Let $f$ be a $C^\infty$ function on $(c,d)$ ,and let $O=\cup_{n\in \mathbb{Z}^+} (a_n,b_n)$ where $(a_n,b_n)$ are disjoint open interval in $(c,d)$ and $O$ is dense in $(c,d)$. Suppose for each $n\in \mathbb{Z}^+$ ,$f$ coincides with a polynomial on $(a_n,b_n)$. Is it necessary that $f$ coincide with a polynomial on $(c,d)$?

  • 1
    $\begingroup$ Note this question with a similar flavor: mathoverflow.net/questions/34059/… $\endgroup$ – Steven Landsburg Apr 14 '12 at 20:57
  • 3
    $\begingroup$ I suspect one can build a $C^\infty(\mathbb{R})$ function supported on the unit interval, which is polynomial on each component of the complement of the Cantor set. $\endgroup$ – Pietro Majer Apr 14 '12 at 21:20

I believe the answer is "no". The key lemma is:

Lemma. Let $f: [c,d] \to {\bf R}$ be smooth, let $I$ be a compact subinterval of $(c,d)$, $q$ be an interior point of $I$, let $n \geq 1$, and let $\varepsilon > 0$. Then there exists a smooth perturbation $g: [c,d] \to {\bf R}$ of $f$ which agrees with $f$ outside of $I$, is a polynomial on a neighbourhood of $q$ of degree at least $n$, but is not a polynomial on all of $I$, and differs from $f$ by at most $\varepsilon$ in $C^n[c,d]$ norm.

If we apply this lemma iteratively for $n=1,2,\ldots$ with $\varepsilon = \varepsilon_n := 2^{-n}$ and $n=0,1,2,\ldots$, starting with $f_0 = 0$, and setting $q = q_n$ to be the first rational (in some enumeration of the rationals) on which $f_n$ is not locally polynomial, one obtains a sequence $f_1, f_2, f_3,\ldots$ of smooth functions on $[c,d]$ which form a Cauchy sequence in $C^k$ for each $k$, and thus converge in the smooth topology to a limit $f$ which is equal to a polynomial on degree at least $n$ on an interval $I_n$, with the $I_n$ disjoint and covering all the rationals, thus dense in $[c,d]$, giving the claim.

To prove the lemma, recall from the Weierstrass approximation theorem that the polynomials are dense in $C^0[c,d]$; integrating this fact repeatedly we see that they are dense in $C^n[c,d]$ as well. So we can approximate $f$ to arbitrary accuracy by a polynomial $h$ in the $C^n$ norm; by a small perturbation one can ensure that $h$ has degree at least $n$. Now using a smooth partition of unity, one can create a merged function $g$ that equals $h$ near $q$ and equals $f$, which can be made arbitrarily close in $C^n$. By modifying $g$ a little bit in $I$ away from $q$ one can ensure that $q$ is not polynomial on all of $I$.

The problem here is superficially similar to that in the previous question If $f$ is infinitely differentiable then $f$ coincides with a polynomial , but the latter has qualitative control at every single point (allowing the powerful Baire category theorem to come into play), whereas here one only has qualitative control on a dense set, which is a far weaker statement, and one which allows for a great deal of flexibility.

| cite | improve this answer | |
  • $\begingroup$ In fact,i create this question while think about the question mathoverflow.net/questions/34059/…. I proved the function $f$ locally a polynomial and wondered if I can use that property to solve that problem. $\endgroup$ – Ben Apr 15 '12 at 5:21
  • $\begingroup$ Can you clarify your construction? What interval I do you use the lemma for? $\endgroup$ – Matthias Ludewig Apr 15 '12 at 21:23
  • $\begingroup$ At each stage, f_n is polynomial on some union of disjoint intervals. One picks a q = q_n outside of these intervals, then one can take I to be an interval around q which does not intersect any of the existing intervals. This way, f_{n+1} is polynomial on all the intervals that f_n was, plus an additional new interval containing q. $\endgroup$ – Terry Tao Apr 15 '12 at 21:43

No. Say that a function $f(x)$ is locally a polynomial at $x_0$ if there is some neighborhood of $x_0$ upon which $f(x)$ coincides with a polynomial. In

W. Donoghue, Jr., Functions which are polynomials on a dense set. J. London Math. Soc. 39 (1964), 533–536,

the author constructs an example of a $C^\infty$ function that is not a polynomial but is nonetheless locally a polynomial on a dense set. Since the set of points at which a function is locally a polynomial is open, hence a countable union of disjoint intervals, the answer to your question follows.

| cite | improve this answer | |

The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\infty$, non-analytic solution, of a simple first order linear delay differential equation.

