Let $f$ be an infinitely differentiable function on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how.

I thought of using Weierstrass approximation theorem, but couldn't succeed.

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    $\begingroup$ This seems like a homework problem in a 1st year course on calculus. $\endgroup$ Jul 31, 2010 at 21:51
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    $\begingroup$ This is a jewel, I will try to recall the solution. $\endgroup$ Jul 31, 2010 at 22:05
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    $\begingroup$ @Ryn: no, this is a classic little problem. @Michael: the problem is correct as stated. $\endgroup$ Jul 31, 2010 at 22:39
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    $\begingroup$ This is basically a double-starred exercise in the book "Linear Analysis" by Bela Bollobas (second edition), and presumably uses the Baire Category Theorem. Since it is double-starred, it is probably very hard!! Solutions are not given, and even single starred questions in that book can be close to research level. However, the version in that book has $f$ on the whole real line, and $f^{(m)}(x) = 0$ for ALL $m>n$. So are you sure your question is correct, since it's assuming a lot less but coming to roughly the same conclusion? $\endgroup$
    – Zen Harper
    Jul 31, 2010 at 23:32
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    $\begingroup$ I agree with Andrew L.'s opinion(but not the more extreme part of it). If such hard questions are given as homework for a first year calculus course, then there will be complaints about the instructor, and indeed about the department. It is my modest contention that anyone who criticizes a question as homework should be able to substantiate it by giving a short solution in the comments. This doesn't take much effort. What I am preaching is just a variant of "All right, but let the one who has never sinned throw the first stone!". Before closing a question as homework, first solve it. $\endgroup$
    – Anweshi
    Aug 1, 2010 at 13:16

10 Answers 10


The proof is by contradiction. Assume $f$ is not a polynomial.

Consider the following closed sets: $$ S_n = \{x: f^{(n)}(x) = 0\} $$ and $$ X = \{x: \forall (a,b)\ni x: f\restriction_{(a,b)}\text{ is not a polynomial} \}. $$

It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and $$ (a,b)\cap X\subset S_n $$ for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S_m$ for all $m\ge n$ and $x\in (a,b)\cap X$.

Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d< n$. (Since either $c$ or $e$ is in $X$.)

So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.

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    $\begingroup$ Thank you! Filling in all the details to this outline is a fantastic exercise in basic real analysis and topology. It strikes me as a great "capstone" to a relevant course. It went through at least 20 relevant topics/ideas: (in roughly decreasing order of complexity) Baire Category Theorem, Heine-Borel, infs/sups (so LUB property of R), compactness, Cauchy/convergent sequences/completeness, (infinite) differentiability, continuity, connectedness, perfect sets, limit points (from the sides), induction, isolated points, open/closed sets, interiors, derivatives of polynomials, and boundedness. $\endgroup$ May 8, 2011 at 22:48
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    $\begingroup$ Is a similar statement true for functions of n>1 variables? $\endgroup$
    – Alex W
    Jan 14, 2019 at 17:40
  • $\begingroup$ @AlexW In that case, the natural hypothesis would be that for each $x_0 \in \mathbb{R}^n$ there exists a multi-index $\alpha=\alpha(x_0)$ such that $\left. \frac{\partial^{|\alpha|} f}{\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \cdots \partial x_n^{\alpha_n}}\right\vert_{x_0}=0$. Essentially the same argument works, by appropriately replacing intervals $(a,b)$ as in this answer with open balls. In particular there are only countably many multi-indices so Baire's theorem applies. $\endgroup$ May 1, 2020 at 16:02
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    $\begingroup$ @MathematicsStudent1122 Let $f(x_1,x_2)=e^{x_2}$, $\alpha=(1,0)$. Then $f^{(\alpha)}(x)=0$ for all $x\in\mathbb{R}^2$ but $f$ is not a polynomial. $\endgroup$
    – Alex W
    May 2, 2020 at 12:04
  • $\begingroup$ @AlexW You're right, thank you! $\endgroup$ May 2, 2020 at 16:44

Note that The Fabius function is nowhere analytic but admits a dense set of points where all but finitely many derivatives vanish.


