Topo. Comp. Crit. for the Ran. of a BLO to be Clo., X and Y are Ban. Sp.

Topological Complement Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces

Recall from the Topological Complements of Normed Linear Subspaces that if $X$ is a normed linear space and $M \subset X$ is a subspace then a topological complement of $M$ is an algebraic complement $M' \subset X$ that is closed.

We will now look at a very nice criterion which tells us that if $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a bounded linear operator, then $T(X)$ is closed if $T(X)$ has a topological complement.

Theorem 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. If $T(X)$ has a topological complement then $T(X)$ is closed.

• Proof: Suppose that $T(X)$ has a topological complement $M' \subset Y$. Then $M'$ is closed and:

(1)

\begin{align} \quad T(X) \cap M' = \{ 0 \} \end{align}

(2)

\begin{align} \quad Y = T(X) \cup M' \end{align}

• Consider the product space:

(3)

\begin{align} \quad X \times M' = \{ (x, y) : x \in X, m' \in M' \} \end{align}

• We define a norm on $X \times M'$ for all $(x, m') \in X \times M'$ by:

(4)

\begin{align} \quad \| (x, m') \| = \| x \| + \| m' \| \end{align}

• We prove that this is indeed a norm. Suppose that $\| (x, m') \| = 0$. Then $\| x \| + \| m' \| = 0$ which happens if and only if $x = 0$ and $m' = 0$, so $(x, m') = (0, 0)$. Clearly $\| (0, 0) \| = 0$. Therefore:

(5)

\begin{align} \quad \| (x, m') \| = 0 \quad \mathrm{iff} \quad (x, m') = (0, 0) \end{align}

• Let $\lambda \in \mathbb{C}$. Then:

(6)

\begin{align} \quad \| \lambda (x, m') \| &= \| (\lambda x, \lambda m') \| \\ &= \| \lambda x \| + \| \lambda m' \| \\ &= |\lambda| \| x \| + |\lambda| \| m' \| \\ &= |\lambda| (\| x \| + \| m' \|) \\ &= |\lambda| \| (x, m') \| \end{align}

• Lastly, if $(x, m'), (y, n') \in X \times M$ then:

(7)

\begin{align} \quad \| (x, m') + (y, n') \| &= \| (x + y, m' + n') \| \\ &= \| x + y \| + \| m' + n' \| \\ &\leq (\| x \| + \| y \|) + (\| m' \| + \| n' \|) \\ &\leq (\| x \| + \|m'\|) + (\| y \| + \| n' \|) \\ &\leq \| (x, m') \| + \| (y, n') \| \end{align}

• So indeed, $X \times M$ with this norm is a normed linear space. We aim to show that $X \times M'$ is a Banach space. We are given that $X$ is a Banach space and that $M' \subset Y$ is closed and hence a Banach space. Let $(x_n, m'_n)_{n=1}^{\infty}$ be a Cauchy sequence in $X \times M$. Then $(x_n)_{n=1}^{\infty}$ converges to some $x \in X$ and $(m'_n)_{n=1}^{\infty}$ converges to some $m' \in M'$. So:

(8)

\begin{align} \quad \lim_{n \to \infty} \| (x_n, m'_n) - (x, m) \| = \lim_{n \to \infty} \| (x_n - x, m'_n - m') \| = \lim_{n \to \infty} \| x_n - x \| + \lim_{n \to \infty} \|m'_n - m' \| = 0 \end{align}

• So $(x_n, m'_n)_{n=1}^{\infty}$ converges to $(x, m')$. So $X \times M'$ is a Banach space.

• We now define a linear operator $S : X \times M' \to Y$ for all $(x, m') \in X \times M'$ by:

(9)

\begin{align} \quad S(x, m') = T(x) + m' \end{align}

• We verify that $S$ is a linear operator. Let $(x, m'), (y, n') \in X \times M'$ and let $\lambda \in \mathbb{C}$. Then:

(10)

\begin{align} \quad S((x, m') + (y, n')) &= S(x + y, m' + n') \\ &= T(x + y) + (m' + n') \\ &= T(x) + T(y) + m' + n' \\ &= (T(x) + m') + (T(y) + n') \\ &= S(x, m') + S(y, n') \end{align}

(11)

\begin{align} \quad S(\lambda (x, m')) &= S(\lambda x, \lambda m') \\ &= T(\lambda x) + \lambda m' \\ &= \lambda T(x) + \lambda m' \\ &= \lambda (T(x) + m') \\ &= \lambda S(x, m') \end{align}

• Furthermore, $S$ is bounded since for all $(x, m') \in X \times M'$ we have that:

(12)

\begin{align} \quad \| S(x, m') \| &= \| T(x) +m' \| \\ &\leq \| T(x) \| + \| m' \| \\ &\leq \| T \| \| x \| + \| m' \| \\ &\leq (\| T \| \| x \| + \| T \| \| m' \|) + (\| x \| + \| m' \|) \\ & \leq (\| T \| + 1) [\| x \| + \| m' \|] \\ & \leq (\| T \| + 1) \| (x, m') \| \end{align}

• Observe further that since $Y = T(X) \oplus M'$ we have that $S$ is surjective. Therefore, $X \times M'$ and $Y$ are both Banach spaces, $S : X \times M' \to Y$ is a bounded linear operator, and $S(X \times M') = Y$ is closed. By the theorem on the IFF Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces page there exists an $N > 0$ such that for every $y \in Y$ there exists an $(x, m') \in X \times M'$ such that:

(13)

\begin{align} \quad S(x, m') = T(x) + m' = y \quad (*) \quad \mathrm{and} \quad \| (x, m') \| \leq N \| y \| \quad (**) \end{align}

• Then by $(*)$ we have that $m' = y - T(x)$. But then $y \in T(X)$ and $T(x) \in T(X)$, so:

(14)

\begin{align} \quad m' \in T(X) \cap M' = \{ 0 \} \quad (\dagger) \end{align}

• Therefore $m' = 0$. So $T(x) = y$. But then from $(**)$ we have that $\| (x, m') \| \leq N \| y \|$, so $\| x \| + \| m' \| \leq N \| y \|$, i.e.:

(15)

\begin{align} \quad \| x \| \leq N \| y \| \quad (\dagger \dagger) \end{align}

• By applying the theorem referenced above again, we see by $(\dagger)$ and $(\dagger \dagger)$ then $T(X)$ is closed. $\blacksquare$