When we talk about dense subalgebra of continuous function on a compact interval, we usually think of smooth functioon or polynomials. Can we find another dense subalgebra which does not contain any non-constant polynomials? I know Stone-Weierstrass theorem gives us the condition of subalgebra being dense iff it separate points, but I haven't seen any concrete examples.

  • 1
    $\begingroup$ Yes, I have clarified the question $\endgroup$
    – Ken.Wong
    Jun 29, 2021 at 17:56
  • $\begingroup$ And "dense" means dense in the sense of the product topology on $\mathbb R^{[a,b]}$? $\endgroup$
    – rschwieb
    Jun 29, 2021 at 18:01
  • 3
    $\begingroup$ Probably not what you are after, but you can use some homeomorphism $[a,b]\cong[0,1]$ to convert (nonconstant) polynomials on $[0,1]$ to nonpolynomials on $[a,b]$. $\endgroup$ Jun 29, 2021 at 18:16

3 Answers 3


The algebra of polynomials is just the algebra generated by the single function "$x$", i.e. the inclusion map from your interval into $\mathbb{R}$. So you can pick some other continuous function instead and take the subalgebra it generates. As long as your function is injective, this subalgebra separates points and so is dense by Stone-Weierstrass, and typically this subalgebra will not contain any nonconstant polynomials. For example, you could take the subalgebra generated by the exponential function, which concretely consists of functions of the form $$x\mapsto\sum_{k=0}^na_ke^{kx}$$ for some constants $a_k$.

A related example which actually is used a lot (in connection with Fourier analysis) is the algebra of trigonometric polynomials, i.e. the subalgebra generated by the sine and cosine functions (which will jointly separate points as long as your interval has length less than $2\pi$). If you consider complex-valued functions instead of real-valued functions, this can also be described as the subalgebra generated by $x\mapsto e^{ix}$ and $x\mapsto e^{-ix}$ (so, it is just like the exponential example above except that there is a factor of $i$ in the exponents and $k$ can range over negative integers too).

  • $\begingroup$ Thank you for the answer and detail explanation! $\endgroup$
    – Ken.Wong
    Jun 29, 2021 at 18:41

If $\phi$ is any homeomorphism of $[a,b]$ then the set of all functions $p\circ\phi$ for $p$ a polynomial is dense in $C[a,b]$. (Because if $f$ is continuous then so is $f\circ\phi^{-1}$, and if $p_n\to f\circ\phi^{-1}$ uniformly then $p_n\circ\phi\to f$ uniformly.)


You can find subalgebras that not only contain no polynomials, but in fact have (essentially) no functions differentiable (essentially) anywhere. I have no idea what it might be useful for, and proving this ended up being surprisingly tricky. However, this is part of a general phenomenon that there can be surprisingly rich algebraic structures in the parts of a function space that are usually discarded as "ill-behaved."

Brownian motion is a relatively wacky function. However, once you fix a time $t$, then the marginal for that time has a continuous distribution supported everywhere. In particular, for any fixed $t_1$ and $t_2$, the probability that $B(t_1)<0<B(t_2)$ is nonzero. So there exists a Brownian motion satisfying that condition. Since we can repeat the above argument for every possible pair $(t_1,t_2)$, there exists a set of Brownian motions that separates points.

Moreover, any nonconstant polynomial in these Brownian motions cannot be differentiable a.e.: recall that any function differentiable at a point $t$ has quadratic variation density $0$ around that point; see this question for a proof. Now let $\vec{B}$ be a vector of i.i.d. Brownian motions and nonconstant $p(\vec{x})\in\mathbb{R}[\vec{x}]$ (where $\vec{x}$ and $\vec{B}$ have the same dimension). Then Itô's formula tells us $$dp(\vec{B}(t))=(\nabla p)(\vec{B}(t))\cdot d\vec{B}(t)+\frac{1}{2}(\Delta p)(\vec{B}(t))\,dt$$ The second term vanishes in the quadratic variation, so $$\int_a^b{\mathcal{Q}[dp(\vec{B}(t))]}=\int_a^b{\|(\nabla p)(\vec{B}(t))\|_2^2\,dt}$$ By the linked post above, $p(\vec{B}(\cdot))$ is differentiable on at most $$\{t:\|(\nabla p)(\vec{B}(t))\|_2^2=0\}\tag{1}$$ (Technically, we've only shown that for $t\in[a,b]$. But now send $a\to0^+$ and $b\to1^-$.)

