If $T:X \rightarrow Y$ is a bounded linear operator that $T^{-1}(0)=\{0\}$ then
$\mathcal{R}(T)=\{Tx\}_{x \in X}$ is closed $\Leftrightarrow$ there is
no sequence $(x_n)$ with $||x_n||=1$ such that $Tx_n \rightarrow 0$
I have $\Rightarrow$ but no idea for $\Leftarrow$.
Best Answer
Suppose that there is no sequence $(x_n)$ with $\|x_n\|=1$ such that $Tx_n\to 0$. Then $T$ is bounded from below, i.e. there exists a constant $C>0$ such that $$\|Tx\| \ge C \|x\|$$ If this was not true then we would necessarily have that $$ \inf_{x\in X} \left\|T\left(\frac{x}{\|x\|}\right)\right\|=0$$ (if this infimum was positive it would be a choice for the constant $C$ in the previous inequality)
In other words, $$ \inf_{\|x\|=1} \left\|Tx\right\|=0$$ But this means that we would have a sequence $(x_n)$ with $\|x_n\|=1$ such that $\|Tx_n\|\to 0$. Contradiction!
The bound from below implies that the range of $T$ closed: take a sequence $(Tx_n)\in\mathcal{R}(T)$ such that $Tx_n\to y$. Then $$ \|x_n-x_m\|\le C^{-1}\| Tx_n-Tx_m \| \to 0 $$ as $n,m\to\infty$, so $(x_n)$ is a Cauchy sequence that converges to some $x\in X$ because $X$ is a Banach space. By continuity of $T$ we have that $y=\lim_{n\rightarrow\infty} Tx_n = Tx\in\mathcal{R}(T)$.