If $f$ is infinitely differentiable then $f$ coincides with a polynomial

148

149

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.

crskhr

Posted 2010-07-31T21:37:22.197

Reputation: 1 992

4

Also asked here: http://mathoverflow.net/questions/64246/

– S. Carnahan – 2011-05-08T01:58:18.690

in cooking up my own solution to this classic problem which I've already encountered elsewhere, I came to wonder wether there are any closed, denumerable sets of reals that have no isolated points... Anyone know the answer to this? – Olivier Bégassat – 2011-05-08T04:56:24.140

nevermind, I think I have one. – Olivier Bégassat – 2011-05-08T04:59:25.990

2

Olivier: you are referring to the notion of "perfect set", and there are no nonempty countable perfect sets. The argument is essentially a Baire category argument. See http://pirate.shu.edu/~wachsmut/ira/topo/proofs/pfctuncb.html

– Todd Trimble – 2011-06-12T14:51:33.430

I don't understand as to why this question was voted $\text{off-topic}.$ – crskhr – 2011-07-31T07:08:48.543

4This is stated as an exercise in Rudin's "Principles", and yes, it gets assigned as freshman's homework (at least where I come from). Originally it was formulated and solved in 1930's by Banach, if I remember correctly. I will try to dig up a more precise reference. – Margaret Friedland – 2011-11-01T13:59:42.967

@Margaret: Thanks, I would like to know the history of this problem as well. I will be thankful if you could provide me with the history. – crskhr – 2011-11-01T17:03:16.773

Chandrasekhar: why have you now changed the question to read $f^n$ instead of $f^{(n)}$? These mean different things – Yemon Choi – 2011-11-01T20:08:12.500

@yemon: OK, i fixed it back. I thought both mean the same. – crskhr – 2011-11-02T07:58:22.107

Margaret, I looked through Rudin's Principles of Mathematical Analysis (3rd edition) but didn't find this exercise. What page is it on? Also, where do you come from (that this exercise is assigned to freshmen)? – Todd Trimble – 2012-03-28T17:55:48.287

1My memory failed me on this problem. The only thing I am now sure about is that it was assigned as a freshman homework when I was a student at Jagiellonian U. in Krakow, Poland, a little more than 20 years ago (no, I did not solve it then). Rudin was used as a supplementary text, but I must have seen it printed somewhere else (or was it added in the Polish translation?). Proving various properties of function spaces via Baire category was a Polish specialty in 1930's, but, as indicated by @juan, this particular theorem apparently has different origin. Sorry for the confusion. – Margaret Friedland – 2012-07-05T16:55:10.953

@TheMathemagician It's not too far away from a homework for first year calculus or analysis, which doesn't mean that it's easy, but only that it could be written in elementary language within a considerable length. In my university, the fact is the derivative of a differentiable function is continuous at a point, is left as a homework exercise, which is also, in fact, a consequence of Baire's category theorem. – Frank Science – 2016-02-19T09:28:15.123

@FrankScience, I suppose that one can derive any true statement from anything, but is there a sense other than this formal one in which the continuity of a differentiable function is a consequence of Baire's category theorem? – LSpice – 2017-08-11T01:55:21.233

@S.Carnahan, the question you reference gives the stronger hypothesis suggested by @‍ZenHarper. – LSpice – 2017-08-11T01:56:54.473

1@LSpice I don't know whether I misunderstood your response. What I said is that, given a differentiable function (on $\mathbb R$, say) $f$, then the derivative $f'$ (not $f$) should be continuous at some point $x_0\in\mathbb R$. I don't think that it's tautological. – Frank Science – 2017-08-11T06:53:33.920

@FrankScience, no, I misunderstood you, not the other way around. I misread it as the totally elementary statement "a differentiable function is continuous." Sorry! – LSpice – 2017-08-11T14:39:43.787

6This seems like a homework problem in a 1st year course on calculus. – Ryan Budney – 2010-07-31T21:51:57.457

11This is a jewel, I will try to recall the solution. – Andrey Gogolev – 2010-07-31T22:05:56.740

Chandru, my suspicion is that instead of "for each $x \in [0,1]$, there is an integer $n \in \mathbb{N}$, it said there is an integer $n \in \mathbb{N}$ such that for each $x \in [0,1]$. – Michael Hardy – 2010-07-31T22:34:29.803

......OK, maybe this is subtler than I thought........ – Michael Hardy – 2010-07-31T22:38:41.043

14@Ryn: no, this is a classic little problem. @Michael: the problem is correct as stated. – Qiaochu Yuan – 2010-07-31T22:39:54.317

7This 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? – Zen Harper – 2010-07-31T23:32:09.500

4In view of Ryan's and Zen's comments, can the author, please, indicate the origin of the question? – Victor Protsak – 2010-07-31T23:52:14.157

Like Zen, I have seen a version of this question before set as an "exercise" - the tricky part, which I never solved on my own, is what to do once you've done the "obvious" Baire category part. Here I say "obvious" in the context of it being one of several BaireCat flavoured exercises in a batch, not "so obvious that everyone should have thought of it" – Yemon Choi – 2010-08-01T01:06:23.063

73@Ryan "This seems like a homework problem in a 1st year course on calculus." If you can find a regularly offered calculus course at more then 4 universities in the world that have the same number of students in the class who have even a clue how to solve this problem,I'll sing the American National Anthem naked on YouTube.I'm dead serious. – The Mathemagician – 2010-08-01T02:59:31.487

8Better be serious than dead. By the way, does desecrating the anthem carry the same consequences as burning the flag? – Victor Protsak – 2010-08-01T05:02:24.033

18I 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. – Anweshi – 2010-08-01T13:16:10.177

@Qiaochu, You're quite right; I was hasty. – Michael Hardy – 2010-08-01T17:39:56.093

Answers

110

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.

