I'm trying to prove a result mentioned in this thread.
Theorem: Let $E, F$ be Banach spaces and $f:E \to F$ linear continuous. Then $f$ is bounded from below, i.e., $$ \exists c>0, \forall x \in E: |f(x)| \ge c |x|, $$ if and only if $f$ is injective and has closed range.
Could you have a check on my attempt?
Proof:
- Assume $f$ is bounded from below.
We have $f(x) = 0$ implies $c|x|=0$, so $\operatorname{ker} f = \{0\}$. Hence $f$ is injective. Let $y_n \in f(E)$ and $y \in F$ such that $y_n \to y$. There is a unique $x_n \in E$ such that $y_n = f(x_n)$. Then $$ |y_n-y_m| = |f(x_n) - f(x_m)| = |f(x_n-x_m)| \ge c|x_n-x_m|. $$
We have $(y_n)$ is convergent and thus Cauchy, so $(x_n)$ is Cauchy. Because $E$ is complete, there is $x\in E$ such that $x_n \to x$. Clearly, $f(x_n)\to f(x)$ or $y_n \to f(x) \in f(X)$. Hence $f$ has closed range.
- Assume $f$ is injective and has closed range.
Let $G := f(E)$. Because $G$ is closed, $G$ is a Banach subspace of $F$. Then $f:E \to G$ is bijective continuous between Banach spaces. By open mapping theorem, $f^{-1}$ is continuous. The claim then follows.
