Edit: The original question contained some errors that lead to some comments and answers that do not apply anymore. I thank those that took time to read the question and pointed out the errors. I tried to be more precise this time.
Let $V$ be the set of smooth (infinitely differentiable) functions $f:[0,+\infty)\rightarrow\mathbb{R}$ that satisfy the following conditions:
- $f(0)=0$
- $f'(x)\geq 0, \forall x$
- $f''(x)\leq 0$, $\forall x$
- $\lim_{x\rightarrow+\infty}f(x)$ is a real number
I'd like to find a dense subset $A\subseteq V$ (w.r.t. to the $L^2$ norm, say), where $A$ is a family of functions depending only on some real-valued parameters.
My first guess was to take the family $\{a\tanh(sx): a,s\geq0\}$, but after some numerical computations I saw it doesn't work: e.g. $\arctan(x)\in V$ cannot be arbitrarily approximated by functions of the form $a\tanh(sx)$.
My questions are:
- Can we find a dense $A$ as described above (depending only on some real parameters).
- In case that the answer to the previous question is "yes", can the hyperbolic tangents family I described be easily modified to become a dense subset?
- What if we change the definition domain from $[0,+\infty)$ to $[0,B]$ for some real number $B>0$? Does this compactness condition help in being able to find a dense subset?
Edit: The following argument has been rendered moot by an edit to the question.
The claim is vacuously true; $V$ is empty.
From the limit condition at $\infty$, there is some $u>0$ with $f(u)>0$. Then, by the Mean Value Theorem, there is some $v\in (0,u)$ with $f'(v) = \frac{f(u)}{u}>0$. Now, for all $x>v$, $f(x)\ge f(v) + (x-v)f'(v)$, which goes to $\infty$ as $x\to\infty$. Our function, which was supposed to have a finite limit at $\infty$, instead goes to $\infty$. There's nothing there - including that $f_{a,s}$ is never in $V$. In fact, for positive $a$ and $s$, the second derivative of $f_{a,s}$ is negative on $(0,\infty)$.