Andrey Gogolev

Posted 2010-07-31T21:37:22.197

Reputation: 2 855

24Thank 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. – Josh Swanson – 2011-05-08T22:48:22.807

24

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

Gerald Edgar

Posted 2010-07-31T21:37:22.197

Reputation: 25 496

21

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?

juan

Posted 2010-07-31T21:37:22.197

Reputation: 4 895

15

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.

Andrea Ferretti

Posted 2010-07-31T21:37:22.197

Reputation: 8 136

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. – Will Sawin – 2011-11-01T05:38:11.903

1Those functions have all derivatives 0 in a point, not on a whole interval – Andrea Ferretti – 2011-11-03T18:54:24.150

@AndreaFerretti, just for me to understanding well: are you proving that if for all $x$ there is a natural $n_x$ such that $n\geq n_x$ implies $f^{(n)}(x)=0$ then $f$ is polynomial? I just want to realize what implies $E_m\subset E_n$ in your proof. – matgaio – 2016-09-04T06:03:50.923

1E_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 – Andrea Ferretti – 2016-09-04T10:09:08.617

@AndreaFerretti of course, thank you. – matgaio – 2016-09-04T22:14:55.767

Hi--

Thanks a lot. Now, does this remain true if we replace $[0,1]$ by $\mathbb{R}$ or $[a,b]$ – crskhr – 2010-08-01T13:21:30.563

3Yes, of course. The proof is the same. – Andrea Ferretti – 2010-08-01T16:21:34.413

8

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

Fosco Loregian

Posted 2010-07-31T21:37:22.197

Reputation: 3 752

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 $I$.

Richard Hevener

Posted 2010-07-31T21:37:22.197

Reputation: 258

2

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...).

user45639

Posted 2010-07-31T21:37:22.197

Reputation:

1PP 58-59 of the 1st edition. – Gerry Myerson – 2014-02-24T22:00:53.847

@smuaug: Thanks for the info. I shall check it ASAP :) – crskhr – 2014-02-26T07:56:03.910

1

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.

i707107

Posted 2010-07-31T21:37:22.197

Reputation: 1 932

-2

I would have ventured this answer in the comments if I had enough reputation to comment. Alas, a perhaps forced and (probably) wrong answer will have to suffice.

I think it can be done without the Baire Category Theorem but with some tricks from Complex Analysis.

If $f$ is infinitely differentiable, then it is (at least) continuously differentiable on $[0,1]$. Now $f$ (I assume from the context) takes on values in $\mathbb{R}$, so no one stops me from saying that it takes on values in $\mathbb{C}$. So, $f:[0,1]\to\mathbb{C}$ is continuously differentiable, which probably means that it is analytic/holomorphic (they're the same in the world of complex analysis) in a neighborhood $G$ of $[0,1]$. Now, since the $n$-th derivative of $f$, $f^{(n)}$ vanishes on $[0,1]$ and the zeros of a nonconstant holomorphic function are isolated, this tells us that $f^{(n)}\equiv 0$ on $G$, and so $f^{(k)}\equiv 0$ for all $k\geq n$.

Let $f(z)=\sum_{k=0}^\infty a_kz^k$ be the power series expansion of $f$ at $0$; the coefficients $a_k$ are given by: $a_k=\frac{1}{k!}f^{(k)}(0)$, so we may write $f(z)=\sum_{k=0}^\infty \frac{z^k}{k!}f^{(k)}(0)$, and since $f^{(k)}(0)=0$ for all $k\geq n$, we have $f(z)=\sum_{k=0}^n \frac{z^k}{k!}f^{(k)}(0)$, hence, $f$ is a polynomial of degree at most $n$.

Admittedly, my argument is flawed in the sense that I'm not 100% sure if $f$ being continuously differentiable in $[0,1]$ implies analyticity in a neighborhood. If $f$ were analytic in a neighborhood of $[0,1]$ a priori then my argument goes through, however.

Alex

Posted 2010-07-31T21:37:22.197

Reputation: 1

5Continuous differentiability does not imply analyticity. For example, a function that is twice but not three times differentiable is not analytic at any point of non-three-times-differentiability. – LSpice – 2017-08-11T02:03:28.237

2The step "... which probably means..." does not hold up, as @LSpice has pointed out – Yemon Choi – 2017-08-11T02:30:33.547

@LSpice Does continuous differentiability on $[0,1]$ imply continuous differentiability in a neighborhood of $[0,1]$ in $\mathbb{C}$? If so, then continuous differentiability in that neighborhood implies analyticity since the two are equivalent in complex analysis. (See John B. Conway's Functions of One Complex Variable I, pgs. 34, 72-73.) – Alex – 2017-08-11T14:15:28.670

2@AlexBates, it is not a question of how you would prove it. The implication simply does not hold. (Consider the function $x \mapsto x\lvert x\rvert$ (or a suitable shift), whose derivative is $x \mapsto 2\lvert x\rvert$.) I think that you may be confusing the continuous differentiability of a function of a real variable, which is a very floppy condition, with that of the continuous differentiability of a function of a complex variable, which is so rigid that it is, as you say, equivalent to analyticity. – LSpice – 2017-08-11T14:38:48.973