Square Integrable and Continuous

Question

I have come across the notation $L^2(\Omega) \cap C(\Omega)$; while I believe understand the resulting behavior, I can't get my head around the machinery. When I think about $f \in C(\Omega)$ I am thinking out a single continuous function. However, for $f \in L^2(\Omega)$ I really mean a representative in an equivalence class.

The upshot:

• $C(\Omega)$ seems like a collection of functions (maybe this is my flaw)

• $L^2(\Omega)$ is a collection of equivalence classes

because of this I am have a hard time understanding the meaning of $L^2(\Omega) \cap C(\Omega)$ and as a result what it means for $f \in L^2(\Omega) \cap C(\Omega)$.

Can someone help shed some light on how to think of this? Thanks!

Answer 1

It is a common abuse of notation to interchange between equivalence classes and representatives thereof when dealing with the $L^p$ spaces. I would say that $L^2(\Omega) \cap C(\Omega)$ might be a collection of functions (square integrable continuous functions) or a collection of equivalence classes (members of $L^2$ which are a.e. equal to a continuous function) depending on which is more useful in context. For example, if you want to take limits, you would want to think of them as functions, because limits (or even existence thereof) depend on the representative chosen. If all you want to do is integration, then equivalence classes are probably more convenient.

Answer 2

It's the collection of continuous functions that also appear in some equivalence class in $L^2$. I.e., "continuous, square-integrable functions".

So a particular instance function exists in an equivalence class and for the normal universe of activities in the neighborhood of Lebesgue integration any member of that equivalence class is substitutable in usage.

Answer 3

For specificity, let's suppose we are talking about real functions. Then $$L^2(\Omega)\cap C(\Omega)=\{f:\Omega\to\mathbb{R}\mid f \text{ is continuous on }\Omega \text{ and } \int_\Omega f^2<\infty\}.$$

For example, if $\Omega=[1,\infty)$ and $f(x)={1\over x}$, then $f\in L^2(\Omega)\cap C(\Omega)$, but $g(x)={1\over \sqrt{x}}$ is not since $g\not\in L^2(\Omega)$.

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$C(\Omega)$ is a set of functions.

$L^2(\Omega)$ is an equivalence class of functions since changing the value of a function on a set of measure zero does not change the value of the integral.