bounded linear operator space is banach

[[concept]]

Theorem

Let $V$ and $W$ be normed spaces. If $W$ is a Banach space, then the bounded linear operator space $B (V, W)$ is a Banach space.

Note

this proof is similar to the one for complete metric spaces have banach continuous bounded function spaces.

And, as the note indicated in Lecture 1, this is the basic outline for showing a space is Banach.

To show that this space is Banach, we will use the characterization that banach spaces have all absolutely summable series are summable.

Proof

Suppose ${T_{n}} \subset B (V, W)$ is a sequence of bounded linear operator space such that

C = \sum_{n} | | T_{n} | | < \infty

ie, the sequence is absolutely summable. We want to show that the $\sum_{n} T_{n}$ is summable.

\sum_{n} T_{n}

is summable

Let $v \in V$ and $m \in N$ . Then

\begin{aligned} \sum_{n = 1}^{m} | | T_{n} v | | & \leq \sum_{n = 1}^{m} | | T_{n} | | \cdot | | v | | \\ \leq | | v | | \sum_{n} | | T_{n} | | \\ = C | | v | | \end{aligned}

ie, the sequence of (nonnegative, real) partial sums ${\sum_{n = 1}^{m} | | T_{n} v | |}$ is bounded and therefore converges.

Thus the series $\sum_{n} T_{n} v$ is absolutely summable in $W$ . Since $W$ is Banach, (and banach spaces have all absolutely summable series are summable), we know that $\sum_{n} T_{n} v$ is summable.

So define our candidate limit $T : V \to W$ as

T v := lim_{m \to \infty} \sum_{n = 1}^{m} T_{n} v

We now need to show that this is a bounded linear operator.

T

is linear

For all $λ_{1}, λ_{2} \in K$ and $v_{1}, v_{2} \in V$ , we have

\begin{aligned} T (λ_{1} v_{1} + λ_{2} v_{2}) & = lim_{m \to \infty} \sum_{n = 1}^{m} T_{n} (λ_{1} v_{1} + λ_{2} v_{2}) \\ (*) & = lim_{m \to \infty} [λ_{1} \sum_{n = 1}^{m} T_{n} v_{1} + λ_{2} \sum_{n = 1}^{m} T_{n} v_{2}] \\ = λ_{1} T v_{1} + λ_{2} T v_{2} \end{aligned}

Note

For $(*)$ , we did not prove that the sum of the limit is the limit of the sums, but it is the same proof as the case in $R$ replacing absolute value with the appropriate norm instead.

T

is bounded

Let $v \in V$ . Then

\begin{aligned} | | T v | | & = | | lim_{m \to \infty} \sum_{n = 1}^{m} T_{n} v | | \\ = lim_{m \to \infty} | | \sum_{n = 1}^{m} T_{n} v | | \\ (*) & \leq lim_{m \to \infty} \sum_{n = 1}^{m} | | T_{n} v | | \\ \leq lim_{m \to \infty} \sum_{n = 1}^{m} | | T_{n} | | \cdot | | v | | \\ = | | v | | \sum_{n} | | T_{n} | | \\ (* *) & = C | | v | | \\ < \infty \end{aligned}

Where $(*)$ is by the triangle inequality and $(* *)$ is because we assume this series is absolutely summable.

Thus $T \in B (V, W)$

T

is indeed the limit

Now, all we need to show is that $T$ is indeed the limit in the operator norm. ie, that $\sum_{n = 1}^{m} T_{n} \to T$ as $m \to \infty$ .

Let $v \in V$ with $| | v | | = 1$ . Then

\begin{aligned} | | T v - \sum_{n = 1}^{m} T_{n} v | | & = | | lim_{m^{'} \to \infty} \sum_{n = 1}^{m} T_{n} v - \sum_{n = 1}^{m} T_{n} v | | \\ = | | lim_{m^{'} \to \infty} {\sum_{n = m + 1}^{m}}^{'} T_{n} v | | \\ \leq lim_{m^{'} \to \infty} \sum_{n = m}^{m^{'}} | | T_{n} v | | \\ (*) & \leq lim_{m^{'} \to \infty} \sum_{n = m + 1}^{m^{'}} | | T_{n} | | \cdot | | v | | \\ (* *) & = lim_{m^{'} \to \infty} \sum_{n = m + 1}^{m^{'}} | | T_{n} | | \\ = \sum_{n = m + 1}^{\infty} | | T_{n} | | \\ ⟹ | | T - \sum_{n = 1}^{m} T_{n} | | & \leq \sum_{n = m + 1}^{\infty} | | T_{n} | | \to 0 as m \to \infty \end{aligned}

Where $(*)$ is by the triangle inequality and $(* *)$ is because $| | v | | = 1$ . The last line we get because this is the tail of a convergent series that we defined at the beginning.

Thus the bounded linear operator space $B (V, W)$ is Banach

\begin{matrix} ◼ \end{matrix}

Mentions

File	Last Modified
Functional Analysis Lecture 2	2025-06-05
Functional Analysis Lecture 3	2025-06-05

Created 2025-05-30 Last Modified 2025-05-30