The theorem:

Theorem: Let $f(x)$ be $C^\infty$ on $(c,d)$ such that for every point $x$ in the interval there exists an integer $N_x$ for which $f^{(N_x)}(x)=0$; then $f(x)$ is a polynomial.

is due to two Catalan mathematicians:

F. Sunyer i Balaguer, E. Corominas, Sur des conditions pour qu'une fonction infiniment dérivable soit un polynôme. Comptes Rendues Acad. Sci. Paris, 238 (1954), 558-559.

F. Sunyer i Balaguer, E. Corominas, Condiciones para que una función infinitamente derivable sea un polinomio. Rev. Mat. Hispano Americana, (4), 14 (1954).

The proof can also be found in the book (p. 53):

W. F. Donoghue, Distributions and Fourier Transforms, Academic Press, New York, 1969.

I will never forget it because in an "Exercise" of the "Opposition" to became "Full Professor" I was posed the following problem:

What are the real functions indefinitely differentiable on an interval such that a derivative vanish at each point?


Let me add one more solution. It is not really different from the accepted one, but it includes all details. The problem is that a student without sufficient experience will not even see necessity to fill details.

Theorem. If $f\in C^\infty(\mathbb{R})$ and for every $x\in\mathbb{R}$ there is a nonnegative integer $n$ such that $f^{(n)}(x)=0$, then $f$ is a polynomial.

The following exercise shows that the result cannot be to easy.

Exercise. Prove that there is a function $f\in C^{1000}(\mathbb{R})$ which is not a polynomial, but has the property described in the above theorem.

Proof of the theorem. Let $\Omega\subset\mathbb{R}$ be the union of all open intervals $(a,b)\subset\mathbb{R}$ such that $f|_{(a,b)}$ is a polynomial. The set $\Omega$ is open, so $$ \Omega=\bigcup_{i=1}^N (a_i, b_i)\, , \qquad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ where $a_i<b_i$ and $(a_i, b_i)\cap (a_j, b_j) = \emptyset$ for $i\neq j$, $1\leq N\leq\infty$. Observe that $f|_{(a_i, b_i)}$ is a polynomial (Why?)*. We want to prove that $\Omega=\mathbb{R}$. First we will prove that $\overline{\Omega}=\mathbb{R}$. To this end it suffices to prove that for any interval $[a,b]$, $a<b$ we have $[a,b]\cap\Omega\neq\emptyset$. Let $$ E_n=\{x\in\mathbb{R}:\, f^{(n)}(x)=0\}\, . $$ The sets $E_n\cap [a,b]$ are closed and $$ [a,b]=\bigcup_{n=0}^\infty E_n\cap [a,b]\, . $$ Since $[a,b]$ is complete, it follows from the Baire theorem that for some $n$ the set $E_n\cap [a,b]$ has nonempty interior (in the topology of $[a,b]$), so there is $(c,d)\subset E_n\cap [a,b]$ such that $f^{(n)}=0$ on $(c,d)$. Accordingly $f$ is a polynomial on $(c,d)$ and hence $$ (c,d)\subset\Omega\cap [a,b]\neq\emptyset. $$ The set $X=\mathbb{R}\setminus\Omega$ is closed and hence complete. It remains to prove that $X=\emptyset$. Suppose not. Observe that every point $x\in X$ is an accumulation point of the set, i.e. there is a sequence $x_i\in X$, $x_i\neq x$, $x_i\to x$. Indeed, otherwise $x$ would be an isolated point, i.e. there would be two intervals $$ (a,x),\, (x,b)\subset\Omega,\ x\not\in\Omega\, . \qquad \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ The function $f$ restricted to each of the two intervals is a polynomial, say of degrees $n_1$ and $n_2$. If $n>\max\{ n_1, n_2\}$, then $f^{(n)}=0$ on $(a,x)\cup (x,b)$. Since $f^{(n)}$ is continuous on $(a,b)$, it must be zero on the entire interval and hence $f$ is a polynomal of degree $\leq n-1$ on $(a,b)$, so $(a,b)\subset\Omega$ which contradicts (2).

