On the relationship between L^p spaces and C_c functions for p = infinity

Very quick post on the relationship between \mathcal{L}^p, \mathcal{C}_c(X) and \mathcal{L}^\infty. I will assume you already know what I am talking about, I’ll just be sharing some intuition on what those mean, but won’t bother with details. It’s more a reminder for me rather than something that intends to be useful, actually, but there’s almost nothing on the Internet about this!

When we discover that \mathcal{C}_c(X) (continuous functions with compact support) is dense in \mathcal{L}^p, we also discover that it does not hold if p = \infty and \mu(X) = \infty.

What that intuitively means is that if you take away functions in \mathcal{C}_c(X) from \mathcal{L}^p, you take away something fundamental for \mathcal{L}^p: you are somehow taking away a net that keeps the ceiling up.

The fact that it becomes false for limitless spaces (\mu(X) = \infty) and p = \infty means that the functions in \mathcal{L}^\infty do not need functions in \mathcal{C}_c(X) to survive.

This is reasonable: functions in \mathcal{L}^\infty are not required to exist only in a specific (compact) region of space, whereas functions in \mathcal{C}_c(X) do. Functions in\mathcal{L}^\infty are simply bounded – their image keeps below some value, but can go however far they want in x direction. Very roughly speaking, they have a limit on their height, but not on their width.

What we find out, however, is that the following chain of inclusions holds:

\mathcal{C}_c(X) \subset \mathcal{C}_\infty(X) \subset \mathcal{L}^\infty

That’s reasonable! Think about it:

  • Functions in \mathcal{C}_c(X) live in a well defined area of space – a confined area of space.
  • Functions in \mathcal{C}_\infty(X) are allowed to live everywhere, with the constraint that they become more and more negligible the farther and farther we go. Not required to ever be zero though.
  • Functions in \mathcal{L}^\infty are simply required to have an upper bound (a finite one, obviously).

I’m not saying this is simple (advanced analysis is at least as difficult as pitching a nail with a needle as hammer), but after careful thinking, it’s just the way it should be, given the definitions.

  • Was this Helpful ?
  • yes   no

Leave a Reply

Your email address will not be published. Required fields are marked *