If $f\in C^2(\mathbb R)$ is a probability density function, then $f'(x)\to0$ implies $f(x)\to0$ as $|x|\to\infty$

Question

Let $f:\mathbb R\to[0,\infty)$ be twice continuously differentiable and assume $$\int f(x)\:{\rm d}x=1.$$

How can we conclude that as $|x|\to\infty$ either

1. $f(x)\to0$ and $f'(x)\to0$,
2. $f(x)\to0$ and $f'(x)\not\to0$ or
3. $f(x)\not\to0$ and $f'(x)\not\to0$?

If I'm not missing anything, the only other case would be $f(x)\not\to0$ and $f'(x)\to0$. So, I guess we need to show that this is impossible. Since $f'(x)\to0$ as $|x|\to\infty$ does not necessarily imply that $\lim_{|x|\to\infty}f(x)$ even exists, this must have something to do with the integrability condition, right? If this is not sufficient, do we need to impose further conditions?

Answer 1

If $f:\mathbb{R} \to [0,\infty)$ is integrable (over $\mathbb{R}$) and $f'(x) \to 0$ as $|x| \to +\infty$, then it follows that $f(x) \to 0$.

For proof by contradiction, assume WLOG that $f(x) \not\to 0$ as $x \to +\infty$. There exists $\beta > 0$ and a sequence $x_n \to +\infty$ such that $f(x_n) \geqslant \beta$. Selecting a subsequence if necessary, we can assume that $x_{n+1} > x_n + \beta.$

Since $f'(x) \to 0$, there exists $a > 0$ such that $-\frac{1}{2} < f'(x) < \frac{1}{2}$ for all $x \geqslant a$, and for all sufficiently large $n \geqslant N$ we also have $x_n > a$.

Thus,

$$\tag{*}\int_a^\infty f(x) \, dx \geqslant \sum_{n=N}^\infty\int_{x_n}^{x_{n+1}}f(x) \, dx > \sum_{n=N}^\infty\int_{x_n}^{x_n + \beta}f(x) \, dx.$$

By the mean value theorem for $x \in (x_n,x_n + \beta$) there exists $c_n \in (x_n,x)$ such that

$$f(x) = f(x_n) + f'(c_n)(x - x_n)> \beta - \frac{1}{2}\beta = \frac{\beta}{2}$$

Therefore, we have

$$\int_{x_n}^{x_n + \beta}f(x) \, dx > \frac{\beta^2}{2},$$

and using (*) we obtain a contradiction,

$$\int_a^\infty f(x) \, dx > \sum_{n=N}^\infty \frac{\beta^2}{2} = + \infty$$