The space $X=\mathbb{R}\setminus\Omega$ is complete. Since $$ X=\bigcup_{n=1}^\infty X\cap E_n\, , $$ the second application of the Baire theorem gives that $X\cap E_n$ has a nonempty interior in the topology of $X$, i.e. there is an interval $(a,b)$ such that $$ X\cap (a,b)\subset X\cap E_n\neq\emptyset\, . \qquad \ \ \ \ \ \ \ \ \ \ \ \ \ (3) $$ Accordingly $f^{(n)}(x)=0$ for all $x\in X\cap (a,b)$. Since for every $x\in X\cap (a,b)$ there is a sequence $x_i\to x$, $x_i\neq x$ such that $f^{(n)}(x_i)=0$ it follows from the definition of the derivative that $f^{(n+1)}(x)=0$ for every $x\in X\cap (a,b)$, and by induction $f^{(m)}(x)=0$ for all $m\geq n$ and all $x\in X\cap (a,b)$.

We will prove that $f^{(n)}=0$ on $(a,b)$. This will imply that $(a,b)\subset\Omega$ which is a contradiction with (3). Since $f^{(n)}=0$ on $X\cap (a,b)=(a,b)\setminus\Omega$ it remains to prove that $f^{(n)}=0$ on $(a,b)\cap\Omega$. To this end it suffices to prove that for any interval $(a_i, b_i)$ that appears in (1) such that $(a_i, b_i)\cap (a,b)\neq\emptyset$, $f^{(n)}=0$ on $(a_i, b_i)$. Since $(a,b)$ is not contained in $(a_i, b_i)$ one of the endpoints belongs to $(a,b)$, say $a_i\in (a,b)$. Clearly $a_i\in X\cap (a,b)$ and hence $f^{(m)}(a_i)=0$ for all $m\geq n$. If $f$ is a polynomial of degree $k$ on $(a_i, b_i)$, then $f^{(k)}$ is a nonzero constant on $(a_i, b_i)$, so $f^{(k)}(a_i)\neq 0$ by continuity of the derivative. Thus $k<n$ and hence $f^{(n)}=0$ on $(a_i,b_i)$. $\Box$

Exercise. As the previous exercise shows the theorem is not true if we only assume that $f\in C^{1000}$. Where did we use in the proof the assumption $f\in C^\infty(\mathbb{R})$?

*It suffices to prove that $f$ is a polynomial on every compact subinterval $[c,d]\subset (a_i, b_i)$. This subinterval has a finite covering by open intervals on which $f$ is a polynomial. Taking an integer $n$ larger than the maximum of the degrees of these polynomials, we see that $f^{(n)}=0$ on $[c,d]$ and hence $f$ is a polynomial of degree $<n$ on $[c,d]$.

  • $\begingroup$ I like this answer a lot, but I am confused about the sentence "The set $\Omega$ is open so $\Omega = \bigcup_{i=1}^\infty (a_i, b_i)$ where $a_i < b_i$ and $(a_i, b_i) \cap (a_j, b_j) = \emptyset$ for $i \neq j$}. Is the last part (empty intersections of intervals) correct? It seems a bit at odds with the claim further down that $\Omega = \mathbb{R}$. $\endgroup$
    – Vincent
    May 31, 2019 at 8:37
  • $\begingroup$ @Vincent I edited my proof, see formula (1). Is it okay now? $\endgroup$ Jun 1, 2019 at 12:40
  • $\begingroup$ Huh no, now it is even more confusing. I understand that open sets are a union of intervals, but not that they are a union of disjoint intervals. Take the case $\Omega = \mathbb{R}$, I don't see any way of writing this as a union of non-overlapping intervals (let alone a finite number of such intervals as the new formula (1) suggests). But on closer inspection I think the condition $(a_i, b_i) \cap (a_j, b_j) = \emptyset$ for $i \neq j$ which is causing my concern is not being used further down the proof, is it? $\endgroup$
    – Vincent
    Jun 2, 2019 at 17:42
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    $\begingroup$ @Vincent $\mathbb{R}=(-\infty,\infty)$. I never said that the intervals are finite. $\endgroup$ Jun 3, 2019 at 0:50
  • $\begingroup$ If we assume $f\in C^{1000}$, then the sets $E_n$ for $n>1000$ are not guaranteed to be closed. For the example, consider $f(x)=0$ if $x<0$, $f(x) = x^{1001}$ if $x\geq 0$. $\endgroup$ Mar 5, 2021 at 21:18

For what it's worth, I post my solution. I assume $f \colon \mathbb{R} \to \mathbb{R}$, which makes no difference but lets me use one less symbol.