The set (1) has measure $$l=\int_0^1{1_{0=\|(\nabla p)(\vec{v})\|_2^2}(\vec{B}(t))\,dt}$$ where $1_S(x)=\begin{cases}1&x\in S\\0&x\notin S\end{cases}$. We can write this in terms of the pushforward measure $\lambda_1$ for $\vec{B}|_{[0,1]}$, known as the local time: $$l=\int_{\mathbb{R}^n}{1_{0=\|(\nabla p)(\vec{v})\|_2^2}(\vec{v})\,\lambda_1(d^n\vec{v})}$$ But as long as $p$ is nonconstant, the gradient-squared is not the zero polynomial, and thus has a zero set of positive codimension. But $\lambda_1$ is absolutely continuous with respect to $n$-dimensional Lebesgue measure, which considers low-dimensional sets to have measure zero. And so the integral vanishes.

The idea is to take this set of Brownian motions and use it to generate our subalgebra $\mathcal{A}$. The challenge is that I've tacitly used theorems that hold a.s., and we have uncountably many conditions. So we need to reduce things to countable size.

The first step is to ensure we don't need too many Brownian motions. Recall that Brownian motion is (a.s.) Hölder-$\left(\frac{1}{2}-\epsilon\right)$. I claim:

Lemma. Let $D$ be dense in $[0,1]$ and $\mathcal{A}$ an algebra such that, for all distinct $x,y\in D$, there exists $f\in\mathcal{A}$ that is Hölder-$\alpha$ ($\alpha<1$) but such that $$|f(x)-f(y)|>|x-y|\tag{1}$$ Then $\mathcal{A}$ separates points on $[0,1]$.
Proof. Fix $a\neq b\in[0,1]$. Then, for any $\epsilon$, there exists distinct $x,y\in D$ such that

  • $|x-a|<\epsilon|a-b|$,
  • $|y-b|<\epsilon|a-b|$, and
  • there exists $f$ satisfying (1) with Hölder constant $K$.

Thus \begin{align*} |f(a)-f(b)|&\geq|f(x)-f(y)|-|f(x)-f(a)|-|f(y)-f(b)| \\ &>|x-y|-2K(\epsilon|a-b|)^{\alpha} \\ &\geq(1-2\epsilon)|a-b|-2K(\epsilon|a-b|)^{\alpha} \\ &\to|a-b|\qquad\text{ (as }\epsilon\to0^+\text{)} \end{align*} Since $|a-b|>0$, for small enough $\epsilon$, the corresponding $f$ (and thus $\mathcal{A}$) separates $a$ and $b$. $\square$

Since we can take $D=\mathbb{Q}\cap[0,1]$, we only need countably-many Brownian motions to ensure $\mathcal{A}$ separates points.

The second step is to review our proof that nonconstant polynomials in Brownian motion are nondifferentiable a.e. There, the only probabilistic fact we used was that the local time $[0,1]$ for $n$ Brownian motions $\mathbb{R}$ is absolutely continuous with $n$-dimensional Lebesgue measure. So that claim does not even require countably-many conditions: it needs exactly one!

Finally, build the algebra inductively: enumerate $D^2$ as $\{(q_n,r_n)\}_{n=1}^{\infty}$ and let $A_0=\emptyset$. If $A_{n-1}$ separates $q_n$ and $r_n$, then $A_n=A_{n-1}$. Otherwise, consider a random Brownian motion $B$. Since the intersection of a probability-$1$ and a positive-probability event has positive probability, the probability that "(1) $B$ separates $q_n$ and $r_n$ and (2) the local time for an enumeration of $A_{n-1}\sqcup\{B\}$ is absolutely continuous" is positive. So there exists some $B$ in that event; we choose it and let $A_n=A_{n-1}\sqcup\{B\}$. Repeat for all $n\in\mathbb{Z}^+$ and let $A_{\infty}=\bigcup_n{A_n}$. Then $\mathcal{A}=\mathbb{R}[A_{\infty}]$. $\mathcal{A}$ is an algebra separating points in which every nonconstant function is differentiable on at most a set of measure zero. And yet Stone-Weierstrass tells us $\overline{\mathcal{A}}=C([0,1])$.


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