show that $T$ is injective and has closed range $\iff \exists c > 0, \, \forall x \in E, \, \, \|T(x)\| \geq c \|x\|$

Question

let $E, F$ be two Banach spaces and $T : E \to F$ be a continuous linear operator.

show that :

$$T \text{ is injective and has closed range } \iff \exists c > 0, \, \forall x \in E, \, \, \|T(x)\| \geq c\|x\|$$

my Attempt :

"$\implies$"

case 1 : $x = 0$ is trivial

case 2 : $x \neq 0$

assume that $\forall c > 0, \, \exists z \neq 0 \in E, \frac{\| T(z)\|}{\|z\|} < c$

this means that $\|T(z)\| = 0 \iff T(z) = 0 \iff z \in \ker T \iff z =0$, since $T$ is injective.

$z$ can't be zero so it's a contradiction.

"$\impliedby$"

let $x, y \in E$ such that $T(x) = T(y)$

then :

there exists $c > 0$ such that : $T(x -y ) = \|T(x-y)\|0 \geq c \| x - y \|$

therefore $x = y$ so $T$ is injective.

let $(y_n)_n \subset T(E)$, for one we know that $y_n \to y \in F$ and that there exists $(x_n)_n \subset E$ such that $Tx_n = y_n$ and $x_n \to x \in E$

both the norm and $T$ are continuous :

$$ 0\leq \|y_n - Tx\|=\|T(x_n - x)\| \leq \|T\| \| x_n - x \|$$

by the squeeze theorem then $y_n \to y = Tx \in T(E)$ so $T(E)$ is closed

could someone please check if my work is correct, I have some doubts because I haven't used all assumptions.

Answer 1

In case 2 of "$\implies$" you have mis-interpreted $$\forall c>0\,\exists z \ne 0\,\in E\,\left(\frac {\|T(z)\|}{\|z\|}<c\right)$$ as $$\exists z\ne 0\,\in E\,\,\forall c>0\,\left(\frac {\|T(z)\|}{\|z\|}<c\right)$$ which renders the argument invalid. By way of comparison let $f(z)=z$ for all $z\in \Bbb R.$ Then $$\forall c>0\,\exists z\ne 0\,(|f(z)|<c)$$ but it is NOT true that $$\exists z\ne 0\,\forall c>0\,(|f(z)|<c).$$ You MUST use the "Banach" (completeness) property of $E$ and $F$. For example let $E$ be the set of every real sequence $(r_n)_{n\in \Bbb N}=(r_n)_n$ such that $\{n\in \Bbb N:r_n\ne 0\}$ is finite. With norm $\|(r_n)_n\|=\max \{|r_n|: n\in \Bbb N\}.$ Let $F=E$ and let $T((r_n)_n)=(r_n/n)_n.$ Then $T:E\to F$ is a continuous linear bijection but $\inf \{\|T(e)\|: \|e\|=1\}=0.$ However this normed linear space $E$ is NOT a Banach space.

Addendum. Another example, to show that the closed range of $T$ must be used: Let $E=F=l_{\infty}.$ Let $T((x_n)_n)=(x_n/n)_n.$ Then $(1/\sqrt n\,)_n\in \overline {T[E]}\setminus T[E]$ and $\inf \{\|T(e)\|:\|e\|=1\}=0.$