Question

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.

Answer 1

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.

Answer 2

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, $f(x)\neq 0\;\forall x\in J$.

Answer 3

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.

Answer 4

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

Answer 5

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 $I$.

Answer 6

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?

Answer 7

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.