2

What is the cardinality of the space of functionals $F$ on $L^2$ (square integrable functions)? (I.e. the space of possible $F : (f \in L^2(\mathbb{R^n})) \to \mathbb{R}$.)

Is it the same as that of $R^n$. (If bigger are there some additional constraints on the functionals such that they are the same.)

Background: This is obviously a simple but important math question. I need to know the answer because I'm looking whether it is possible to construct a bijection for a practical purpose. This is obviously impossible if the spaces are not of the same cardinality to begin with. There are plenty of similar (on-topic) questions on for example the cardinality of ($L^p$) functions (example what is the cardinality of set of all smooth functions in $L^1$?).

Kvothe
  • 233
  • 1
  • 12
  • Wow I am completely lost why this question has been closed. Is more information needed to assess the possible cardinalities. Is it unclear what I meant by functionals? I think it is pretty standard that it means a map from a function space to a real or complex number. Specifically I meant the space of Functionals $F : \mathbb{L}^2 \to \mathbb{R}$. – Kvothe Dec 16 '21 at 12:06
  • 1
    Why is it closed? For example, you don't tell us what $\mathbb{L}^2$ is. – J. De Ro Dec 16 '21 at 12:16
  • Thanks! My mistake! I thought it was standard notation. I see it is normally written just $L^2$. I am talking about square integrable functions. – Kvothe Dec 16 '21 at 12:52
  • 1
    But square integrable functions on what? You can consider this on any measure space, and the underlying cardinality of the measure space makes a huge difference for the answer of your question! – J. De Ro Dec 16 '21 at 13:14
  • @QuantumSpace, I would be interested in the cases $\mathbb{R}^n \to \mathbb{R}$ and $\mathbb{C}^n \to \mathbb{R}$ . From this answer https://math.stackexchange.com/a/1384613/408562 I understand that any $L^p$ space has at most cardinality $\mathbb{R}$. I of course don't know what that would imply for the space of functionals on these spaces but I would be surprised if it would be different depending on whether the functions are on say $R$ or $R^n$. But perhaps more exotic things would make a difference. (Or I am just completely wrong about this.) – Kvothe Dec 16 '21 at 13:20
  • 1
    $\dim X\le\dim X^\ast$ for any normed linear space $X$ and its dual space $X^\ast$ – FShrike Apr 08 '22 at 16:26
  • 2
    It would help convince Readers to reopen your Question if you shared even some partial result you found in searching the "plenty of similar (on-topic) questions." Some knowledge of linear algebra is useful for this problem, and I'd be interested to see some indication of what background you bring to the problem (you mentioned that it came up for you more than once). Please add such info in the body of the Question now that it has been undeleted. – hardmath Apr 08 '22 at 19:34
  • 4
    Are you asking about the set of all functions $L^2(\mathbb{R}^n)\to\mathbb{R}$? Or only linear functions? Or only bounded linear functions? – Eric Wofsey Apr 08 '22 at 20:49
  • @EricWofsey, just to be clear I am asking about the space of Functionals that can act on these functions. Yes I was thinking of all $L^2$ functions without further constraints although I am also interested in answers for related cases with additional constraints on either the functional or the functions on which they act. – Kvothe Apr 09 '22 at 11:41

1 Answers1

3

No, the cardinality of functions from $L^2(\mathbb R^n)$ to $\mathbb R$ is much bigger than the cardinality of $\mathbb R^n$ (for natural number $n$). So a bijection is not possible.

As a matter of definition, the set of functions from set $B$ to set $A$ is called $A^B$. Building on that we further define $|A|^{|B|} = |A^B|$. There is something to be proved about this definition, e.g. that it is well-defined regardless of the choice of representative sets $A,B$ for equivalence classes $|A|,|B|$ respectively, and that it agrees with our arithmetic notion in the case of finite cardinals. But for our immediate purposes we will assume that this is a sensible definition.

Your problem is then to compare $|\mathbb R|^{|L^2(\mathbb R^n)|}$ with $|\mathbb R^n| = |\mathbb R|^n = |\mathbb R|$. Written this way it appears that the former is much bigger than the latter since $L^2(\mathbb R^n)$ is an infinite family of equivalence classes of functions. But our intuition is sometimes not reliable about infinite cardinals, so let's exercise care in drawing that conclusion.

A useful step in comparing these cardinals is the proposition that if $2\le |A| \le |2^B|$ and $B$ infinite, then $|A^B|=|2^B|$.

To apply this to the first cardinal, we check whether $|\mathbb R| \le |2^{L^2(\mathbb R^n)}|$. Since $|\mathbb R| = |2^{\mathbb N}|$, and since:

$$ |\mathbb N| \le |L^2(\mathbb R^n)| $$

it is true that:

$$ |\mathbb R| = |2^{\mathbb N}| \le |2^{L^2(\mathbb R^n)}| $$

Therefore $|\mathbb R|^{|L^2(\mathbb R^n)|} = |2^{L^2(\mathbb R^n)}|$.

Finally since $|\mathbb R| \le |L^2(\mathbb R^n)|$, we can use Cantor's theorem to conclude:

$$ |\mathbb R| \lt |2^{\mathbb R}| \le |2^{L^2(\mathbb R^n)}| = |\mathbb R|^{|L^2(\mathbb R^n)|} $$

Your desire to have a bijection between $\mathbb R^n$ and the space of functions from $L^2(\mathbb R^n)$ to $\mathbb R$ might be met by some approximation scheme. This is a problem of analysis and approximation theory, and the right approach depends on the topology needed for your intended application.

hardmath
  • 37,015