Let $k$ be a commutative ring and $A$ a commutative $k$-algebra. Call $D(A) := \mathrm{Hom}_k(A,k)$ the dual of $A$ as a $k$-module, and $DD(A) := \mathrm{Hom}_k(D(A),k)$ the dual of the latter. Let $\Phi\colon A\to DD(A)$ be the canonical map $a \mapsto (u\mapsto u(a))$.

Define a multiplication on $DD(A)$ as follows: if $\xi,\eta \in DD(A)$, define $\xi\bullet\eta$ to be $D(A) \ni u \mapsto \eta(y\mapsto \xi(x\mapsto u(xy))) \in k$. This is clearly $k$-bilinear, and furthermore $\Phi(a) \bullet \eta = \eta \bullet \Phi(a)$ is $u \mapsto \eta(y \mapsto u(ay))$ (for $a \in A$ and $\eta \in DD(A)$); in particular, $\Phi(a)\bullet\Phi(b) = \Phi(ab)$. Clearly this is "the correct" multiplication on $DD(A)$.

I'm sure the following will come as a surprise to others as it did to me: this product is not necessarily commutative. For a counterexample, consider $A = k[t]$ the ring of polynomials over a finite field $k$: then $D(A) = k^{\mathbb{N}}$ as a $k$-vector space, and $DD(A)$ contains at least the elements $\Lambda_{\mathscr{F}}\colon u \mapsto \lim_{\mathscr{F}} u$ where $\mathscr{F}$ is an ultrafilter on $\mathbb{N}$ and their linear combinations (apparently these don't exhaust $DD(A)$: see here; but this doesn't matter); now one can easily check that if $\mathscr{F}$ and $\mathscr{G}$ are ultrafilters on $\mathbb{N}$ then $\Lambda_{\mathscr{F}} \bullet \Lambda_{\mathscr{G}} = \Lambda_{\mathscr{F}+\mathscr{G}}$ where $\mathscr{F}+\mathscr{G} = \{U \subseteq \mathbb{N} : \{j \in \mathbb{N} : U-j \in \mathscr{F}\} \in \mathscr{G}\}$ is the standard addition on $\beta\mathbb{N}$ defined here (§3.2 "Addition on the Stone–Čech compactification of the naturals") and which is not commutative (see, e.g., Hindman & Strauss, Algebra in the Stone-Čech Compactification (1998), §4.2).

So here's my question: What nice conditions on the $k$-algebra $A$ guarantee that $DD(A)$ is commutative? I'm pretty sure that $A$ being finite (i.e., of finite type as a $k$-module) is sufficient, but even this I don't have an appropriate reference for (e.g.: in Vasconcelos, Arithmetic of Blowup Algebras (1994), prop. 1.1.15, the author does not bother to define the multiplication on $DD(A)$).

Contrariwise, does someone have a counterexample to $DD(A)$ being commutative that does not require ultrafilters or some use of the axiom of choice?

Edit: I believe the following gives a positive answer ($DD(A)$ is commutative) when $k$ is a noetherian integral domain and $A$ is a finite $k$-algebra. Indeed, when $k$ is a noetherian integral domain with fraction field $F$, if $M$ is a $k$-module of finite type, then we can write a presentation $k^s \to k^r \to M \to 0$ (with $r,s$ natural numbers), and by comparing the obvious $0 \to D_k(M) \otimes_k F \to F^r \to F^s$ and $0 \to D_F(M \otimes_k F) \to F^r \to F^s$ (where $D_k(M) := \mathrm{Hom}_k(M,k)$ as a $k$-module), we see that the natural map $D_k(M) \otimes_k F \to D_F(M \otimes_k F)$ is an isomorphism — and also, $D_k(M)$ is a $k$-submodule of this. Dualizing twice (and using the fact that $D_k(M)$ is a $k$-module of finite type, being a submodule of $k^r$), we see that $D_k D_k(M) \otimes_k F$ is isomorphic to $D_F D_F(M\otimes_k F) = M\otimes_k F$ (finite dimensional vector space over a field!), and $D_k D_k(M)$ is a $k$-submodule of it. Now if $M = A$ is a finite $k$-algebra, one can check that the multiplication on $D_k D_k (A)$ is indeed the one obtained by restricting the multiplication on $D_F D_F(A\otimes_k F) = A\otimes_k F$ to it: but this multiplication is commutative. So $D_k D_k (A)$ is a commutative $k$-algebra (indeed, a subalgebra of $A\otimes_k F$).

This is the case I was interested in, but I'm leaving the question open, since maybe someone can say something more general or more interesting about the question.