TITLE: Is a differentiable probability density function bounded?
QUESTION [5 upvotes]: Let $g$ be a lebesgue probability density function, differentiable, such that $\sup_x|g'(x)| < \infty.$
Is $g$ bounded?

REPLY [5 votes]: Let $M := \sup_x \lvert g'(x)\rvert$. Let $n \geqslant 2$. Suppose there exists an $x_n$ with $g(x_n) = n$. Let $x_{n-1}$ be a closest point to $x_n$ with $g(x_{n-1}) = n-1$ (since $g$ is continuous, and $\int g(x)\,dx = 1$, there exist points with $g(x) = n-1$, as $\liminf\limits_{x\to\pm\infty} g(x) = 0$). Without loss of generality, $x_{n-1} < x_n$. Let 
$$
  y_n = \min \{ x : x > x_{n-1}, g(x) = n\}.
$$ 
Then $n-1 \leqslant f(x) \leqslant n$ for $x_{n-1} \leqslant x \leqslant y_n$, and
$$1 = g(y_n) - g(x_{n-1}) = \int_{x_{n-1}}^{y_n} g'(x) \,dx \leqslant (y_n - x_{n-1})\cdot M,$$
hence $\lambda(\{x : n-1\leqslant g(x) \leqslant n\}) \geqslant \frac{1}{M}$, where $\lambda$ is the Lebesgue measure. But since $\int g(x)\,dx = 1$ we have 
$$
  (n-1)\cdot \lambda(\{x : n-1\leqslant g(x) \leqslant n\}) \leqslant \int g(x)\,dx = 1,
$$ 
hence $\frac{1}{M} \leqslant \frac{1}{n-1}$ or $n \leqslant M + 1$.