# Existence of certain bounded approximate identity

In trying to follow the proof of Proposition 4.11 in

M. C. White, Injective modules for uniform algebras, Proc. London Math. Soc. 73 (1996) 155--184

there is a part which seems unclear.

Let $I$ be a left ideal in a unital Banach algebra $A$. Assume $I$ is weakly complemented as a Banach $A$-module in $A$. That is, the s.e.s $$0\xleftarrow{} I^*\xleftarrow{i^*} A^*\xleftarrow{\pi^*} (A/I)^*\xleftarrow{} 0$$ splits in the category mod-$A$. Let $\sigma$ be a right inverse $A$-morphism of $\pi^*$.

In Proposition 4.11 of the paper it is stated that $I$ has a bounded right approximate identity $(e_\alpha)$ such that $$\sup_\alpha\Vert 1-e_\alpha\Vert\leq C$$ for $C=\Vert\sigma\Vert$. I understand the proof until the moment the author claims (see the end of the page 12 )

"...so $e\in I^{\perp\perp}$. As $I^{\perp\perp}$ is the weak-star closure of $I$ in $A ' '$, we may choose a net in $I$ which tends to $1-e$ in the $\sigma(A ' ', A ')$-topology and which is bounded by $\Vert 1- e\Vert$. This net is a weak bounded approximate identity..."

(The author uses $X'$ for the dual of a Banach space $X$.)

As far as I can see, he constructs a net of the form $(1-e_\alpha)_{\alpha}$ such that

1) $\sup_{\alpha}\Vert 1-e_\alpha \Vert \leq \Vert 1-e\Vert$

2) the net $(1-e_\alpha)_{\alpha}$ converges to $1-e$ in the $\sigma(A ' ', A ')$ topology

3) $(e_\alpha)_{\alpha}$ is contained in $I$

4) $(e_\alpha)_{\alpha}$ is weak bounded approximate identity

Paragraphs 1) and 2) are just the Goldstine theorem. Paragraph 4) is a simple computation. The main problem is paragrpah 3), I can't show that $(e_\alpha)_{\alpha}$ is contained in $I$.

But even if we prove somehow that $(e_\alpha)\subset I$. There is one more step in the proof

"...a weak bounded approximate identity has a norm bounded approximate identity, with the same bound..."

I agree that the statement is true, but if we look carefully it says that we would have genuine approximate identity $(f_\beta)$ with $\sup_{\beta}\Vert f_\beta\Vert\leq\sup_\alpha\Vert e_\alpha\Vert$, though we want $\sup_\beta\Vert 1-f_\beta\Vert\leq \sup_\alpha\Vert 1-e_\alpha\Vert$

That are steps I do not understand.

• Small correction: it is claimed that $I$ has a right BAI Jun 13 '14 at 15:13
• Also, what point in Michael's proof don't you understand? On MO it is better to ask for specifics, rather than saying "I don't understand this argument, can someone explain it to me" Jun 13 '14 at 15:14
• Ok, I'll add the details in 10 minutes Jun 13 '14 at 15:20
• Hmm. Well, for a start $I^{\perp\perp}$ can be identified with $I''$ so there is no problem getting a bounded net in $I$. Doing so at the same time as (1) is not entirely obvious to me right now Jun 13 '14 at 15:55
• @Norbert: If any bounded net $(x_i)_i$ in a Banach space $X$ converges to an $x$ in the second dual $X^{**}$ in the weak$^*$ topology, then there is a net $(y_j)_j$ in the convex hull of $\{ x_i : i\}$ such that $(y_j)_j$ converges to $x$ in the weak$^*$ topology and satisfies $\limsup_j \| y_j \| = \| x \|$. This gives you $(e_\alpha)$ satisfying (1) - (4). Jun 15 '14 at 23:40

Recall that $I^{\perp\perp}$ is the weak${}^*$ closure of $I$ in $A^{**}$, so by Goldstine theorem we can choose a net $(e_\nu'')_{\nu\in N''}\subset I$ such that it weak${}^*$ converges to $e$. Clearly $(1-e_\nu'')_{\nu\in N''}$ converges to $1-e$ in the same topology. By lemma 1.1 from this paper there exists a net in $\operatorname{conv}(1-e_\nu'')_{\nu\in N''}=1-\operatorname{conv}(e_\nu'')_{\nu\in N''}$ that weak${}^*$ converges to $1-e$ with norm bound $\Vert 1-e\Vert$. Denote this net as $(1-e_\nu')_{\nu\in N'}$, then it is easy to check that $(e_\nu')_{\nu\in N'}$ weak${}^*$ converges to $e$ and a weak right approximate identity for $I$. By proposition 33.2 in Approximate identities and factorization in Banach algebras by Doran R. S., Wichmann J. there is a net $(e_\nu)_{\nu\in N}\subset\operatorname{conv}(e_\nu')_{\nu\in N'}$ which is a right bounded approximate identity for $I$. For any $\nu\in N$ the vector $1-e_\nu$ is in $\operatorname{conv}(1-e_\nu')_{\nu\in N'}$, then taking into account the norm bound on $(1-e_\nu')_{\nu\in N'}$ we get $$\sup_{\nu\in N}\Vert 1-e_\nu\Vert \leq\Vert 1-e\Vert$$