How to bound a function

Question

I want to prove that there exists a constant $c>0$ such that

$$e^{\Big(tanh(\frac{t}{2})-\frac{t}{2}\Big)b^2}\le e^{-ct^3b^2}$$ for all $b\in\mathbb{R}$ and for all $t\le 1$.

Please help me to do so. Thanks

Answer 1

The exponential function is monotonically increasing, so you want $c$ such that $$ \left( \tanh(t/2) - t/2 \right) b^2 \leq -c t^3 b^2 \text{.} $$
Since $b$ is real, either $b = 0$, giving equality in the above, or $b^2 > 0$ and we can divide it from both sides, giving $$ \tanh(t/2) - t/2 \leq -c t^3 \text{.} $$
Recall the power series for $\tanh(t/2) = \frac{t}{2} - \frac{t^3}{24} + \frac{t^5}{240} - \frac{17t^7}{40320} + \cdots$, so $$ \tanh(t/2) - t/2 \in \left( \frac{-t^3}{24}, \frac{-t^3}{24} + \frac{t^5}{240} \right)$$ for all $t$ in the radius of convergence. So take $c = 1/24$, since $$ \tanh(t/2) - t/2 \leq -\frac{1}{24} t^3 $$ within that radius. The radius of convergence is $|t| < \pi$, so the above resolves the question for $t \in (-\pi, 1]$.

At $t = -\pi$, $\tanh(-\pi/2)- \frac{-\pi}{2} = 0.65364\dots{} < 1.2919\dots{} = \frac{-1}{24}(-\pi)^3$. For $t < -\pi$, we have $\frac{\mathrm{d}}{\mathrm{d}t} \tanh(t/2) - t/2 = -1/2 + \frac{1}{2}\mathrm{sech}^2(t/2) > -1/2$ and $\frac{\mathrm{d}}{\mathrm{d}t} \frac{-t^3}{24} < \frac{-\pi^2}{8}< -1$, so the ordering of the two expressions at $t = -\pi$ does not reverse anywhere to the left of $-\pi$. That is, the inequality holds on $(-\infty, -\pi]$ as well.

(Note: There are many ways to show what is essentially a line, $\tanh(t/2) - t/2 < -t/2$ is less than the cubic, $-c t^3$, on $(-\infty, \pi]$. I've just picked a relatively elementary method above: the cubic is always descending faster than $\tanh(t/2) - t/2$ and they have not crossed by $t= -\pi$.)