Bounded monotonic increasing functions

Question

Let $f:[0,\infty)\to\mathbb{R}$ be a real valued monotonic increasing function that is twice differentiable on $(0,\infty)$. Is it always true that $\exists y\in\mathbb{R}$ such that $\forall x>y$ we have $f''(x)\leq 0$ whenever $\lim\sup_{x\to\infty} f(x)<\infty$?

My attempt: If we assume $\lim\sup_{x\to\infty} f(x)<\infty$ then $\exists \alpha\in\mathbb{R}$ and $x_0\in\mathbb{R}$ such that $f(x)\leq \alpha$ for all $x\geq x_0$. On the other hand if we fail to have some $y\in\mathbb{R}$ such that for all $x\geq y$ we have $f''(x)\leq 0$ then we must necessarily have that for any $y\in\mathbb{R}$ there exists open intervals $(y-\varepsilon,y+\varepsilon)$ where $f''(x)>0$ for all $x\in(y-\varepsilon,y+\varepsilon)$. Then we get that $$f(x)>f'(y)(x-y)+f(y)\hspace{0.3cm}\forall x\in[y,y+\varepsilon)$$ If $\varepsilon\equiv \infty$ then $f(x)\to\infty$ as $x\to\infty$ violating the boundedness. If $\varepsilon<\infty$ then we must necessarily have consecutive intervals where $f$ is concave up followed by intervals where it is concave down such that it still is monotonic increasing function. It seems that again it should violate the boundedness but I am not sure about it. Would the violation again occur? How could we formally verify or nullify this claim?

Answer 1

Take for $f(x)$ the function $\displaystyle f(x)=\int_0^x (\sin(t))^2 \exp(-t)dt$. Then $f$ has a limit as $x\to +\infty$, $f^{\prime}$ is $\geq 0$ for all $x$, but as $f^{\prime\prime}(x)=(2\cos(x)-\sin(x))\sin(x)\exp(-x)$, no such an $y$ exists.

Answer 2

No, even with stronger regularity assumptions.

https://math.stackexchange.com/a/788818/66096

Answer 3

The statement you gave is unfortunately false. The function can continue to increase in little "spurts" indefinitely, where the second derivative keeps switching sine. For an easy construction, first pick a monotonic increasing function that's not bounded -- take $g(x) = x+\sin(x)$. Now apply some function to constrain it to a bound, such as $h(x) = -1/(1+x)$. Then $f = h\circ g$ will be a counterexample. In this case, we have that

$$g(x) > 0 \implies f(g) = \frac{-1}{1+g} < 0$$

so it's clearly bounded. Since $g$ is monotonic increasing, and $h$ is as well, so must be $f$. But the second derivative will clearly oscillate.