I'm solving conway's functional analysis.
Let $\mathcal{A}$ be a Banach algebra but do not assume it has an identity. If $I$ is a left ideal of $\mathcal{A}$, we say $I$ is modular left ideal if there is $u$ in $\mathcal{A}$ s.t. $\mathcal{A}(1-u) \equiv \{a-au:a\in\mathcal{A}\} \subset I$.$\qquad$(a) if $u$ is right modular unit(i.e. $u \equiv au \mod I $) for left ideal and $u \in I$, then $I = \mathcal{A}$
I solved this one because for $a \in \mathcal{A}$, $a = au + i$ for $i \in I$ and using $I$ is left ideal.
(b) Maximal modular left ideals are maximal left ideals.
I tried this one that assuming $M$ is Maximal modular left ideal. Then $M + Au$ is also modular left ideal so $M= M+Au$ or $A = M+Au$, so, in former case $M= Au$, and second case, $M = \{a-au:a\in\mathcal{A}\}$, but I cannot proceed to these two are maximal left ideals.
(c) If $I$ is proper modular left idea, then $I$ is contained in a maximal left ideal.
I solved this one via Zorn's lemma
(d) If $I$ is a proper modular left ideal and $u$ is a modular right unit for $I$, then $||u-x||\geq 1$for all $x$ in $I$ and $cl I$ is a proper modular left ideal.
I think this is very similar to the lemma that if $||1-a||<1$ then it has inverses. In my guess, $u$ is similar to identity in $A/I$(can I define this one? $I$ is not ideal(but left ideal)) so it can be solved with (a), but no idea because it is only one-side identity.
(e) Every maximal modular left ideal of $\mathcal{A}$ is closed
(e) is direct result of (d).
I need help of problem (b) and (d). Since I only studied [Dummit and Foote's Abstract Algebra] so my background of Algebra is too weak. Thanks.