  1. Let $A_n = \{ x \in R \mid f^{(n)}(x) = 0 \}$, $E_n$ the interior of $A_n$. Clearly $E_n \subset E_m$ for $n < m$, and by Baire $E_n$ is eventually not empty.

  2. Each $E_n$ is a countable union of open segments. It is easy to see that in passing from $E_n$ to $E_{n+1}$ new segments can appear, but those already in $E_n$ remain unchanged. Moreover two such segments are never adiacent.

  3. By this remark is it enough to prove that $\bigcup E_n = \mathbb{R}$. Indeed if this holds and $E_n \neq \emptyset$, then $E_n = \mathbb{R}$, which implies the thesis. Otherwise the points in the boundary of $E_n$ don't appear in the union.

  4. Let $E = \bigcup E_n$, $B$ its complementary set, and assume by contradiction $B \neq \emptyset$. $B$ is itself a complete metric space, hence can apply Baire to it. So for some $k$ we find that $A_k \cap B$ has non-empty interior in $B$. This means that there is an interval $I$ such that $B \cap I \subset A_k$ (and $B \cap I \neq \emptyset$).

  5. From remark 2, $B$ has no isolated points. The contradiction that we want to find is that $I \setminus B \subset A_k$. Indeed from this it follows that $I \subset A_k$, hence $E_k \cap B \neq \emptyset$.

  6. By construction $I \setminus B$ is a union of intervals which appear in some $E_n$. Take such an interval $J$, say $J \subset E_N$ (where $N$ is minimal), and let $x$ be one end point of $J$ (which is not on the boundary of $I$). Then $x \in I \cap B \subset A_k$, so $f^{(k)}(x) = 0$. Moreover $x$ is not isolated in $B$, so it is the limit of a sequence $x_i$ of points in $B$.

  7. By the same argument $f^{(k)}(x_i) = 0$. Between two point where the $k$-th derivative vanish lies a point where the $k+1$-th does, so by continuity we find $f^{(k+1)}(x) = 0$. Similarly we find $f^{(m)}(x) = 0$ for all $m \geq k$. On $J$ $f$ is a polynomial of degree $N$; it follows that $N \leq k$, and we conclude that $J \subset E_k$. Since $J$ was arbitrary we conclude that $I \setminus B \subset E_k$, which we have shown to be a contradiction.

  • $\begingroup$ Hi-- Thanks a lot. Now, does this remain true if we replace $[0,1]$ by $\mathbb{R}$ or $[a,b]$ $\endgroup$
    – C.S.
    Aug 1, 2010 at 13:21
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    $\begingroup$ Yes, of course. The proof is the same. $\endgroup$ Aug 1, 2010 at 16:21
  • $\begingroup$ In step $3$, what about functions of the form $e^{-1/x}$. They can have a derivative $0$ on an interval and all future ones zero on the boundary. $\endgroup$
    – Will Sawin
    Nov 1, 2011 at 5:38
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    $\begingroup$ Those functions have all derivatives 0 in a point, not on a whole interval $\endgroup$ Nov 3, 2011 at 18:54
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    $\begingroup$ E_n is the interior of A_n. For a point in E_n you have a whole interval where the nth derivative vanishes identically, hence all subsequent derivatives vanish $\endgroup$ Sep 4, 2016 at 10:09

Maybe unuseful, but it remains true if you consider $f\in C^\infty(\mathbb R,\mathbb R)$.

Try showing that

Lemma. Let $I\subseteq \mathbb R$ be a nonempty interval and $f\in C^{\infty}(I)$. If $f$ is not a polynomial on $I$, then there exists a compact subset $J\Subset I$ in which $f$ is not a polynomial. Moreover, $f(x)\neq 0\;\forall x\in J$.


In Andrey Gogolev's answer the following two assertions appear:

"It is clear that $X$ is a non-empty . . . set" and "Now consider any maximal interval $(c,e) \subset ((a,b) - X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$."

