Assume $I$ is an ideal of unitization $A_+$ of Banach algebra $A$. Assume $A_+/I$ is weakly complemented in $A_+$. That is s.e.s
$$
0\xleftarrow{} I^*\xleftarrow{i^*} A_+^*\xleftarrow{\pi^*} (A_+/I)^*\xleftarrow{} 0
$$
splits. Let $\sigma$ be a right inverse $A$-morphism of $\pi^*$. In proposition 4.11 [here](http://www.mas.ncl.ac.uk/~nmcw/papers/ua21.pdf) it is stated that $I$ has bounded 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..."



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 statemant 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.