A proof of a basic conclusion in operator algebra

Question

There is a quotation below: (in a book named "C*-algebras Finite-Dimensional Approximations")

Lemma 2.3.4. Let $A$ be a Banach space, $\mathbb{B}(A)$ be the space of all bounded linear maps from $A$ to $A$ and $C\subset B(A)$ be any convex set. Then the point-weak and point-norm closures of $C$ coincide

Proof.

Step 1. Let $T\in \mathbb{B}(A)$ be a map such that there exists a net $\{T_{i}\}_{i\in I}\subset C$ with the property that for every $a\in A$ and functional $f\in A^{*}$, $$f(T_{i}(a))\rightarrow f(T(a)).$$ We must show that for each finite set $F=\{a_{1},...,a_{k}\}\subset A$ and $ \epsilon>0$ there exists $S\in C$ such that $||S(a_{j})-T(a_{j})||<\epsilon$ for $1\leq j\leq k$.

Step 2. This is a standard application of the Hahn-Banach Theorem since the net of maps (k-fold direct sum) $$T_{i}\oplus...\oplus T_{i}$$ converges point-weakly to $$T\oplus...\oplus T$$ in the Banach space $\mathbb{B}(A\oplus...\oplus A)$ (take $l^{p}$-norm on $A\oplus...\oplus A$).

Thus the element $$T(a_{1})\oplus...\oplus T(a_{k})$$ belongs to the weak closure of $$\{T_{i}(a_{1})\oplus...\oplus T_{i}(a_{k}\}_{i\in I}$$ and hence, by the Hahn-Banach Theorem, also to the norm closure of the convex hull.

Step 3. This implies that we can find a single convex combination of the $T_{i}$ which is simultaneously close to $T$ on all of $F$.

I have several incomprehensions on the proof above:

1. Why does the author consider the k-fold direct sum of $T_{i}$ in the step 2?

2. How to use Hahn-Banach Theorem, in step 2, to get the norm closure of the convex hull?

Answer 1

It is a basic consequence of the Hahn-Banach theorem that the weak and norm closures of a convex set agree. This is the key result used in the argument.

Here they want to use that fact, but they have the problem that they need to be able to choose a net of operators in $C$. That's the role of the finite sets $F$: they will be the indices of the net: for each $F$, their argument produces a $T_F$ (denoted $S$ in the argument).

To find such operator, they want to use the result in the first paragraph, but now they have to make it work on $a_1,\ldots,a_k$. A typical trick in such situation is to use the direct sum to work with a single element containing all the information.

Answer 2

1. They want to consider the values of $T$ at several points. Giving $T(a_1),T(a_2),...,T(a_k)$ is the same as giving $T^{\oplus k}$ at $(a_1,a_2,...,a_k)$. It is essentially the definition of Cartesian product.