If $f$ is infinitely differentiable then $f$ coincides with a polynomial 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.
 A: 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, $f(x)\neq 0\;\forall x\in J$.
A: 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 $I$.
A: 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...).
A: Note that The Fabius function is nowhere analytic but admits a dense set of points where all but finitely many derivatives vanish.
A: 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?
A: 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 $[c,d]$.
A: 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:


*

*$P_{n+1}^c$ 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 $R$)

*$(a,b)$ 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. 
A: 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.
A: 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.


*

*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.

*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.

*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.

*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 $B \cap I \neq \emptyset$).

*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 $E_k \cap B \neq \emptyset$.

*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 $B$.

*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.
