# “Insanely increasing” $C^\infty$ function with upper bound

Let $$C^\infty$$ denote the collection of functions $$f:\mathbb{R}\to\mathbb{R}$$ such that for every positive integer $$n$$, the $$n$$-th derivative of $$f$$ exists. For $$f\in C^\infty$$ we set

• $$f^{(0)} = f$$, and
• $$f^{(n+1)} = \big(f^{(n)}\big)'$$ for all non-negative integers $$n$$.

Is there $$f\in C^\infty$$ with the following properties?

1. for all $$x\in (-\infty, 0]$$ we have $$f(x)=0$$.
2. for all non-negative integers $$n$$ and $$x\in (0,\infty)$$ we have $$f^{(n+1)}(x) > f^{(n)}(x)$$
3. there is a function $$h:\mathbb{R}\to\mathbb{R}$$ such that for all non-negative integers $$n$$ and all $$x\in\mathbb{R}$$ we have $$f^{(n)}(x)\leq h(x)$$.

(Additional question for curiosity, answering it is not needed for acceptance of answer: can $$h$$ be chosen to be continous? Or even $$h\in C^\infty$$?)

• @GerhardPaseman For 2 only, something based on $e^{cx}$ with $c>1$ would work – მამუკა ჯიბლაძე Nov 26 '18 at 18:17
• @GerhardPaseman Well there are tricks for 1, actually I don't readily see how to handle 3 – მამუკა ჯიბლაძე Nov 26 '18 at 18:21
• If $h$ is continuous (or even $L^\infty_{\text{loc}}$), then Taylor's theorem implies that $f$ is everywhere real analytic. So your function cannot possibly exist, when combining (1) and (2). So at the very least near $0$ you must have $h$ is allowed to blow-up. – Willie Wong Nov 26 '18 at 19:12
• @Neal: yes. condition 2 implies $f$ has at most one zero in $(0,\infty)$, similarly $f'$ hast at most one zero in $(0,\infty)$. Suppose $f'$ is not everywhere positive on $(0,\infty)$, there exists then some small interval $(0,a)\subset (0,\frac12)$ on which $f' < 0$. By the mean value theorem we must be able to find $b\in (0,a)$ such that $0> f(b) > f'(b)$, leading to a contradiction. So $f'$ has to be everywhere positive. – Willie Wong Nov 26 '18 at 20:30
• If f' is positive then so are all higher derivatives, and then one can take h to be monotone increasing, hence also in $L^\infty_{loc}$, so it seems Willie's arguments show that such a function does not exist. – Terry Tao Nov 26 '18 at 20:34

Combining my comments with that of Terry Tao's:

1. First we show that $$f^{(n)} > 0$$ on $$(0,\infty)$$ for all $$n \geq 1$$. The argument is given for $$f'$$, but, extends easily to all $$n \geq 1$$.
Start by noticing that $$f'(a) = 0 \implies f''(a) > 0$$ so that $$f'$$ changes sign at most once, and that if it is not everywhere positive on $$(0,\infty)$$ there must be an initial interval $$(0,\epsilon)$$ on which $$f' < 0$$. Assume WLOG that $$\epsilon < \frac12$$. On $$(0,\epsilon)$$, we have that $$f(x) \geq x \inf_{y\in (0,x)} f'(y)$$ by the mean value theorem. This is incompatible with $$f'(y) > f(y)$$. Hence we conclude that $$f'$$ is always positive on $$(0,\infty)$$.

2. Step 1 implies that $$f^{(n)}$$ is increasing for all $$n \geq 1$$. Therefore if $$h$$ is a function as in condition (3) of the question, so is the increasing function $$\tilde{h}(y) = \inf_{[y,\infty)} h$$ This function is locally bounded, and implies (by Taylor's inequality) that $$f$$ is real analytic.

3. Real analytic functions can't be vanishing on $$(-\infty,0)$$ and be non-trivial.

By Willie Wong's comment and answer, $$f^{(n)}>0$$ on $$(0,\infty)$$ for all $$n=0,1,\dots$$. Hence, $$f^{(n)}\ge0$$ on $$\mathbb R$$. So, by Bernstein's theorem on completely monotone functions (used "right-to-left"; see the "footnote" for details), $$f(x)=\int_0^\infty e^{tx}\mu(dt)$$ for some nonzero nonnegative measure $$\mu$$ on $$[0,\infty)$$ and all real $$x\le1$$. So, $$f>0$$ on $$(-\infty,1]$$, which contradicts condition 1 in the OP. Thus, there exists no $$f$$ satisfying conditions 1 and 2. (Condition 3 is not needed for the non-existence.)

"Footnote": The condition $$f^{(n)}\ge0$$ on $$\mathbb R$$ for all $$n=0,1,\dots$$ implies $$(-1)^n g^{(n)}\ge0$$ on $$[0,\infty)$$ for all $$n=0,1,\dots$$, where the function $$g\colon[0,\infty)\to\mathbb R$$ is defined by reading $$f$$ "right-to-left": $$g(y):=f(1-y)$$ for $$y\in[0,\infty)$$. So, $$g$$ is completely monotone and hence, by Bernstein's theorem, $$g(y)=\int_0^\infty e^{-ty}\nu(dt)$$ for some nonnegative measure $$\nu$$ and all real $$y\ge0$$. Moreover, the measure $$\nu$$ is nonzero, because otherwise we would have $$g=0$$ (on $$[0,\infty)$$), that is, $$f=0$$ on $$(-\infty,1]$$, which would contradict the condition that $$f^{(n)}>0$$ on $$(0,\infty)$$ for all $$n=0,1,\dots$$. So, $$f(x)=g(1-x)=\int_0^\infty e^{-t+tx}\nu(dt)=\int_0^\infty e^{tx}\mu(dt)$$ for all real $$x\le1$$, where $$\mu(dt):=e^{-t}\nu(dt)$$, and the measure $$\mu$$ is nonnegative and nonzero.

• This looks like very neat and concise argument but could you explain what do you mean by "right-to-left"? I failed to figure this out from the Wikipedia page, sorry. Definition of complete monotonicity there involves alternating signs for odd/even order derivatives... – მამუკა ჯიბლაძე Nov 30 '18 at 11:17
• @მამუკაჯიბლაძე : Thank you for your comment. I have added a "footnote" with details on the meaning of the "right-to-left". – Iosif Pinelis Nov 30 '18 at 14:38
• For the record: Functions with $f^{(n)} \ge 0$ are also known as absolutely monotonic functions, see e.g., mathworld.wolfram.com/AbsolutelyMonotonicFunction.html (and connect to CM functions as noted in Iosif's answer above) – Suvrit Nov 30 '18 at 16:56