Cardinality of the space of functionals on $L^2$ functions 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 $L^1$?).
 A: 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.
