22
$\begingroup$

Suppose $f: \mathbb{R} \rightarrow \mathbb{R} $ is a continuous function, and there exists some $a \in \mathbb{R}$ where all derivatives of $f$ exist and are identically $0$, i.e. $f'(a) = 0, f''(a) = 0, \ldots$ Must $f$ be a constant function? or if not, are there examples of non-constant $f$ that satisfy these properties?

What if the hypothesis is changed so that the derivatives of $f$ are identically $0$ on an open interval, i.e. $f'(A) = 0, f''(A) = 0, \ldots$ for some open interval $A$. Are the answers still the same?

$\endgroup$
3
  • 2
    $\begingroup$ Hint: Taylor series. $\endgroup$ – Gautam Shenoy Nov 18 '12 at 18:56
  • 2
    $\begingroup$ @GautamShenoy, I think this hint is misleading. See the flat function in Brian's answer. Its Taylor Series about x=0 is simply 0, yet the function is not constant on any non-trivial interval. $\endgroup$ – kuzzooroo Dec 8 '15 at 23:28
  • $\begingroup$ Oh. I misread some $a \in \mathbb{R}$ for all $a \in \mathbb{R}$. Kindly ignore my hint. $\endgroup$ – Gautam Shenoy Dec 9 '15 at 6:15
25
$\begingroup$

Yes, such functions do exist; they’re called flat functions. The simplest example that I know is the one given at the link:

$$f(x)=\begin{cases} e^{-1/x^2},&\text{if }x\ne 0\\ 0,&\text{if }x=0\;, \end{cases}$$

which is flat at $x=0$.

You can modify this example to get one that is flat on the interval $[0,1]$ but not constant on $\Bbb R$:

$$f(x)=\begin{cases} e^{-1/x^2},&\text{if }x<0\\ 0,&\text{if }0\le x\le 1\\ e^{-1/(x-1)^2},&\text{if }x>1\;. \end{cases}$$

In effect I’ve just cut the function at $x=0$ and moved the righthand half $1$ unit to the right, filling in the gap with the zero function.

$\endgroup$
11
$\begingroup$

Cauchy's function $f(x)=e^{-1/x^2}$ for $x\ne0$ and $f(0)=0$ has all derivatives at $0$ equal to $0$, but the function is not constant on any interval, thus answering your first question.

For your second question, of course if a function has first derivative equal to $0$ on an interval then the function is constant on that interval.

$\endgroup$
2
  • $\begingroup$ Hi. May I know why $f^{(i)}(0) = 0$ for all natural numbers $i?$ For example, $f'(x) = \frac{2f(x)}{x^3}.$ How to conclude that $f'(0) = 0?$ $\endgroup$ – Idonknow Mar 3 '19 at 9:52
  • $\begingroup$ see: math.stackexchange.com/questions/491227/… $\endgroup$ – Ittay Weiss Mar 3 '19 at 17:57
8
$\begingroup$

As others have pointed out, the canonical counter-example is the function $f(x)=e^{-1/x^2}$. But what is special about this function? The answer is that it is very badly behaved near $0$ in the complex plane, because $-1/x^2$ is arbitrarily large and positive along the imaginary axis close to $0$.

$\endgroup$
2
$\begingroup$

The function $f(x)$ defined as follows is non-constant and satisfies your second condition.

$f(x) = x$ if $x \le 0$

$f(x) = 0$ if $0 < x < 1$

$f(x) = x - 1$ if $1 \le x$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.