Suppose that $F$ is the CDF of a standard normal distribution. Hayashi (2000) claims that the following is true $$ |\log F(v)|\leq |\log F(0)|+|v|+|v|^2\quad\text{for all}\quad v. $$ How does one prove something like this please? I think it looks like it's based on a Taylor approximation but what has happened to the error term?
Edit 2: I have plotted $$ |\log F(0)|+|v|+|v|^2-|\log F(v)| $$ for $v\in[-25,25]$. It looks something like this:
$$F(v)=\frac{1}{\sqrt{2\pi}}\int^{v}_{-\infty}\exp(-\frac{x^2}{2})\,dx$$ $$F(0)=\frac{1}{2}$$ the inequality is equivalent to $$|\log F(v)|-|\log(\frac{1}{2})|\leq|v|+|v|^2\Leftrightarrow \log\frac{1}{F(v)}-\log2\leq|v|+|v|^2\Leftrightarrow\log\frac{1}{2F(v)}\leq|v|+|v|^2$$ exponentiating both sides yields $$\frac{1}{2F(v)}\leq\exp(|v|+|v|^2)\Leftrightarrow 2F(v)\geq \exp(-|v|-|v|^2)$$ Consider the function $$G(v)=2F(v)-\exp(-|v|-|v|^2)=2F(v)-\exp(-|v|-v^2)$$ then the problem is equivalent to showing that $$G(v)\geq0$$ for all $v\in\mathbb{R}$. First consider $v\geq0$ then we have $$G(v)=2F(v)-\exp(-v-v^2)\Rightarrow G'(v)=2F'(v)+(1+2v)\exp(-v-v^2)$$$$=\frac{2}{\sqrt{2\pi}}\exp(-\frac{v^2}{2})+(1+2v)\exp(-v-v^2)$$ $$=\exp(-\frac{v^2}{2})[\frac{2}{\sqrt{2\pi}}+(1+2v)\exp(-v-\frac{v^2}{2})]\geq0$$ for all $v\geq0$. Therefore $$G(v)\geq G(0)=0$$ for all $v\in [0,\infty)$. Secondly let $v<0$ then we obtain $$G(v)=2F(v)-\exp(v-v^2)\Rightarrow \frac{dG}{dv}=2\frac{dF}{dv}-(1-2v)\exp(v-v^2)$$ $$=\exp(-\frac{v^2}{2})[\frac{2}{\sqrt{2\pi}}-(1-2v)\exp(v-\frac{v^2}{2})]$$ But the function $$g(v)=\frac{2}{\sqrt{2\pi}}-(1-2v)\exp(v-\frac{v^2}{2})\geq g(\frac{-3-\sqrt{17}}{4})>0$$ for all $v\in (-\infty,0)$. Therefore $$\frac{dG}{dv}=\exp(-\frac{v^2}{2})\cdot g(v)>0$$ for all $v\in (-\infty,0)$. Since the derivative is positive in this domain then it must be true that $$G(v)\geq \lim_{v\to-\infty}G(v)=\lim_{v\to-\infty}\{2F(v)-\exp(v-v^2)\}=0$$ for all $v\in (-\infty,0)$.