Let's consider the problem

$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2} $$ where we seek for a function $f\in C^1(\mathbb{R})$ with support in $I:=[0,1]$, and with the condition $f(x)=f(1-x)$.

I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$, corresponding to an eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is very easy to check that $f$ is locally polynomial on the complement of the Cantor set; and it is, of course, $C^\infty$.

Indeed, since $f (3x)=0$ for $ x > \frac{1}{3}$, from the equation we have that $f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$, so that $f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is polynomial of degree $k-1$ on each component of the complement of the closed set $F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k \cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that $\{F_k\} _ {k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose intersection is the Cantor set.

$$*$$ Construction. Consider the linear operator $T$ on $L^\infty(0,1)$ (thought as subspace of $L^\infty(\mathbb{R})$ via extension by zero), defined by

$$Tu(x)= \cases{ \int_0^{3x}u(t)\, dt\qquad \qquad 0 < x < \frac{1}{3}\\\ 0 \qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\\ \int_0^{3(1-x)}u(t)\, dt \qquad\frac{2}{3}< x < 1.} $$ Since $T$ and $T ^ 2$ are integral operators with non-negative kernels, their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x) $, $\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$. Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is invertible. The function

$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _ {[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty \left( \frac{3}{2}\right) ^ k T ^ k \chi _ { [\frac{1}{3}, \frac{2}{3}] } $$


$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\; dt\qquad \mathrm{for\; all}\quad 0 < x < \frac{1}{3}\, ,$$
$$f(x)=1\qquad \mathrm{for\, all}\quad \frac{1}{3} < x < \frac{2}{3}\, ,$$

and $$f(x)=f(1-x)\qquad \mathrm{for\, all}\quad x\in\mathbb{R} \, .$$ The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has to be checked, are $\frac{1}{3}$ and $\frac{2}{3}$. We have

$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3} f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\, ,$$


$$ 2 \int _ 0 ^ \frac{1}{3} f(x)dx = 3\int _ 0 ^ \frac{1}{3} \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x f(t) dt dx=$$

$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt= \int _ 0 ^ 1 tf(t)dt = $$

$$ =\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^ 1 f(t)dt\, . $$


$$ \int _ 0 ^ 1f(t)dt = \frac{2}{3}\, ,$$ which implies the continuity of $f$ at $ \frac{1}{3}$ (and by symmetry at $ \frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \, . $$ As a consequence, $f$ actually satisfies $$f(x)=\frac{3}{2} \int_0^{3x}f(t)\, dt $$ for all $ x < \frac{2}{3}\, ,$ and we conclude that $f$ is in $ C ^ 1(\mathbb{R})$, and has the stated properties.


Here is a Pudding Function $f$, and its integral function, $F(x):=\int_0^x f(t)dt$. Note how the latter coincides with the first third of the former, up to rescaling. Simple considerations based on odd and even symmetry show that $F(1)=\frac{2}{3}$.

| cite | improve this answer | |

The following is wrong, but the comments are really nice :)

Suppose That on $(a_1, b_1)$, $f$ is a polynomial of degree $N$. This means that in its Taylor series at any point $x \in (a_1, b_1)$, every coefficiant past the $(N+1)$st vanishes.

This must also be true for the border points, $x = a_1$ and $x=b_1$. However, because the intervals are dense in $O$, $a_1$ or $b_1$ lies in the closure of some other interval. The rest is something like an induction: It follows that on every set $(a_n, b_n)$, the coefficients past the $(N+1)$st vanish.

This shows that $f$ is a polynomial, by the fundamental theorem of calculous.

| cite | improve this answer | |
  • 2
    $\begingroup$ It seems to me that there is a problem with the part where you say that $a_1$ and $b_1$ lie in the closure of some other interval. For example, say $b_1 = 0$, and $$ O \cap (0,1) = \bigcup_{k=1}^\infty \left(\frac1{k+1},\frac1{k}\right). $$ $\endgroup$ – Aaron Tikuisis Apr 14 '12 at 19:47
  • $\begingroup$ Exactly: the OP's partition of $(c,d)$ allows things as ugly as Aaron's partition, or even $$(0,1)=\bigcup_{k=1}^\infty \bigcup_{j=1}^\infty \left(\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j+1},\frac{1}{k+1}+\frac{1}{k(k+1)}\frac{1}{j}\right)$$ though I'm not sure the process can be continued indefinitely. It's an interesting question, then: how far can this idea be pushed? How many points can there be that do not have a "next neighbour to the right"? Can every interval $(a_n,b_n)$ be made to not have a next neighbour? $\endgroup$ – Emilio Pisanty Apr 14 '12 at 20:47
  • $\begingroup$ Follow up: This answer shows you can. $\endgroup$ – Emilio Pisanty Apr 15 '12 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.