Crit. for a Linear Map to be an Iso. Between Normed Linear Spaces

Isomorphisms Between Normed Linear Spaces

Recall from the Isomorphisms Between Normed Linear Spaces page that if $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ are normed linear spaces then an isomorphism between $X$ and $Y$ is a bijective bounded linear operator $T : X \to Y$ such that there exists constants $m > 0$ and $M > 0$ for which:

(1)

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

for every $x \in X$. We now give an alternative criterion for determining if a bounded linear operator $T$ is an isomorphism from $X$ to $Y$.

Theorem 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. A bounded linear operator $T \in \mathcal B(X, Y)$ is an isomorphism from $X$ to $Y$ if and only if there exists a bounded linear operator $S \in \mathcal B(Y, X)$ such that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$.

• Proof: $\Rightarrow$ Suppose that the bounded linear operator $T \in \mathcal B(X, Y)$ is an isomorphism from $X$ to $Y$. Let $S = T^{-1}$. Then $S : Y \to X$. We aim to show that $S$ is linear.

• Let $y_1, y_2 \in Y$. Then:

(2)

\begin{align} \quad T(S(y_1 + y_2)) = y_1 + y_2 = T(S(y_1)) + T(S(y_2)) = T(S(y_1) + S(y_2)) \end{align}

• Since $T$ is injective, we have that $S(y_1 + y_2) = S(y_1) + S(y_2)$.

• Let $\alpha \in \mathbb{R}$ and let $y \in Y$. Then:

(3)

\begin{align} \quad T(S(\alpha y)) = \alpha y = \alpha T(S(y)) = T(\alpha S(y)) \end{align}

• Again, since $T$ is injective we have that $S(\alpha y) = \alpha S(y)$. Hence $S$ is linear.

• We now show that $S = T^{-1}$ is bounded. Since $T$ is an isomorphism there exists $m > 0$ and $M > 0$ such that for every $x \in X$ we have that:

(4)

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

• Since $T$ is surjective for each $y \in Y$ there is a $x \in X$ for which $y = T(x)$. So for all $y \in Y$ we have that:

(5)

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

• From the first inequality we get that for all $y \in Y$:

(6)

\begin{align} \quad \| S(y) \|_X \leq \frac{1}{m} \| y \|_Y \end{align}

• So $S$ is a bounded linear operator. Lastly, we have that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$ since $S = T^{-1}$.

• $\Rightarrow$ Suppose that there exists a bounded linear operator $S \in \mathcal B(Y, X)$ such that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$.

• Then $T$ and $S$ are bijective. Since $S$ is bounded, for all $y \in Y$ we have that that for some $m > 0$ and for all $y \in Y$ that:

(7)

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

• Since $S$ is bijective, for all $x \in X$ there is a $y \in Y$ such that $x = S(y)$. So:

(8)

\begin{align} \quad \| x \|_X \leq m \| S^{-1}(x) \|_Y = \| T(x) \|_Y \end{align}

• Hence for all $x \in X$ we have that:

(9)

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

• And from the boundedness of $T$, for $M > 0$ we have that for all $x \in X$ that:

(10)

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

• So $T$ is an isomorphism from $X$ to $Y$. $\blacksquare$