Let $s= s_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))$ where $\zeta_0 = \left(\frac{3\pi}{4}\right)^{2/3}$ and $s_0 = b^{-2/3}\zeta_0.$ Note that here $b$ is a parameter.

We define $\omega(s) = \exp\left(\frac{2}{3}s_{+}^2\right)$ where $s_{+} = \max(0,s).$ I want to show that for some constant $C$ we have that, $$\omega(s) \leq C\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right)\leq C\exp\left(-\frac{1}{2\sqrt{b}}\right)\Sigma_{b}^{-1},$$ where $\Sigma_{b}^{-1}=\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right).$

For the second inequality does not make sense to me as $\exp\left(-\frac{1}{2\sqrt{b}}\right)\leq 1$, but I might be wrong.

For the first inequality, $$\frac{2}{3}s_{+}^{2}\leq \frac{2}{3}|s|^2\leq \frac{4}{3}|s_0|^2+\frac{8}{3}(\zeta_0^{-1}b^{-1/3} + C\zeta_0^{-1}b^{2/3})$$ $$=\frac{4}{3}\left(\frac{3\pi}{4b}\right)^{4/3}+\frac{8}{3}\left(\left(\frac{4}{3\pi\sqrt{b}}\right)^{2/3} + C\left(\frac{4b}{3\pi}\right)^{2/3}\right)$$

where I used the inequality $|a+b+c|^2\leq 2|a|^2+4(|b|^2+|c|^2).$ I cannot see how I reduce this expression to derive the first inequality. Any hints would be much appreciated.


1 Answer 1


The second inequality never holds.

The first inequality is equivalent to $\frac 23s_+^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$ for some constant $D$. Since $\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\ge -\frac 1{4\pi}$ for each $b\ge 0$, if $D\ge -\frac 1{4\pi}$ then the inequality holds for $s_+=0$. So it remains to consider an inequality

$\frac 23s^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$

$\frac 23\left(b^{-2/3}\zeta_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))\right)^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$

For big $b$ this inequality can fail. For instance, when $\mathcal{O}(\sqrt{b})$ term equals $-1-\sqrt{b}$, then when $b$ tends to infinity, the left hand side of the inequality tends to infinity, whereas the right hand side is bounded. For small $b$ the inequality fails too. Indeed, when $b$ tends to zero, the left hand side of the inequality grows as $b^{-4/3}$, whereas the right hand grows as $b^{-1}$.


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