1. ## multiplicative linear functionals

show that there are more multiplicative linear functionals on $\displaystyle H^{\infty}(\mathbb{D})$ then the point evaluations in $\displaystyle \mathbb{D}$

where $\displaystyle H^{\infty}(\mathbb{D})$ is the set of bounded analytic function on the unity disc $\displaystyle \mathbb{D}$

2. Originally Posted by Mauritzvdworm
show that there are more multiplicative linear functionals on $\displaystyle H^{\infty}(\mathbb{D})$ then the point evaluations in $\displaystyle \mathbb{D}$

where $\displaystyle H^{\infty}(\mathbb{D})$ is the set of bounded analytic function on the unit disc $\displaystyle \mathbb{D}$
The multiplicative linear functionals on $\displaystyle H^{\infty}(\mathbb{D})$ that are not point evaluations in the open unit disk $\displaystyle \mathbb{D}$ form what is called the corona of the spectrum of $\displaystyle H^{\infty}(\mathbb{D})$. As the name suggests, they are somehow associated with the unit circle (the boundary of $\displaystyle \mathbb{D}$), but they are not just point evaluations at the points of the circle.

It would be convenient, for example, if we could define a multiplicative linear functional $\displaystyle \phi$ on $\displaystyle H^{\infty}(\mathbb{D})$ by $\displaystyle \phi(f) = \textstyle\lim_{r\nearrow1}f(r)$, taking the limit of f along the positive real axis as r increases to 1. But unfortunately this limit may not exist. There is no easy way to specify points in the corona. However, you can see that such points must exist, as follows.

The theory of commutative unital Banach algebras tells us that there is a topology on the set of multiplicative linear functionals that makes this set into a compact space. For the point evaluations on the algebra $\displaystyle H^{\infty}(\mathbb{D})$, this topology coincides with the usual topology of $\displaystyle \mathbb{D}$. But the open unit disk is not compact, so there must be additional multiplicative linear functionals that form a 'completion' of this set. However, the completion is not the usual one under which the completion of the open disk is the closed disk. Its structure is very much more complicated.

3. Originally Posted by Opalg
However, the completion is not the usual one under which the completion of the open disk is the closed disk. Its structure is very much more complicated.
Just out of curiosity is this completion like the following completion of a metric space:

Let $\displaystyle X$ be a metric space and fix $\displaystyle x_0\in X$. For every $\displaystyle x\in X$ define $\displaystyle f_x:X\mapsto \mathbb{R}$ by $\displaystyle y\mapsto d(y,x)-d(y,x_0)$.

It can be shown that if $\displaystyle F:X\mapsto \mathcal{C}\left[X,\mathbb{R}\right]$ given by $\displaystyle x\mapsto f_x$ is an isometry and so $\displaystyle F(x)$ is an isometric image in $\displaystyle \mathcal{C}\left[X,\mathbb{R}\right]$.

So, we define the completion $\displaystyle X^*$ of $\displaystyle X$ to be $\displaystyle X^*=\overline{F(X)}$?

4. Originally Posted by Drexel28
Just out of curiosity is this completion like the following completion of a metric space:

Let $\displaystyle X$ be a metric space and fix $\displaystyle x_0\in X$. For every $\displaystyle x\in X$ define $\displaystyle f_x:X\mapsto \mathbb{R}$ by $\displaystyle y\mapsto d(y,x)-d(y,x_0)$.

It can be shown that if $\displaystyle F:X\mapsto \mathcal{C}\left[X,\mathbb{R}\right]$ given by $\displaystyle x\mapsto f_x$ is an isometry and so $\displaystyle F(x)$ is an isometric image in $\displaystyle \mathcal{C}\left[X,\mathbb{R}\right]$.

So, we define the completion $\displaystyle X^*$ of $\displaystyle X$ to be $\displaystyle X^*=\overline{F(X)}$?
That is one way of constructing a completion for a metric space. But in general a space has many different completions. To take a very simple example, the open unit interval (0,1) can be completed by adding two points (0 and 1, obviously). It can also be completed by adding just one point, via the mapping $\displaystyle x\mapsto e^{2\pi ix}$, which wraps the interval round a circle and then only the single point 1 needs to be added.

Any locally compact space has a whole family of compactifications, ranging from the one-point compactification (as described above for the unit interval) to the Stone–Čech compactification. The corona of the algebra of bounded analytic functions on the disk comes somewhere between those two extremes.

5. Originally Posted by Opalg
That is one way of constructing a completion for a metric space. But in general a space has many different completions. To take a very simple example, the open unit interval (0,1) can be completed by adding two points (0 and 1, obviously). It can also be completed by adding just one point, via the mapping $\displaystyle x\mapsto e^{2\pi ix}$, which wraps the interval round a circle and then only the single point 1 needs to be added.

Any locally compact space has a whole family of compactifications, ranging from the one-point compactification (as described above for the unit interval) to the Stone–Čech compactification. The corona of the algebra of bounded analytic functions on the disk comes somewhere between those two extremes.
It is kind of coincidental since I am doing the Stone-Cech compactification of Tychonoff spaces right now.

Now, one more question haha. If you were to do a compactification of this space would you do it via the embedding into a product of intervals or would you (more likely considering this is FA) use the maximal ideal space?

6. Originally Posted by Drexel28
If you were to do a compactification of this space would you do it via the embedding into a product of intervals or would you (more likely considering this is FA) use the maximal ideal space?
The maximal ideals of a commutative unital Banach algebra are exactly the kernels of the multiplicative linear functionals. So yes, it's the maximal ideal space structure that's relevant here.