Prove that $\frac12 (e^{\alpha t^2}-1) -F(t)$ is bounded from below

Solution 1:

Set $g(t):= \frac{1}{2}(e^{\alpha t^2}-1)-F(t)$ $(t \in \mathbb{R})$. Then $g'(t) = (\alpha t -1)e^{\alpha t^2} +1$. Thus $g'(t) \to \pm \infty$ $(t \to \pm \infty)$. Hence $g(t) \to \infty$ $(t \to \pm \infty)$. Since $g$ is continuous $\min g(\mathbb{R})$ exists.

Edit: There are $a < 0$ and $b > 0$ such that $g'(t) \le -1$ $(t \le a)$ and $g'(t) \ge 1$ $(t \ge b)$. So, by the Mean Value Theorem $g(t) -g(b) \ge t-b$ $(t \ge b)$. Thus $g(t) \to \infty$ $(t \to \infty)$. The limit $g(t) \to \infty$ $(t \to -\infty)$ can be proved the same way. Next, $g(0)=0$. Since $g(t) \to \infty$ $(t \to \pm \infty)$ there is a compact interval $[-c,c]$ such that $g(t)>0$ for $|t|>c$. Hence $\min g(\mathbb{R}) = \min g([-c,c])$.