Second IFF Crit. for the Ran. of a BLO to be C. when X, Y are Ban. Sps.

# Second IFF Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces

Recall from the IFF Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces page the following important result: If $X$ and $Y$ are **both** Banach spaces and $T : X \to Y$ is 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 $x \in X$ with $T(x) = y$ and $\| x \| \leq M \| y \|$.

We will prove a second if and only if criterion for the range of a bounded linear operator to be closed when $X$ and $Y$ are Banach spaces that is sometimes easier to apply.

Theorem 1: Let $X$ and $Y$ 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 $x \in X$ with $\| T(x) \| < 1$ there exists an $x' \in X$ with $T(x) = T(x')$ and $\| x' \| < M'$. |

**Proof:**$\Rightarrow$ Suppose that $T(X)$ is closed. By the IFF criterion for the range of a bounded linear operator to be closed when $X$ and $Y$ are Banach spaces, we have that 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 $\| x' \| \leq M \| y \|$.

- Now every $y \in T(X)$ corresponds to some $x \in X$, so equivalently, there exists a positive constant $M \in \mathbb{R}$, $M > 0$ such that for every $x \in X$ there exists an $x' \in X$ such that $T(x') = T(x)$ and $\| x' \| \leq M \| T(x) \|$.

- Take any $x \in X$ such that $\| T(x) \| < 1$ and choose $\delta > 0$ such that:

\begin{align} \quad \| T(x) \| \leq \frac{1}{1 + \delta} \end{align}

- Then from the remarks made above with $M' = M$, there exists an $x' \in X$ such that $T(x) = T(x')$ and:

\begin{align} \quad \| x' \| \leq M \| T(x) \| \leq \frac{M}{1 + \delta} < M = M' \end{align}

- $\Leftarrow$ Suppose that there exists a positive constant $M' \in \mathbb{R}$, $M' > 0$ such that for every $x \in X$ with $\| T(x) | < 1$ there exists an $x' \in X$ such that $T(x) = T(x')$ and $\| x' \| < M'$.

- Let $x \in X$ be such that $T(x) \neq 0$. Then clearly $\| T(x) \| < 2 \| T(x) \|$. So then $\displaystyle{\frac{\| T(x) \|}{2 \| T(x) \|} < 1}$ and so there exists an $x' \in X$ such that:

\begin{align} \quad \frac{1}{2 \| T(x) \|} T(x) = T(x') \quad , \quad \| x' \| \leq M' \end{align}

- Let $x'' = 2 \| T(x) \| x'$. Then:

\begin{align} \quad T(x'') = T(2 \| T(x) \| x') = 2 \| T(x) \| T(x') = T(x) \end{align}

- And moreover by setting $M = 2M'$ we have that:

\begin{align} \quad \| x'' \| = 2 \| T(x) \| \| x' \| \leq 2M' \| T(x) \| = M \| T(x) \| \end{align}

- So there exists a positive constant $M \in \mathbb{R}$, $M > 0$ such that for every $T(x) \in T(X)$ there exists an $x'' \in X$ such that $T(x'') = T(x)$ and $\| x'' \| \leq M \| T(x)$. So by the IFF criterion for the Range of the bounded linear operator to be closed when $X$ and $Y$ are Banach spaces we have that $T(X)$ is closed. $\blacksquare$