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This morning I was trying to imagine what a function would look like if all it's derivatives were zero at a point $a$ (assuming it is $C^\infty$). My first thought was that it should be identically zero in a neighborhood of $a$, but this is only true if the function is analytic. So my question is this:

What can we say in general about a smooth function (i.e. has derivatives of all orders) whose derivatives are all zero at a single point $a$?

It seems like the function $f$ should approach zero at $a$ faster than any polynomial because of Taylor's theorem. In other words, for every $k\geq 1$, there exists some remainder function $r_k(x)$ with $$\frac{f(x)}{(x-a)^k}=r_k(x)\qquad \left( \text{and}\quad \lim_{x\rightarrow a}\ r_k(x)= 0\right).$$

So can we conclude that $f$ approaches zero faster than any polynomial? If so, how is this made precise? And is this all we can say about these functions? (These questions can all be subordinated to the main question)

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    $\begingroup$ You can't say anything. There are smooth functions on the whole real line which vanish together with all of their derivatives at $x=0$ and are positive everywhere else. The standard example is $f(x) = {\rm exp}(-1/x^2)$ for nonzero $x$ and $f(0) = 0$. These kinds of functions are quite important to create examples of bump functions for partitions of unity. See en.wikipedia.org/wiki/Bump_function $\endgroup$
    – KCd
    Commented Oct 29, 2011 at 19:53
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    $\begingroup$ @KCd I'm aware of the standard non-analytic smooth function, but it's clear that $\exp(-1/x^2)$ approaches $0$ faster than any polynomial as $x\rightarrow 0$. My question is whether or not this is a general property of such functions, or if there are any other properties of such functions. $\endgroup$
    – Chris Brooks
    Commented Oct 29, 2011 at 20:07
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    $\begingroup$ Well you say it exactly as you did: $\lim_{x \to a} \frac{f(x)}{(x-a)^n}=0$ for all $n$. This is equivalent to $\lim_{x \to a} \frac{f(x)}{P(x)}=0$ for all polynomials $p(x)$. $\endgroup$
    – N. S.
    Commented Oct 29, 2011 at 20:18
  • $\begingroup$ You are right, if $f^{(n)}(a)=0$ for every $n \ge 0$, then $f(x)$ approaches $0$ faster than any power of $x-a$, in the sense that for fixed $n$, $\lim_{x\to a}\frac{f(x)}{(x-a)^n}=0$. As you pointed out, an error bound in Taylor's Theorem will do it. But maybe from the way I wrote the limit, you can think of another way! $\endgroup$
    – André Nicolas
    Commented Oct 29, 2011 at 20:23
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    $\begingroup$ One more comment, this condition is equivalent to $\frac{1}{f(x)}$ to go to infinity faster than any polynomial at $a$. I think that this is equivalent to asking that $g(x)=\frac{1}{f(a + \frac{1}{x})}$ grows faster than any polynomial at $\infty$. $\endgroup$
    – N. S.
    Commented Oct 29, 2011 at 20:31

1 Answer 1

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If all the derivatives of $f$ at $0$ vanish, then $\lim_{x \to 0}\ f(x)/x^k=0$ for all $k$, by l'Hospital's theorem used $k$ times.

The converse is not true: $e^{-1/x^2} \sin e^{1/x^2}$ goes to zero faster than any polynomial, but its derivative is not defined continuous at $0$, so the second derivative at $0$ doesn't exist.

If $f$ is infinitely differentiable at $0$, and goes to $0$ faster than any polynomial, then $f$ has all its derivatives at $0$ vanish; this follows from Taylor's theorem. I'll give a bit more detail: Suppose for the sake of contradiction that $f$ is infinitely differentiable and $f^{(k)}(0) = c \neq 0$. Then Taylor's theorem shows that $f(x) = c x^k/k! + r(x)$ where $\lim_{x \to 0} r(x)/x^k=0$. So $\lim_{x \to 0} f(x)/x^k = c/k!$, not $0$, a contradiction.

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    $\begingroup$ Why is that counterexample not differentiable at $0$? It seems to me its derivative is just $0$. $\endgroup$
    – scineram
    Commented Oct 30, 2011 at 18:17
  • $\begingroup$ Sorry, corrected. $\endgroup$
    – David E Speyer
    Commented Oct 30, 2011 at 18:54
  • $\begingroup$ @DavidSpeyer Do you want to comment on this? math.stackexchange.com/questions/75617/… $\endgroup$
    – Phira
    Commented Oct 31, 2011 at 17:29
  • $\begingroup$ I was kind of hoping someone else would. The Miller-Sturmfels book does contain a complete proof, and I completely understand why not all readers follow it. I'd guess that it would take me several hours to write something better than they do, and I don't have time to put in that work right now. $\endgroup$
    – David E Speyer
    Commented Oct 31, 2011 at 17:32

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