These are true, but perhaps not transparently obvious. In attempting to fill the gaps, I developed a variation of the proof which requires neither the observation that $X$ has no isolated points nor any argument about degrees of polynomials. Here is my adaptation, borrowed freely from Gogolev:

I use the symbol "$\bot$" for "contradiction."

Define $I = [0,1]$ and $X = \{x \in I: \forall (a,b) \ni x: f|_{(a,b) \cap I} \; is \; not \; a \; polynomial\}$ .

We first establish the following:

Lemma: Suppose $[c,d] \subset I$ is an interval on which $f$ coincides with a polynomial $p$. Then there exists a maximal subinterval $[cm,dm]$ having the properties $[c,d] \subset [cm,dm] \subset I$ and $f = p$ on $[cm,dm]$. Furthermore, $cm \in X \cup \{0\}$ and $dm \in X \cup \{1\}$.

Proof: Let $cm$ = LUB $\{x: f(x) \neq p(x)\} \cup \{0\}$ and $dm$ = GLB $\{x: f(x) \neq p(x)\} \cup \{1\}$. It is clear that $[cm,dm]$ is maximal. Supppose that $cm \not \in X$ and $cm \neq 0$. Then we can find another interval $(u,v)$ with $cm \in (u,v) \subset I$ on which $f$ coincides with a polynomial $q$. But on $[cm,v]$ we have $f = p = q$, whence $f = p$ on $[u,dm]$. Since $u < cm$, we see that $[cm,dm]$ is not maximal ($\bot$). Therefore, $cm \in X$ or $cm = 0$. Likewise, $dm \in X$ or $dm = 1$.

Now we begin the proof-by-$\bot$ of the main result. Suppose that $f$ is not a polynomial on $I$.

If $X = \emptyset$, we begin with any $[c,d]$, and the lemma tells us that $cm = 0$ and $dm = 1$, so $f$ is a polynomial on $I$ ($\bot$). Thus, $X \neq \emptyset$. Now define $S_n = \{x: f^{(n)}(x) = 0\}$. $X$ and $S_n$ are clearly closed. Applying the Baire category theorem to the covering $\{X \cap S_n\}$ of the complete metric space $X$, we get that there exists an interval $(a,b)$ such that $(a,b) \cap X \neq \emptyset$ and $(a,b) \cap X \subset S_n$ for some $n$. (It is important here that $S_n$ is closed.)

Put $J = (a,b) \cap I$, and let $a1$ and $b1$ be the left and right end-points of $J$. (Observe that it is possible that $a1 = 0$ or $b1 = 1$, so J may not be open.) If $J \subset S_n$, then $f$ is a polynomial on $J$, whence $(a,b) \cap X = (a,b) \cap I \cap X = J \cap X = \emptyset$ ($\bot$). Thus, we can choose a point $t \in J - S_n$. Now $t \not \in X$, since $(a,b) \cap X \subset S_n$. Therefore, we can find an interval $(c,d) \ni t$ such that $f$ coincides with a polynomial $p$ on $(c,d) \cap I$. Furthermore, $f = p$ on the closure of $(c,d) \cap I$, which is an interval of the form $[c1,d1] \subset I$. Apply the lemma to $[c1,d1]$ to obtain a maximal interval $[cm,dm]$ having the stated properties. Since $t \not \in S_n$ and considering $p$, we see that $cm \not \in S_n$. Suppose $cm > a1$. Then we have $a \le a1 < cm \le c1 \le t < b$, so $cm \in (a,b)$. From the lemma, $cm \in X$, since $cm > a1 \ge 0$. Thus, $cm \in (a,b) \cap X \subset S_n$ ($\bot$). Therefore, $cm \le a1$. Likewise, $dm \ge b1$. Thus, $f$ is a polynomial on $J \subset [a1,b1] \subset [cm,dm]$, whence, as above, $(a,b) \cap X = \emptyset$ ($\bot$). We are at last forced to conclude that $f$ must indeed be a polynomial on $I$.


You can also find a solution of this gem p.65, in "A primer of real functions", third edition, by R.P. Boas, Jr (which is a very nice little book...).

