Find a function $\gamma: (0, \infty) \to (0, \infty)$ satisfying the following conditions:
i) $\gamma \not \in L^1$,
ii) $\dot \gamma(t) \le 0$ for every $t>0$ and $t \dot \gamma \in L^1$;
iii) There exists a sequence $\{t_n\}$ such that $t_n \to \infty$ and $\ddot \gamma(t_n)<0$, for all $n$.
Thank you very much.
First consider that your condition i) is not a restriction at all… if you find a function $\gamma$ that fulfill condition ii) and iii) but $\gamma\in L^1$ then nececarrily it also holds $$\lim_{t\to\infty} \gamma(t) = 0$$
But then also $$\gamma_c = \gamma + c$$ fulfill condition ii) and iii) for each $c\in\Bbb R\setminus \{0\}$ due to properties of the derivative but $$\lim_{t\to\infty} \gamma_c(t) = c \not= 0$$ so $\gamma_c \not\in L^1$.
To get a solution for your problem take $$\dot \gamma(t) = \frac{\sin(t) - 1}{t^3}\cdot 1_{[1,\infty)}$$
Then $\dot\gamma(t) \le 0$ and $t\dot\gamma$ is integrable due to $$|t\dot\gamma| \le \frac{1}{t^2}\cdot 1_{[1,\infty)}$$
On the other hand $$\ddot \gamma(x) = \frac{-3\sin(x) + x\cos(x) + 3}{x^4}$$
Taking $t_n = \left(1 + 2n\right)\pi$ gives you the sequence you are looking for by $$\ddot \gamma(t_n) = \frac{t_n\cos(t_n) + 3}{t_n^4} < \frac{-3.14 + 3}{t_n^4} < 0$$
Together with the first remark $\dot\gamma$ has an antiderivative that is not in $L^1$ and you are done…