Consider a probability density function $\phi(\cdot)$ that is positive everywhere. For any a>0, $\phi(x)/\phi(x+a)$ is strictly increasing in x. I think a function like this must be continuous, first strictly increasing, and then strictly decreasing. Could anyone tell me how to prove it? Thanks.



An implication of SMLRP
If $\phi(x)/\phi(x+a)$ is strictly increasing, then you get
$$
\frac{\phi(x)}{\phi(x+a)} < \frac{\phi(x+a)}{\phi(x+2a)}
$$
for any $x$.
From this you get that the function $-\log(\phi)$ is strictly "midpoint-convex", and any Lebesgue-measurable midpoint-convex function is in fact convex; see for example these [two](https://math.stackexchange.com/questions/71019/example-of-a-function-such-that-varphi-left-fracxy2-right-leq-frac-va)
[questions](https://math.stackexchange.com/questions/83383/midpoint-convex-and-continuous-implies-convex) for discussion.


From that you can get that $-\log(\phi)$ first decreases then increases, so that $\phi$ first increases and then decreases, as you want.


