Isomorphic Topological Vector Spaces

Isomorphic Topological Vector Spaces

Definition: Let $E$ and $F$ be topological vector spaces. A bijective bicontinuous linear operator $f : E \to F$ is called an Isomorphism of $E$ onto $F$. If an isomorphism from $E$ to $F$ exists, then $E$ and $F$ are said to be Isomorphic Topological Vector Spaces.

Recall that a bijective function $f$ is said to be bicontinuous if both $f$ and its inverse, $f^{-1}$, are continuous.

Isomorphic Normed Vector Spaces

When $E$ and $F$ are specifically normed vector spaces (which are of course examples of topological vector spaces), we have another criterion for determining when a bijection $f : E \to F$ is an isomorphism.

Proposition 1: Let $E$ and $F$ be normed vector spaces and let $f : E \to F$ be a bijection. Then $E$ and $F$ are isomorphic normed vector spaces if and only if there exists constants $M, N > 0$ such that $M \| x \| \leq \| f(x) \| \leq N \| x \|$ for all $x \in E$.
  • Proof: $\Rightarrow$ Suppose that $E$ and $F$ are isomorphic normed vector spaces. Let $f : E \to F$ be an isomorphism of $E$ onto $F$. Then both $f$ and $f^{-1}$ are continuous. Since $f$ is continuous, there exists an $\alpha > 0$ such that:
(1)
\begin{align} \quad \| f(x) \| \leq \alpha \| x \| \end{align}
  • Since $f^{-1}$ is continuous, there exists a $\beta > 0$ such that $\| f^{-1}(y) \| \leq \beta \| y \|$ for all $y \in F$. But for each $y \in F$ there exists an $x \in X$ with $y = f(x)$. Thus $\| x \| = \| f^{-1}(f(x)) \| \leq \beta \| f(x) \|$ for all $x \in X$. Dividing by $\beta > 0$ yields:
(2)
\begin{align} \quad \frac{1}{\beta} \| x \| \leq \| f(x) \| \end{align}
  • So set $M := \frac{1}{\beta}$ and $N := \alpha$. Then $M \| x \| \leq \| f(x) \| \leq N \| x \|$ for all $x \in E$. $\blacksquare$
  • $\Leftarrow$ Suppose that there exists an $M, N > 0$ such that $M \| x \| \leq \| f(x) \| \leq M \| x \|$. Then $f$ is continuous. For each $x \in E$ there exists exactly one $y \in F$ such that $x = f^{-1}(y)$. Thus from the inequality above we have that $M \| f^{-1}(y) \| \leq \| y \|$ and so $\displaystyle{\| f^{-1}(y) \| \leq \frac{1}{M} \| y \|}$. Thus $f^{-1}$ is continuous too. So $f$ is isomorphic normed vector spaces. $\blacksquare$
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