Functions with the property: $f(\infty)=0$ and $f''(\infty)=\infty$ Please help me with an interesting question.
There are functions $f: D \subset \mathbb{R} \to \mathbb{R}$
such that for some $a \in \mathbb{R}$  $f(a) = 0$ and $f''(a) = +\infty$;
e.g., $f(x)=\sqrt{x-a}$ or $f(x)=(x-a)^{3/2}$.
Do there exist such functions $f: D \subset \mathbb{R} \to \mathbb{R}$ for $a=+\infty$;
that is, $\lim_{x \to +\infty} f(x) = 0$ and $\lim_{x \to +\infty} f''(x) = +\infty$ ?
I think that there are no such functions but I can't prove this.
And I have another question.
Do there exist complex functions $f: D \subset \mathbb{C} \to \mathbb{C}$
such that $f(\infty)=0$ and $f''(\infty)=\infty$ ?
Thank you very much in advance!
 A: I will answer the real-valued case with $D = \mathbb{R}$.
In the case $f'$ is absolutely continuous, you can argue as follows.
If $\lim_{x \to \infty} f''(x) = +\infty$, then, there is $K > 0$, such that
$$f''(x) \ge 1 \quad\text{for all } x \ge K.$$
This yields
$$f'(x) = f'(K) + \int_K^x f''(y) \, dy \ge f'(K) + (x-K)$$
and
$$f(x) = f(K) + \int_K^x f'(y) \, dy \ge f(x) + f'(K) \, (x-K) + \dfrac12(x - K)^2.$$
Now, you get $\lim_{x\to\infty}f(x) = +\infty$, too.
Edit: If $f'$ is assumed to be only continuous, you can construct a counter-example using Devil's staircase:
Let $d : [0,1] \to [0,1]$ be the devil's staircase. In particular, $d$ is continuous, $d(0) = 0$, $d(1) = 1$ and $d' = 0$ almost everywhere on $[0,1]$. Denote $I = \int_0^1 d(x) \, dx$. Now, set
$$g(x) = \int_0^x d(y) \, dy - I \, x^2.$$
Then, $g'(x) = d(x) - 2\,I \, x$ is continuous, $g'' = -2\,I$ almost everywhere on $[0,1]$, $g(0) = g(1) = 0$ and $|g| \le 2$.
If you now construct $f$ from suitable scaled (both horizontally and vertically) copies of $g$, you get a function with the following properties: $f$ and $f'$ are continuous, $f''$ exists almost everywhere, $\lim_{x\to\infty}f'' = +\infty$ (where you ignore the points where $f''$ is not defined), and $\lim_{x\to\infty}f(x) = 0$.