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    $\begingroup$ PP 58-59 of the 1st edition. $\endgroup$ Feb 24, 2014 at 22:00
  • $\begingroup$ @smuaug: Thanks for the info. I shall check it ASAP :) $\endgroup$
    – C.S.
    Feb 26, 2014 at 7:56

I remember solving this in one whole week, but after a while, forgot how I did. I actually tried to remember but couldn't, so I tried this again. Spending five days, I got the solution. Compared to other solutions posted here, mine is more brute force approach. Mine has the same line of argument with the proof of Baire Category Theorem.

Problem: $f\in C^{\infty}(\mathbb{R})$, for all $x\in \mathbb{R}$, there exists $n_x\in \mathbb{N}$ such that $f^{(n_x)}(x)=0$. Show that $f$ is a polynomial.

My solution: Suppose $f$ is not a polynomial.

Let $A_n = \{x\in\mathbb{R}|f^{(n)}(x) = 0\}$. Each $A_n$ is a closed set, so it can be decomposed as $A_n=P_n\cup C_n$, where $P_n$ is perfect set, $C_n$ is at most countable. Note that $\cup_n A_n = \mathbb{R}$, and $P_n\subset P_{n+1}$ for all $n$. We derive a contradiction by showing that $\cap_n P_n^c$ is uncountable. (This is a contradiction since $\cap_n P_n^c\subset \cup_n C_n$).

Let $(a,b)$ be any maximal interval of a $P_n^c$(which exists since we assumed $f$ is not a polynomial). Then $P_{n+1}$ cannot contain intervals $(a,s)$ or $(t,b)$, otherwise, $f^{(n)}$ be constant on those intervals, and the constant should be zero, which contradicts maximality of $(a,b)$.

Thus, we have either one of two cases:

  1. $P_{n+1}^c$ has at least two maximal intervals inside $(a,b)$. Call one of them by $L$, and one of the others by $R$. (let all members of $L$ be less than any members of $R$)

  2. $(a,b)$ remains a maximal interval of $P_{n+1}^c$.

Let $I_{n+1}$ be 'either $L$ or $R$' in Case 1, '$(a,b)$' in Case 2.

We continue finding maximal interval $I_{m+1}$ of $P_{m+1}^c$ inside $I_m$ where $m\geq n$.

Considering choices of $I_m$ for $m\geq n$, and taking intersections $\cap_{m\geq n} I_m$, we can generate uncountably many members of $\cap_n P_n^c$.

Remark:: If Case 1 occurs infinitely many times, consider $LR$ sequences that both have $L$ and $R$ infinitely many times.

If Case 1 only occurs finitely many, then the interval sequence $I_m$ is stationary.


Some have mentioned the Fabius function as one close to being a counterexample. I think the following example might be of interest. The function $f$ is almost everywhere locally a polynomial but not one itself. It is constructed similarly to the Fabius function. Define the sequence of real-valued functions of domain $[0,1]$, $(f_n)_0^\infty$ as follows. \begin{align} f_0(x)&=1\\ f_{n+1}(x) &= \frac92\int_0^x u_n(t) \, dt\\ \text{where } u_n(x) &= \begin{cases} f_n(3x) \text{ if } 0\leq x<\frac{1}{3}\\ 0\text{ if } \frac{1}{3}\leq x <\frac{2}{3}\\ -f_n(3x-2) \text{ if } \frac{2}{3}\leq x \leq 1 \end{cases} \end{align} Then $(f_n)$ converges uniformly to a function $f:[0,1]\to\mathbb{R}$. Some of its properties are the following.

  • $f$ is infinitely differentiable.
  • For every $x\in [0,1]\backslash\mathcal{C}$ (where $\mathcal{C}$ denotes the Cantor set) there exists a neighborhood $(a,b)\ni x$ on which $f$ coincides with a polynomial. In fact said neighborhood is widest when $a = \max(y\in\mathcal{C}:y<x)$ and $b = \min(y\in\mathcal{C}:y>x)$. Nicely, that polynomial gets matched by all $f_n$ on $(a,b)$ when $n\geq N$, for some $N$ (depending on $x$).
  • If $x\in\mathcal{C}$ and $x$ is a unilateral limit point of $\mathcal{C}$ then by continuity the derivatives of $f$ at $x$ will eventually vanish. If $x$ is a bilateral limit point of $\mathcal{C}$ then the derivatives of $f$ at $x$ are all non-zero.

So a close call. The function looks like this.


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