If a probability density function (pdf) has bounded derivative, is the pdf itself bounded?

Question

Suppose a probability density function (pdf) $f$ is differentiable almost everywhere and continuous and has a bounded derivative. Is the pdf itself bounded?

Answer 1

Your conjecture appears to be true, and I think I have succeeded in proving it below. To prove this, we can give the proposal a definite statement for a real-valued random variable, and prove the result in this simple context. (It should be simple to extend this to multivariate cases.) The proof first shows that if the density is unbounded then it must have an infinite limit at an internal point of the domain. If the derivative is bounded almost everywhere then the density cannot change much within a neighbourhood of this point and so it must be infinite within this neighbourhood, which would lead to an infinite integral for the density.

Theorem: Consider a density function $f: \mathbb{R} \rightarrow \mathbb{R}_+$. If $f'$ exists and is bounded almost everywhere on the domain, then $f$ is bounded.

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Proof: We will use a proof-by-contradiction. Suppose ---contrary to the theorem--- that all the antecedent conditions of the theorem hold, but $f$ is unbounded. Since $f$ is a density function it is bounded from below by $f \geqslant 0$, so it must be unbounded from above. Moreover, since $f$ is a density function, it approaches zero in its left and right tails, so it must approach infinity at an internal point, from either the left or the right (or both). Without loss of generality, we will suppose that the function approaches infinity from the left, so we have: $$\lim_{x \uparrow a} f(x) = \infty \quad \quad \quad \text{for some } a \in \mathbb{R}.$$

Since the codomain of $f$ is the set of real numbers we have $f(a) \neq \infty$, so this is a point where the function approaches infinity, but is not equal to infinity, so $f$ is not continuous at $a$. Since $f$ is not continuous at $a$ it follows that $f'$ does not exist at $a$. Since $f'$ exists and is bounded almost everywhere on the domain, but it does not exist at $a$, this means that it must exist and be bounded on the interval $[a-\epsilon, a)$ for some $0<\epsilon<\infty$. Let $0<U<\infty$ be the upper bound of the derivative on this interval. Then for all $0 < \varepsilon < \epsilon$ we can apply the fundamental theorem of calculus to get: $$\lim_{x \uparrow a} f(x) - f(a-\varepsilon) = \lim_{x \uparrow a} \int \limits_{a-\varepsilon}^x f'(r) dr \leqslant \lim_{x \uparrow a} U \cdot (x - (a-\varepsilon)) = U \cdot \varepsilon.$$

For all $0 < \varepsilon < \epsilon$ we can rearrange this inequality to obtain: $$f(a-\varepsilon) \geqslant \lim_{x \uparrow a} f(x) - U \cdot \varepsilon = \infty - \text{finite} = \infty.$$

This means that we must have $f(a-\varepsilon) = \infty$ for all $0 < \varepsilon < \epsilon$. This contradicts the stated codomain of the function, and it also means that: $$\int \limits_{a-\epsilon}^a f(r) dr = \infty,$$ which contradicts the fact that the probability density must integrate to one. This proves the theorem. $\blacksquare$

Answer 2

The statement is correct but the proof given by Ben was wrong, since a probability density $f$ can be unbounded when approaching $\pm\infty$.

Note that continuity of $f$ already implies that it is bounded on every compact set. Thus, we simply must exclude that there is a sequence $x_n$ s.t. $\vert x_n\vert \to\infty$ with $f(x_n)\to\infty$. Now, seeking a contradiction, suppose that such a sequence exists.

First observe that $f$ is Lipschitz since $\vert f(x)-f(y)\vert \leq \int_x^y \vert f^\prime(z)\vert dz\leq C\vert x-y\vert $, where $\vert f^\prime\vert \leq C$. Moreover, we get for all $0<\delta<1$ that $0<f(x_n\pm\delta) = f(x_n\pm\delta)\pm f(x_n)\geq f(x_n)-C$. Thus $f(y)\in [f(x_n)-C,f(x_n)+C]$ for all $y\in [x_n-\delta,x_n+\delta]$ and we get $$ \int_{x_n-1}^{x_n+1} f(y)dy\geq 2(f(x_n)-C)\to\infty$$ which is a contradiction to $f$ being a density.