Criterion for the Range of a BLO to be Closed when X is a Ban. Space

# Criterion for the Range of a BLO to be Closed when X is a Banach Space

 Theorem 1: Let $X$ be a Banach space, $Y$ be a normed linear space, and let $T : X \to Y$ be a bounded linear operator. If there exists a constant $M \in \mathbb{R}$, $M > 0$ such that for every $y \in T(X)$ there exists an $x \in X$ such that: 1) $T(x) = y$. 2) $\| x \| \leq M \| y \|$. Then the range of $T$, $T(X)$ is a closed subspace of $Y$.
• Proof: To show that $T(X)$ is closed in $Y$ we will show that $T(X)$ contains all of its accumulation points.
• Let $(y_n)_{n=1}^{\infty}$ be a sequence of points in $Y$ that converge to some point $y \in Y$, that is:
(1)
\begin{align} \quad \lim_{n \to \infty} y_n = y \end{align}
• We want to show that $y \in T(X)$.
• Now since $(y_n)_{n=1}^{\infty}$ converges to $y$, every subsequence of $(y_n)_{n=1}^{\infty}$ converges to $y$. In particular, since $(y_n)_{n=1}^{\infty}$ converges it is Cauchy. So we may assume that for all $n \in \mathbb{N}$:
(2)
\begin{align} \quad \| y_n - y_{n+1} \| < \frac{1}{2^n} \end{align}
• Now for each $n \in \mathbb{N}$ we have that $y_n, y_{n+1} \in T(X)$. Therefore $(y_n - y_{n+1}) \in T(X)$. By the hypotheses in the statement of the theorem, there exists an element $s_n \in X$ with $T(s_n) = y_n - y_{n+1}$ and moreover:
(3)
\begin{align} \quad \| s_n \| \leq M \| y_n - y_{n+1} \| < \frac{M}{2^n} \end{align}
• We define a sequence of points $(x_n)_{n=1}^{\infty}$ in $X$ for each $n \in \mathbb{N}$ by:
(4)
\begin{align} \quad x_n = \sum_{k=1}^{n} s_n \end{align}
• We now show that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence. We have that:
(5)
\begin{align} \quad \| x_m - x_n \| = \left \| \sum_{k=n+1}^{m} s_k \right \| \leq \sum_{k=n+1}^{m} \| s_k \| \leq \sum_{k=n+1}^{m} M \| y_k - y_{k+1} \| < \sum_{k=n+1}^{m} \frac{M}{2^k} \leq \sum_{k=n+1}^{\infty} \frac{M}{2^k} \end{align}
• The righthand side of the inequality above goes to $0$ as $n \to \infty$. This shows that the quantity $\| x_m - x_n \|$ can be made arbitrary small for all $m, n \geq N$ for $N \in \mathbb{N}$ chosen sufficiently large. Therefore $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $X$.
• Since $X$ is a Banach space, $(x_n)_{n=1}^{\infty}$ converges to some point $x \in X$, that is:
(6)
\begin{align} \quad \lim_{n \to \infty} x_n = x \end{align}
• And so since $x \in X$, $T(x) \in T(X)$. But then we have that (by using the continuity of $T$ since $T$ is a bounded linear operator):
(7)
\begin{align} \quad T(x) = T \left (\lim_{n \to \infty} x_n \right ) = \lim_{n \to \infty} T \left ( \sum_{k=1}^{n} s_n \right ) = \lim_{n \to \infty} \sum_{k=1}^{n} T(s_k) = \sum_{k=1}^{\infty} (y_n - y_{n+1}) = y_1 - \lim_{n \to \infty} y_{n+!} = y_1 - y \end{align}
• Therefore:
(8)
\begin{align} \quad y = y_1 - T(x) \end{align}
• But we have that $y_1 \in T(X)$ and $T(x) \in T(X)$. So $y \in T(X)$. Hence $T(X)$ is closed. $\blacksquare$