Representing a Banach space as a function space When I think about a vector space, I like to see them as spaces of parameters. Although finite dimensional spaces have been generalized to Banach spaces, I see function spaces as the best generalization, at least heuristically, for a space of parameters. Given the function space $L^p(X,\Omega,\mu)$, for simplicity $L^p(X)$, the set of Dirac's deltas at each point works as a kind of canonical base, so when we choose a function, we are specifying somehow $|X|$ number of parameters; $|X|$ is the cardinality of $X$. My first question is: what are sufficient and necessaries conditions for a Banach space to be a function space?
To be more precise:


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*When is a Banach space $B$ isometric to some space $L^p(X)$? (isometric representation)

*When is there a linear isomorphism between $B$ and some $L^p(X)$, that is, a linear operator $A:B\to L^p(X)$ such that $c\lVert x\rVert_B\le \lVert Ax\rVert_{L^p}\le C\lVert x\rVert_B$? (continuous representation)

*When is there a space $X$ and a continuous operator $A:C_c(X)\to B$, such that $A$ is 1-1 and the image of $A$ is dense? $C_c(X)$ is the space of compactly supported functions on $X$; we can endow it with the usual topology. (weak representation)


The definition I like the most is the third, I think it's the weaker. The first two definitions involve $L^p$ spaces, and they are not the sole function spaces out there, so these definitions are incomplete. Can you imagine a weaker way of representing a Banach space?
If $B$ is a Hilbert space, take an orthonormal expansion, so every vector is $f=\sum_{i\in I} f(i)e_i$ and we have an isometric representation with $L^2(I)$, but this is not the most appealing representation, and it leaves more questions. What can we say about $X$?
If you take $L^2(\mathbb{R}^n)$, you can use a partition of unity and Fourier series to write every function $f$ as a sum $\sum_{i\in I} f(i)e_i$, where $I$ is countable and the $e_i$ form an orthonormal base. But, what is then the difference between $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$? Doesn't the structure of the space $X$ matter? I feel uneasy.
Many times in my work I have encountered operators like $A:L^p(X)\to L^q(Y)$, where $X$ and $Y$ are topological spaces of dimensions $N$ and $M$, respectively; for simplicity, suppose that $A$ is injective. Heuristically I apply the finite dimensional logic, for example take $N=M$, and I think: the cokernel of $A$ is like a "tiny" function space, compared with $X$ or $Y$, over a topological space $Z$ of dimension $<M$. And I'm "always" right, I can even guess properties of the function space $Z$, like $\text{dim}(Z)$, from topological properties of $X$ and $Y$. Then, what does it happen with $L^2(\mathbb{R}^n)$? Can we, under additional assumption, prove a kind of invariance of domain theorem for function spaces? It may help the reader to think about the operator $\iota:C_c(\mathbb{R})\to C_c({\mathbb{R}^2})$, given by $\iota f(x,y)\mapsto f(x)\varphi(y)$, where $\varphi$ is supported near to zero, and the quotient $C_c({\mathbb{R}^2})/\iota C_c(\mathbb{R})$; we can think that the dimension of the cokernel is "$\mathbb{R}^2-\mathbb{R}\approx \mathbb{R}^2$".
 A: The third question always has a positive answer. Let $E$ be a Banach space and let $B_{E^*$}$ be the unit ball of the dual space $E^*$ endowed with the weak-$\ast$ continuous topology. Then, any $x \in E$ has a representation as a continuous function $\hat{x} \in C(B_{E^*})$, given by $\hat{x}(\lambda) = \lambda(x)$. Recall that $B_X$ is compact by Banach-Alaouglu.
I do not known a definitive answer to the other questions (identify the closed subspaces of $L^p$). Using Rademacher techniques/Khintchine inequalities you can definitely embed $\ell^2$ inside $L^p$ for $p  < \infty$.
A: One known way of representing any real Banach space as a function space is as follows: Let $X$ be a real Banach space. Let $A(B_{X^{\ast}})$ denote the Banach space of all real-valued affine weak$^{\ast}$-continuous functions on the closed unit ball of $X^{\ast}$, denoted by $B_{X^{\ast}}$, where $B_{X^{\ast}}$ is endowed with the weak$^{\ast}$ topology. Now consider a subspace of $A(B_{X^{\ast}})$ namely $A_{\sigma}(B_{X^{\ast}}) = \{a \in A(B_{X^{\ast}}): a(x^{\ast}) = -a(-x^{\ast})\mbox{ for each }x^{\ast} \in B_{X^{\ast}}\}$. Then $X$ is isometric and linearly isomorphic to $A_{\sigma}(B_{X^{\ast}})$ under the mapping $x \mapsto \hat{x}$, where $\hat{x}(x^{\ast}) = x^{\ast}(x)$, for each $x^{\ast} \in B_{X^{\ast}}$ and $x \in X$.  The reference for this result is the book The Isometric Theory of Classical Banach spaces by H. E. Lacey. https://3lib.net/book/1304350/ffc878
