Does a continuous probability density function (pdf) have zero values on +infinity and -infinity?

Question

Assume a pdf $f(x)$ is continuous along $-\infty$ to $+\infty$. Does this assumption guarantee that $f(+\infty)=f(-\infty)=0$? How to prove? Thanks in advance.

Answer 1

Here is a setting worth keeping in mind when looking for examples. Consider some sequences $(\mu_n)$, $(\sigma^2_n)$ and $(p_n)$ with $\sigma^2_n\gt0$, $p_n\gt0$, and $\sum\limits_np_n=1$, and the function $$ f=\sum_np_n\,g_{\mu_n,\sigma^2_n}, $$ where, for every $(\mu,\sigma^2)$, $g_{\mu,\sigma^2}$ is the gaussian density with mean $\mu$ and variance $\sigma^2$.

Then $f$ is a PDF since $\sum\limits_np_n=1$. However, if $\mu_n\to\infty$ and $\sigma_n\ll p_n$ when $n\to\infty$ (say, $\mu_n=n$, $p_n=1/2^n$ and $\sigma_n^2=1/8^n$), then $$ f(\mu_n)\geqslant p_ng_{\mu_n,\sigma^2_n}(\mu_n)=\frac{p_n}{\sqrt{2\pi\sigma_n^2}}, $$ hence the density $f$, far from converging to zero at infinity, is actually unbounded at infinity.

Answer 2

Refer to the example I constructed here that easily lends itself to a counterexample.