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Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ having the following properties?

  • $f(x) = 0$ for all $x \le 0$.
  • $f(x) = 1$ for all $x \ge 1$.
  • For $0 < x < 1$, $f$ is strictly increasing.
  • $f$ is everywhere $C^\infty$.
  • The sequence of $L^\infty$ norms $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ is bounded.

If we impose only the first four conditions, there is a well-known answer: for $0 < x < 1$, define $f(x)$ by $$ f(x) = \frac{e^{-1/x}}{e^{-1/x} + e^{-1/(1-x)}} = \frac{1}{1 + e^{1/x - 1/(1-x)}} $$ However, the derivatives of this function appear to grow quite rapidly. (I'm not sure how to verify this, but it seems at least exponential to me.)

If such a function does not exist, what is the smallest order of asymptotic growth that the sequence $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ can have?

Community
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David Zhang
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    Does the fact that the sequence $|f^{(n)}|_{\infty}$ is bounded imply that $f$ must be analytic? If so, then there is no such function, as any analytic function that is $0$ on an interval is $0$ everywhere. – JimmyK4542 Dec 27 '15 at 02:00

2 Answers2

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Suppose $f$ satisfies your first four hypotheses. By the mean value theorem, there is some $x_1\in (0,1)$ such that $f'(x_1)=1$. There is then some $x_2\in(0,x_1)$ such that $f''(x_2)=1/x_1$, and then there is some $x_3\in(0,x_2)$ such that $f'''(x_3)=1/x_1x_2>1/x_1^2$, and so on. By induction, we see that we have points $x_n$ such that $f^{(n)}(x_n) \ge 1/x_1^{n-1}$ for all $n \ge 1$. Thus the norms $\|f^{(n)}\|_\infty$ must grow at least exponentially in $n$.

By a more careful argument, you can show that in fact the derivatives must grow factorially. More precisely, if $\|f^{(n)}\|_\infty=M$, we get that for any $x$, $|f^{(n-1)}(x)|\leq M|x|$, and then we get $|f^{(n-2)}(x)|\leq M|x|^2/2$, and so on, up to $|f(x)|\leq M|x|^n/n!$. It follows that if there exists a constant $c>0$ such that $\|f^{(n)}\|_\infty<c^nn!$ for all $n$, then $f(x)=0$ for $|x|<1/c$. But then replacing $f(x)$ with $g(x)=f(x\pm 1/c)$, we get that $f(x)=0$ for all $|x|<2/c$, and iterating this we get that $f$ vanishes everywhere. So the norms $\|f^{(n)}\|_\infty$ must grow faster than $c^nn!$ for all $c>0$.

Eric Wofsey
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  • Excellent answer! But it leaves me wondering, if $c^n n!$ cannot be achieved for any $c > 0$, just what is the slowest achievable rate of asymptotic growth? I suppose there may not be an easy way to answer this. – David Zhang Dec 27 '15 at 10:11
  • I made some edits that I thought are typos. Please check them to be sure. :) – Hosein Rahnama Jul 06 '16 at 10:45
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If the sequence of infinity norms $\{\|f^{(n)}\|_{\infty}\}_{n = 0}^{\infty}$ is bounded, i.e. there exists a constant $M$ such that for all $n \in \mathbb{N}_0$, $|f^{(n)}(x)| \le M$ for all $x \in \mathbb{R}$, then by this theorem $f$ is analytic.

However, if an analytic function is $0$ on any interval (such as $(-\infty,0)$), then its Taylor series about any point in that interval is identically $0$. Then, since $f$ is analytic, the Taylor series converges to $f$ everywhere, i.e. $f \equiv 0$.

Therefore, no such function exists.

JimmyK4542
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