Closed Ranges of BLOs when Y is a Banach Space

Closed Ranges of BLOs when Y is a Banach Space

Theorem 1: Let $X$ be a normed linear space, $Y$ be a Banach space, and let $T : X \to Y$ be a bounded linear operator. If $T(X)$ is closed then there exists a constant $M \in \mathbb{R}$, $M > 0$ such that for every $\epsilon > 0$ and for every $x \in X$ then exists a point $x' \in X$ such that:
a) $\| T(x) - T(x') \| < \epsilon$.
b) $\| x' \| \leq M \| T(x) \|$.
  • Proof: Since $Y$ is a Banach space and $T(X)$ is closed, $T(X)$ is also a Banach space.
  • Let $B_X$ denote the open unit ball in $X$ and let $B_{T(X)}$ denote the open unit ball in $T(X)$.
  • Step 1: We first aim to show that:
(1)
\begin{align} \quad T(X) = \bigcup_{n=1}^{\infty} n \overline{T(B_X)} \quad (*) \end{align}
  • Let $x \in T(X)$ and let $n_x \in N$ be such that $\| x \| \leq n_x$. Then $\frac{1}{n_x} x \in B_X$. So
(2)
\begin{align} \quad T(x) = n_x T \left ( \frac{1}{n_x} x \right ) \in n_x T(B_X) \in n_x \overline{T(B_X)} \subseteq \bigcup_{n=1}^{\infty} n \overline{T(B_X)} \end{align}
  • Therefore $T(X) \subseteq \bigcup_{n=1}^{\infty} n \overline{T(B_X)}$. Now clearly, for each $n \in \mathbb{N}$ we have that $n \overline{T}(B_X) \subseteq T(X)$. Therefore $T(X) \supseteq \bigcup_{n=1}^{\infty} n \overline{T(B_X)}$. So indeed, the equality at $(*)$ holds.
  • Now observe that $T(X)$ is a countable union of closed, nowhere dense sets in $T(X)$. But observe that $T(X)$ has nonempty interior in $T(X)$. Hence there must exist some $m \in \mathbb{N}$ such that $m\overline{T(B_X)}$ has nonempty interior.
  • Let $y_0 \in \mathrm{int} \left ( m\overline{T(B_X)} \right )$. Then there exists an $r > 0$ such that:
(3)
\begin{align} \quad B(y_0, r) \subseteq \mathrm{int} \left ( m\overline{T(B_X)} \right ) \subseteq m \overline{T(B_X)} \end{align}
  • So from the inclusion above we have that whenever $\| y_0 - T(x) \| < r$ that then $T(x) \in m \overline{T(B_X)}$. In particular, if $\| T(x) \| \leq 1$ then:
(4)
\begin{align} \quad rT(x) + y_0 \in m \overline{T(B_X)} \end{align}
  • Let $(s_n)_{n=1}^{\infty}$ and $(t_n)_{n=1}^{\infty}$ be sequences in $B_X$ such that:
(5)
\begin{align} \quad \lim_{n \to \infty} m T(s_n) = y_0 \quad \mathrm{and} \quad \lim_{n \to \infty} m T(t_n) = rT(x) + y_0 \end{align}
  • Let:
(6)
\begin{align} \quad \zeta_n = \frac{1}{2} t_n - \frac{1}{2} s_n \end{align}
  • Then we have that:
(7)
\begin{align} \quad \| \zeta_n \| = \biggr \| \frac{1}{2} t_n - \frac{1}{2} s_n \biggr \| \leq \frac{1}{2} \| t_n \| + \frac{1}{2} \| s_n \| \leq 1 \end{align}
  • Hence $(\zeta_n)_{n=1}^{\infty}$ is a sequence contained in $B_X$. Furthermore:
(8)
\begin{align} \quad \lim_{n \to \infty} 2mT(\zeta_n) = \lim_{n \to \infty} 2m T \left ( \frac{1}{2} t_n - \frac{1}{2} s_n \right ) = \lim_{n \to \infty} m[T(t_n) - T(s_n)] = \lim_{n \to \infty} mT(t_n) - \lim_{n \to \infty} mT(s_n) = rT(x) \end{align}
  • Therefore $rT(x) \in 2m\overline{T(B_X)}$ for every $x$ with $\| x \| \leq 1$. Hence:
(9)
\begin{align} \quad r\overline{B_{T(X)}} \subseteq 2m \overline{T(B_X)} \quad (**) \end{align}
  • Step 3: Let $\epsilon > 0$ be given and let $x \in X$ be such that $T(x) \neq 0$. Then by $(**)$ we have that:
(10)
\begin{align} \quad \frac{r}{\| T(x) \|} \cdot T(x) \in \overline{B_{T(X)}} \subseteq 2m \overline{T(B_X)} \end{align}
  • So there exists an element $z \in B_X$ such that:
(11)
\begin{align} \quad \biggr \| \frac{r}{\| T(x) \|} \cdot T(x) - 2m T(z) \biggr \| & < \frac{\epsilon \cdot r}{\| T(x) \|} \\ \quad \biggr \| T(x) - \frac{2m \| T(x) \|}{r} T(z) \biggr \| & < \epsilon \end{align}
  • So let $x' = \frac{2m \| T(x) \|}{r} z$. Then:
(12)
\begin{align} \quad \| T(x) - T(x') \| < \epsilon \quad (1) \end{align}
  • Furthermore since $z \in B_X$, $\| z \| \leq 1$ and:
(13)
\begin{align} \quad \| x' \| = \biggr \| \frac{2m \| T(x) \|}{r} z \biggr \| = \frac{2m \| T(x) \|}{r} \| z \| \leq \frac{2m}{r} \| T(x) \| \end{align}
  • By letting $M = \frac{2m}{r}$ we have that:
(14)
\begin{align} \quad \| x' \| \leq M \| T(x) \| \quad (2) \end{align}
  • Which completes the proof. $\blacksquare$
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