus $Y$ is isomorphic to $X$.

• Proof of c) Suppose that $X$ is isomorphic to $Y$ and $Y$ is isomorphic to $Z$. Then there exists bijective $S \in \mathcal B(X, Y)$ and a bijective $T \in \mathcal B(Y, Z)$, as well as constants $m > 0$, $M > 0$, $c > 0$, and $C > 0$ such that for all $x \in X$ and for all $y \in Y$:

(5)

\begin{align} \quad m \| x \|_X \leq \| S(x) \|_Y \leq M \| x \|_X \end{align}

(6)

\begin{align} \quad c \| y \|_Y \leq \| T(y) \|_Z \leq C \| y \|_Y \end{align}

• Since $S$ and $T$ are bijective, the function $T \circ S : X \to Z$ is bijective. For every $z \in Z$ there is a $y \in Y$ such that $z = T(y)$. Similarly, for every $y \in Y$ there is $x \in X$ such that $S(x) = y$. So for every $z \in Z$ there is a $x \in X$ such that $z = T(S(x))$. So the first and second equations translate to:

(7)

\begin{align} \quad cm \| x \|_X \leq c \| S(x) \|_Y = c \| y \|_Y \leq \| z \|_Z \leq C \| y \|_Y = C \| S(x) \|_Y \leq CM \| x \|_X\\ \end{align}

• Since $z = T(S(x))$, we have that:

(8)

\begin{align} \quad cm \| x \|_X \leq \| T(S(x)) \|_Z \leq CM \| x \|_X \end{align}

• So $S \circ T$ is an isomorphism from $X$ to $Z$, and thus, $S$ is isomorphic to $T$. $\blacksquare$