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 $\mathcal{B}(V,W)$ 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 $\{T_n\}\subset \mathcal{B}(V,W)$ 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 $m\in \mathbb{N}$. Then

\sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert &\leq \sum_{n=1}^m \lvert \lvert T_{n} \rvert \rvert\cdot \lvert \lvert v \rvert \rvert \\ &\leq \lvert \lvert v \rvert \rvert \sum_{n} \lvert \lvert T_{n} \rvert \rvert \\ &= C \lvert \lvert v \rvert \rvert \end{align}$$ ie, the sequence of (nonnegative, real) partial sums $\left\{ \sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert \right\}$ is bounded and therefore **converges**. Thus the series $\sum_{n}T_{n}v$ is [[Concept Wiki/summable series\|absolutely summable]] in $W$. Since $W$ is [[Concept Wiki/Banach space\|Banach]], (and [[Concept Wiki/banach spaces have all absolutely summable series are summable]]), we know that $\sum_{n}T_{n}v$ is [[Concept Wiki/summable series\|summable]].

So define our candidate limit $T:V\to W$ as $Tv:={\lim }_{m\to \infty }{\sum }_{n=1}^m{T}_{n}v$ We now need to show that this is a bounded linear operator.

$T$ is linear For all ${\lambda }_1,{\lambda }_2\in \mathbb{K}$ and $v_1,v_2\in V$, we have

T(\lambda_{1}v_{1}+\lambda_{2}v_{2}) &= \lim_{ m \to \infty } \sum_{n=1}^m T_{n}(\lambda_{1}v_{1}+\lambda_{2}v_{2}) \\ (*)&= \lim_{ m \to \infty }\left[ \lambda_{1} \sum_{n=1}^m T_{n} v_{1} + \lambda_{2}\sum_{n=1}^m T_{n} v_{2} \right]\quad \\ &= \lambda_{1} Tv_{1} + \lambda_{2} Tv_{2} \end{align}$$ > [!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 $\mathbb{R}$ replacing absolute value with the appropriate norm instead. >

$T$ is bounded Let $v\in V$. Then

\lvert \lvert Tv \rvert \rvert &= \left\lvert \left\lvert \lim_{ m \to \infty } \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ &= \lim_{ m \to \infty } \left\lvert \left\lvert \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ (*)&\leq \lim_{ m \to \infty } \sum_{n=1}^m \lvert \lvert T_{n}v \rvert \rvert \\ &\leq \lim_{ m \to \infty } \sum_{n=1}^m \lvert \lvert T_{n} \rvert \rvert \cdot \lvert \lvert v \rvert \rvert \\ &= \lvert \lvert v \rvert \rvert \sum_{n} \lvert \lvert T_{n} \rvert \rvert \\ (**)&= C \lvert \lvert v \rvert \rvert \\ &< \infty \end{align}$$ Where $(*)$ is by the [[Concept Wiki/triangle inequality]] and $(* *)$ is because we assume this series is [[Concept Wiki/summable series\|absolutely summable]].

Thus $T\in \mathcal{B}(V,W)$

$T$ 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

\left\lvert \left\lvert Tv - \sum_{n=1}^m T_{n}v \right\rvert \right\rvert &= \left\lvert \left\lvert \lim_{ m' \to \infty } \sum_{n=1}^m T_{n}v - \sum_{n=1}^m T_{n}v \right\rvert \right\rvert \\ &= \left\lvert \left\lvert \lim_{ m' \to \infty } \sum_{n=m+1}^m' T_{n}v \right\rvert \right\rvert \\ &\leq \lim_{ m' \to \infty } \sum_{n=m}^{m'} \lvert \lvert T_{n}v \rvert \rvert \\ (*) &\leq \lim_{ m' \to \infty } \sum_{n=m+1}^{m'} \lvert \lvert T_{n} \rvert \rvert \cdot \lvert \lvert v \rvert \rvert \\ (**)&= \lim_{ m' \to \infty } \sum_{n=m+1}^{m'} \lvert \lvert T_{n} \rvert \rvert\\ &= \sum_{n=m+1}^{\infty} \lvert \lvert T_{n} \rvert \rvert \\ \implies \left\lvert \left\lvert T - \sum_{n=1}^m T_{n} \right\rvert \right\rvert & \leq \sum_{n=m+1}^\infty \lvert \lvert T_{n} \rvert \rvert \to 0 \text{ as } m \to \infty \end{align}$$ Where $(*)$ is by the [[Concept Wiki/triangle inequality]] and $(* *)$ is because $\lvert \lvert v \rvert \rvert =1$. 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 $\mathcal{B}(V,W)$ is Banach \tag*{$\blacksquare$}

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