IFF Crit. for the Ran. of a BLO to be Closed when X, Y are Ban. Spaces
IFF Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces
Recall from the Criterion for the Range of a BLO to be Closed when X is a Banach Space page that if $X$ is a Banach Space, $Y$ is a normed linear space, and $T : X \to Y$ is a bounded linear operator, then if there exists a positive constant $M \in \mathbb{R}$, $M > 0$ such that for every $y \in T(X)$ there exists an $x \in X$ such that:
$T(x) = y$.
$\| x \| \leq M \| y \|$
Then $T(X)$ is closed.
Observe that $Y$ is NOT assumed to be a Banach space in the above theorem.
If $X$ and $Y$ are both Banach spaces, then more can be said.
Theorem 1: Let $X$ and $Y$ both be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then the range $T(X)$ is closed if and only if there exists a positive constant $M \in \mathbb{R}$, $M > 0$ such that for every $y \in T(X)$ there exists an $x \in X$ such that $T(x) = y$ and $\| x \| \leq M \| y \|$.
Observe that both $X$ and $Y$ ARE assumed to be Banach spaces.
Proof: $\Rightarrow$ Suppose that $T(X)$ is closed. Then by the theorem on the Closed Ranges of BLOs when Y is a Banach Space page there exists a positive constant $M^* \in \mathbb{R}$, $M^* > 0$ such that for every $\epsilon > 0$ and for every $x \in X$ there is a $y \in T(X)$ such that:
\begin{align} \quad \| T(x) - y \| < \epsilon \quad , \quad \| x \| \leq M^* \| y \| \end{align}
Now let $u_1 \in X$ be such that:
\begin{align} \quad \| y - T(u_1) \| < \frac{\| y \|}{2} \quad , \quad \| u_1 \| \leq M^* \| y \| \end{align}
Let $u_2 \in X$ be such that:
\begin{align} \quad \| (y - T(u_1)) - T(u_2) \| < \frac{\| y \|}{2^2} \quad , \quad \| u_2 \| \leq \frac{M^* \| y \|}{2} \end{align}
We continue on in this manner. Let $u_n \in X$ be such that:
\begin{align} \quad \| [y - T(u_1) - T(u_2) - ... - T(u_{n-1})] - T(u_n) \| < \frac{\| y \|}{2^n} \quad , \quad \| u_n \| \leq \frac{M^* \| y \|}{2^{n-1}} \end{align}
For each $n \in \mathbb{N}$ let:
\begin{align} \quad x_n = \sum_{k=1}^{n} u_k \end{align}
Then we have that:
\begin{align} \quad \| y - T(x_n) \| < \frac{\| y \|}{2^n} \quad (*) \end{align}
\begin{align} \quad \| x_n \| = \biggr \| \sum_{k=1}^{n} u_k \biggr \| \leq \sum_{k=1}^{n} \| u_k \| \leq \sum_{k=1}^{n} \frac{M^* \| y \|}{2^{k-1}} \leq 2M^* \| y \| \quad (**) \end{align}
From $(*)$ we have that $y = \lim_{n \to \infty} T(x_n)$.
We now show that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence. We have that:
\begin{align} \quad \| x_m - x_n \| = \biggr \| \sum_{k=n+1}^{m} u_k \biggr \| \leq \sum_{k=n+1}^{m} \frac{M^* \| y \|}{2^{k}} \leq \sum_{k=n+1}^{\infty} \frac{M^* \| y \|}{2^k} \end{align}
The righthand side of the inequality above goes to $0$ as $n \to \infty$. So indeed, $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $X$ . Since $X$ is a Banach space, it converges to $x$, and so $T(x) = y$. Now lastly from taking the limit as $n \to \infty$ at $(**)$ we have that:
\begin{align} \quad \| x \| \leq 2M^* \| y \| \end{align}
Taking $M = 2M^*$ and we have that for every $y \in T(X)$ there exists an $x \in X$ with $T(x) = y$ and such that $\| x \| \leq M \| y \|$ as desired. $\blacksquare$
$\Leftarrow$ This direction is a consequence of the general theorem mentioned at the top of the page.