I'm doing exercise 2.14.1 in Brezis' book of Functional Analysis. Could you have a check on my attempt?
Let $(E, | \cdot |_E), (F, | \cdot |_F)$ be Banach spaces and $T \in \mathcal{L}(E, F)$. Let $N(T), R(T)$ be the kernel and image of $T$ respectively. Prove that $R(T)$ is closed iff there exists $c>0$ such that $d(x, N(T)) \leq c |T x|_F$ for all $x \in E$.
My proof: Consider the quotient map $$f:E \to E/N(T), x \mapsto \hat x := x+N(T).$$
Because $N(T)$ is closed and $E$ is a Banach space, $E/N(T)$ is also a Banach space w.r.t. the quotient norm $\| \cdot \|$ defined by $$\| \hat x\| := d(x, N(T)), \quad \forall x \in E.$$
Also, $f$ is linear continuous. First, we have the following remark:
Assume $(\hat x_n)$ is a Cauchy sequence in $E/N(T)$ such that $\hat x_n \to \hat x$, i.e., $\| \widehat{x_n - x}\| \to 0$. Hence there is $z_n \in N(T)$ such that $|(x_n-x)-z_n|_E < 1/n$. This means $(x_n-z_n) \to x$. Then $Tx_n = T(x_n-z_n) \to Tx$.
Let $R(T)$ be closed. Then $R(T)$ is a Banach space. We define a map $H:R(T) \to E/N(T)$ as follows. For each $y\in R(T)$, there is some $x\in E$ such that $y=Tx$. Let $H(x):= \hat x$. If $T a =Tb$, then $a-b\in N(T)$ and thus $\hat a = \hat b$. Hence $H$ is well-defined and also linear. Let $(y_n, \hat x_n)$ be a sequence in $\operatorname{graph} H$ that converges to $(y, \hat x)$. By our remark, $Tx_n \to Tx$. Also, $Tx_n = y_n \to y$. Hence $Tx=y$ and thus $(y, \hat x) \in \operatorname{graph} H$. By closed graph theorem, $H$ is continuous. It follows that $c:= \|H\|_{\mathcal L(R(T), E/N(T))}$ satisfies the condition.
Let the inequality hold. Let $(y_n)$ be a sequence in $R(T)$ such that $y_n \to y \in F$. There is $x_n$ such that $Tx_n=y_n$. We have $\|\hat x\| \le c |Tx|_F$ for all $x\in E$, so $(\hat x_n)$ is a Cauchy sequence in $E/N(T)$. Then there is $x \in E$ such that $\hat x_n \to \hat x$. By our remark, $y_n = Tx_n \to Tx$. As such, $Tx = y$ and thus $y\in R(T)$.