What are some "useful" sets of functions (rings under pointwise multiplication) $\mathbb{R} \rightarrow \mathbb{R}$ that are unique factorization domains, other than the polynomial ring?
I will quantify usefulness in my context: I'm only interested in functions that form an algebra over $\mathbb{R}$ (or $\mathbb{C}$) in the space of continuous functions, and preferably also forms a dense algebra in $C([a,b])$ with respect to the usual supremum norm.
This is a very nice question.
I'll boldly assert that outside of algebraic geometry there are probably no examples of spaces of functions that are unique factorization domains!
The reason is that most spaces of functions, be they continuous, differentiable or smooth (=$\mathcal C^\infty$) do not even constitute a domain.
Holomorphic functions on a connected open subset $U\subset \mathbb C$ (or on a connected holomorphic manifold of arbitrary dimension) do constitute a domain $\mathcal O(U)$ , but (unless I'm mistaken) are never unique factorization domains.
This is easy to see for the entire functions $\mathcal O(\mathbb C)$ : irreducible entire functions are of the form $az+b$ and $\sin(z)$ is certainly not a finite product of such affine functions.
In algebraic geometry however there do exist affine varieties $X$ for which the regular functions $\mathcal O(X)$ form a unique factorization domain : for example $\mathbb C$ or $\mathbb C^*$.
And affine varieties for which this is not the case, like the quadric in $\mathbb C^3$ given by the equation $z^2-xy=0$.
Deciding which is which for an affine variety is a very delicate problem.
And to finish on a slightly humorous note, let me remark that the domain of trigonometric polynomials $\mathbb R[\cos(\theta), \sin(\theta)]\subset \mathcal C(\mathbb R)$ is not a unique factorization domain because of the reasonably well known formula $$\cos\theta \cdot \cos\theta=(1-\sin \theta)\cdot (1+\sin \theta) $$
According to this answer, the spaces $H(S^1)$ and $H(\overline{\mathbb{D}})$ are both UFDs, where $H(K)$ is the set of functions that extend to holomorphic functions defined over an open neighborhood of $K$, $\mathbb{D}$ is the closed unit disk and $S^1$ is its boundary.
I'd like to add that, if I got it correctly, in constract with $H(\overline{\mathbb{D}})$, the disk algebra $A(\mathbb{D}):=H(\mathbb{D})\cap C(\overline{\mathbb{D}})$, consisting the holomorphic functions over $\mathbb{D}$ that extend continuously to $\overline{\mathbb{D}}$, is not a UFD: let $f_0$ be an infinite Blaschke product with zeros $\{a_n\}^\infty_{n=1}$, where $a_n\to 1$, then $f_0$ is bounded over $\overline{\mathbb{D}}$. According to this question, $f_0$ is analytic (hence continuous) over $\overline{\mathbb{D}}\setminus\{1\}$, so $f:=(1-z)f_0$ is continuous over $\overline{\mathbb{D}}$, which means that $f\in A(\mathbb{D})$. But $f$ has infinitely many zeros in $\mathbb{D}$.
