Find convex $f$: $\limsup_{n \to \infty} \frac{1}{n} \left| \frac{d^{n+1}}{ dx^{n+1}} e^{f} /\frac{d^{n}}{ dx^n} e^{f} \right|=\infty $

Question

Does there exits a convex function $f$ that satisfies the following property: let \begin{align} a_n= \left| \frac{ \frac{d^{n+1}}{ dx^{n+1}} e^{f(x)}}{ \frac{d^{n}}{ dx^{n}} e^{f(x)}} \right|_{x=1} \end{align} such that \begin{align} \limsup_{n \to \infty} \frac{a_n}{n} =\infty, \end{align} but \begin{align} \liminf_{n \to \infty} \frac{a_n}{n} <1. \end{align}

Answer 1

Enough to show for $x=0$.

Consider $f = \ln(e^x + e^{x^2})$ .

Then $f'' = \frac{e^{x^2}}{(e^x + e^{x^2})^2} \left(2e^{x^2} + (4x^2 - 4x +3)e^x \right) > 0 $. Thus $f$ is indeed a convex function.

I will represent $\frac{d}{dx}$ as $D$.

Indeed, $\boldsymbol{c_n \equiv D^n e^f \,\,|_{x=0}} = D^n e^{x^2}\,\,|_{x=0} + D^n e^x\,\,|_{x=0} = 1 +D^n e^{x^2}\,\,|_{x=0} $.

Now I will define sequence of polynomial as follows.

$$ \tilde{H}_n (x) = e^{-x^2} D^n e^{x^2} \quad(\Rightarrow D^n e^{x^2}\,\,|_{x=0} = \tilde{H}_n (0)) $$

It is similar to Hermit polynomial, but slightly different.

I will prove simple recurrence relation for $\tilde{H}_n (x)$, $$ \tilde{H}_{n+1} (x) = 2x\tilde{H}_{n} (x) + 2n\tilde{H}_{n-1} (x). $$

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Proof) $$ {\tilde{H}_{n}}' (x) = D (e^{-x^2} D^n e^{x^2}) = -2xe^{-x^2} D^{n} e^{x^2} + e^{-x^2} D^{n+1} e^{x^2} \\ = -2x\tilde{H}_{n}(x) + \tilde{H}_{n+1}(x) \,-(1) $$

Also, $$ {\tilde{H}_{n}}' (x) = -2xe^{-x^2} D^{n} e^{x^2} + e^{-x^2} D^{n+1} e^{x^2} = -2xe^{-x^2} D^{n} e^{x^2} + e^{-x^2} D^{n}( 2xe^{x^2})\\ = -2xe^{-x^2} D^{n} e^{x^2} + e^{-x^2} (n \,D(2x)D^{n-1}e^{x^2} + 2x D^{n}e^{x^2} ) \\=2n\,e^{-x^2}D^{n-1}e^{x^2} = 2n{\tilde{H}_{n-1}} (x) \, - (2) $$

By using $(1)$ and $(2)$, we get our equality.

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For convenience, I will define $b_n = \tilde{H}_{n} (0)$.

Then $b_{n+1} =2nb_{n-1}$ (from above equality) and $b_0 = 1,\,\,b_1 = 0\,\,$ gives $$ b_n = \left\{\begin{array}{lr} 0 & \text{for } n \text{ is odd}\\ 2^{n/2} (n-1)!! & \text{for } n \text{ is even}\\ \end{array}\right\}. $$

Now, $$ c_n = b_n\, + \,1 = \left\{\begin{array}{lr} 1 & \text{for } n \text{ is odd}\\ 1+ 2^{n/2} (n-1)!! & \text{for } n \text{ is even}\\ \end{array}\right\} $$ and $$ a_n = \left|\frac{c_{n+1}}{c_n}\right|= \left\{\begin{array}{lr} 1+ 2^{(n+1)/2} (n)!! & \text{for } n \text{ is odd}\\ {1}/({1+ 2^{n/2} (n-1)!!}) & \text{for } n \text{ is even}\\ \end{array}\right\}. $$

Finally we get $$ \limsup_{n \to \infty} \frac{a_n}{n} =\infty \quad \&\quad \liminf_{n \to \infty} \frac{a_n}{n} =0 . $$

($\because$ Since $[1+ 2^{(n+1)/2} (n)!! ]/ n \geq 2^{(n+1)/2} (n-1)(n-2)!!\,$.)