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Let $\mathcal{A}$ be an algebra. $d: \mathcal{A}\to \mathbb{N}$ is a function satisfying 1) $d(S+T)\le d(S)+d(T)$, 2)$d(ST)\le d(S)+d(T)$ and 3) $d(aS)=d(S)$ for all $a\in \mathbb{C}, S,T\in\mathcal{A}$.

I am wondering what else can we say about this function and what kind of information can this function give about the algebra?

For people who may be interested about the background of this function, I am thinking about the defect function in the almost invariant subspaces theory.

Thanks!

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You mean $a\in\mathbb C\setminus\{0\}$, right? Does $\mathbb N$ contain $0$? –  Jonas Meyer May 28 '12 at 22:06
    
@JonasMeyer Yes. That is true. I thought about this and it would be great if I could show this ideal is maximal for some k. So I wonder whether there is some method to show an ideal of an operator algebra to be maximal. Do you have any suggestions? Another thing that is interesting is that the set you defined can be thought of as a basis for a certain topology on the algebra. Do you know something related to this kind of topology? Thanks! –  Hui Yu May 29 '12 at 0:52
    
Nevermind, I'll delete that comment. Thank you for asking. (For those who may be confused: I made a silly thinking error, saying that $\{x:d(x)\leq k\}$ is an ideal for each $k\in \mathbb N$.) I don't have anything to say about topology. –  Jonas Meyer May 29 '12 at 1:27
    
@JonasMeyer what about the maximality of an ideal in a operator algebra? How can one show an ideal is maximal? –  Hui Yu May 29 '12 at 1:34
    
I don't know if I can say anything useful, but it's good to keep in mind that an ideal $I$ is maximal if and only if $\mathcal A/I$ is simple. Also, in a unital Banach algebra, the closure of a proper ideal is a proper ideal, which implies that maximal ideals are always closed. This means that the quotient is also a Banach algebra with norm $\|a+I\|=d(a,I)=\inf\limits_{x\in I}\|a-x\|$. –  Jonas Meyer May 29 '12 at 1:37

1 Answer 1

Pick any $m\in {\mathbb N}$ (I am assuming ${\mathbb N}$ does not include zero). The constant function $d: A \to \{m\}$ satisfies all three conditions. Hence without further information or restrictions, a function satisfying all three conditions might not tell you anything about the original algebra.

By the way, your second condition is not what people usually mean by "submultiplicative".

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