$\frac{1}{t}e^{F(t)} \to \infty \Rightarrow t F'(t) \to \infty$?

Question

Let $F:[0,\infty)\to[0,\infty)$ be a non-decreasing continuously differentiable function satisfying $F(t)\to\infty$ as $t\to\infty$. Consider the following statements:

1. $t F'(t) \to \infty$, as $t\to\infty$.
2. $\frac{F(t)}{\log t} \to \infty$, as $t\to\infty$.
3. For each $\epsilon>0$, $\frac{1}{t}e^{\epsilon F(t)} \to \infty$, as $t\to\infty$.
4. For each $\epsilon>0$, $F'(t)e^{\epsilon F(t)} \to \infty$, as $t\to\infty$.

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I can derive that 1 $\Rightarrow$ 2 $\Rightarrow$ 3 and 1 $\Rightarrow$ 4 $\Rightarrow$ 3, as follows.

1 $\Rightarrow$ 2: Straightforward application of L'Hôpital's rule.

2 $\Rightarrow$ 3: $\frac{1}{t}e^{\epsilon F(t)} = \exp( \epsilon F(t) - \log t ) = \exp\left[ \left(\epsilon \frac{F(t)}{\log t} - 1 \right) \log t \right]$.

1 $\Rightarrow$ 4: Using 1 and 3, $F'(t)e^{\epsilon F(t)} = t F'(t) \cdot \frac{1}{t}e^{\epsilon F(t)}$.

4 $\Rightarrow$ 3: Straightforward application of L'Hôpital's rule.

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So here are my questions:

• Are the above derivations correct?
• Can we prove that 3 $\Rightarrow$ 1? If so, then we get two closed loops 1 $\Rightarrow$ 2 $\Rightarrow$ 3 $\Rightarrow$ 1 and 1 $\Rightarrow$ 4 $\Rightarrow$ 3 $\Rightarrow$ 1, so that all the four statements are equivalent. If not, any counterexamples?

TIA...

Answer 1

Here is a counterexample of 2 $\Rightarrow$ 1:

Let $F(t) = t + \sin(t)$, then $F(t)$ is non-decreasing since $F'(t) = 1 + \cos(t) \ge 0$.

In addition, $\lim_\limits{t\to +\infty}\dfrac{F(t)}{\log t} = +\infty$, but $\lim\limits_{t\to +\infty}tF'(t)$ does not exist since $F'((2n+1)\pi) = 0$.

I believe this is also a counterexample of 3 $\Rightarrow$ 1.