For decreasing $f,g$, do $\int_0^\infty f(t) \,\text dt < \infty$ and $\int_0^\infty g(t) \,\text dt = \infty$ imply $f(t) = o \big( g(t) \big)$?

Question

Let $f,g : \mathbb R_+ \to \mathbb R_+$ be (not necessarily strictly) decreasing functions such that $\int_0^\infty f(t) \,\text dt < \infty$ and $\int_0^\infty g(t) \,\text dt = \infty$.

Does it follow that $f(t) = o\big(g(t) \big)$ as $t \to \infty$?

Thoughts

Intuitively, this seems to make sense. After all, the "tail" of $f$ must be thinner than the the "tail" of $g$.

Certainly, $\int_z^\infty f(t) \,\text dt \overset{z \to \infty}{\longrightarrow} 0$, whereas $\int_z^\infty g(t) \,\text dt = \infty$ for all $z \in \mathbb R_+$.

It's not too hard to show that $\liminf_{t \to \infty} \frac{f(t)}{g(t)} = 0$:

• Suppose not. Then there exists $0 < c < \liminf_{t \to \infty} \frac{f(t)}{g(t)} \leq \infty$. So for some $z_0 \in \mathbb R_+$ we have $c \cdot g(t) < f(t)$ for all $t \geq z_0$. Then $\int_0^\infty f(t) \,\text dt = \int_0^{z_0} f(t) \, \text dt + \int_{z_0}^\infty f(t) \, \text dt \geq \int_0^{z_0} f(t) \, \text dt + c \underbrace{ \int_{z_0}^\infty g(t) \, \text dt}_{=\infty} = \infty$, a contradiction.

But that's not sufficient. We need $\limsup_{t \to \infty} \frac{f(t)}{g(t)} = 0$.

Perhaps this isn't true after all. But I imagine then, that it would be under some weak conditions.

Take $f(x)=\frac{1}{x^2}$ or any other integrable function near $\infty$; if you insist on integration from $0$, patch it as necessary near $0$, it won't change the behaviour around $\infty$. I will work with the integral $\int_1^\infty$ in what follows.

Now, make $g(x)$ so that the integral is infinite:

• $g(x)=1$ on $x\in[1,2)$
• $g(x)=\frac{1}{4}$ on $x\in[2,6)$
• $g(x)=\frac{1}{36}$ on $x\in[6,42)$
• $g(x)=\frac{1}{42^2}$ on $x\in[42,42+42^2)$

etc.

Obviously, at points $1,2, 6, 42\ldots$ the functions $f(x)$ and $g(x)$ coincide, so $\lim\sup_{x\to\infty}\frac{f(x)}{g(x)}=1$. However, the integral of $g(x)$ on each of the intervals $[1,2], [2,6], [6,42],\ldots$ is $1$, so $\int_1^\infty g(x)dx=+\infty$.