Prove that $R(A)$ (the range of $A$) is closed if and only if there is a $C>0$ such that $\|x\|\leq C\|Ax\|$ Let $A\in L(E,F)$,where $L(E,F)$ consists of all the linear continuous operator from $E$ to $F$. Suppose that the null space $N(A)={0}$. Prove that $R(A)$ (the range of $A$) is closed if and only if there is a $C>0$ such that $\|x\|\leq C\|Ax\|$ for any $x \in E $ .($E,F$ are Banach spaces)
 A: Suppose the given condition holds ($ \| x \| \le C\| Ax \|$). Take a sequence $y_n \in R(A)$ converging to $y \in F$. For each $y_n$, there is $x_n \in E$ such that $Ax_n = y_n$. Then we see $$\| x_n - x_m \| \le C \| A(x_n - x_m) \| = C \| Ax_n - Ax_m \| = C\| y_n - y_m \|.$$ Since the latter goes to zero as $n,m \to \infty$, we see that $\{x_n\}$ is Cauchy. Since $E$ is complete, $x_n \to x \in E$ for some element $x$. By continuity of $A$, we must have $Ax = \lim A x_n = \lim y_n = y$ which shows that $y \in R(A)$. Thus $R(A)$ is closed since it contains its limit points.
Conversely, suppose that $R(A)$ is closed. Then $R(A)$ is a Banach space (since it is a closed subspace of a Banach space). Since $N(A) = \{ 0\}$, $A$ is injective so $A : X \to R(A)$ is a continuous (equiv. bounded), bijective operator from one Banach space to another. Thus by the bounded inverse theorem, $A$ has a bounded inverse $A^{-1} : R(A) \to X$. Let $C' > 0$ be a bound on $A^{-1}$. Then we see for any $x \in E$, $$\| x \| = \| A^{-1} Ax\| \le C' \|Ax\|.$$
