Let $f:\mathbb R\to\mathbb R$ $$f(x)=\frac{a}{12}\,(x^4-2\,x^2)+b\,x\,$$ with parameters $a,b>0\,$. $f$ is concave on the interval $\left[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right]$ and convex outside.
I am interested in the convex function $g:\mathbb R\to\mathbb R$ which is as close as possible to $f$. Precisely, I want to minimise the oscilation $$\textrm{osc}(f-g) := \sup_{x\in\mathbb R}(f(x)-g(x)) - \inf_{x\in\mathbb R}(f(x)-g(x)) \;.$$
Is it possible to compute $g$ (or a good approximation of it)? Can we get a good estimate of $\textrm{osc}(f-g)$ as a function of the parameters $a,b$? Can we estimate $\textrm{osc}(f-g)$ when $b$ is large?
It may be useful to distinguish the following two cases.
- If $3^5\,b^2<4\,a^2$, $f$ should have two minimum points $m_1,m_2$ such that $m_1<-\frac{1}{\sqrt{3}}<\frac{1}{\sqrt{3}}<m_2$. In this case intuitively I would build $g$ drawing a line that connects $(m_1+\delta_1,f(m_1+\delta_1))$ and $(m_2+\delta_2,f(m_2+\delta_2))$ for suitable $\delta_1,\delta_2>0$. Which $\delta_1,\delta_2$?
- If $3^5\,b^2>4\,a^2$, $f$ should have only one minimum point $m<-\frac{1}{\sqrt{3}}$. In this case I have less intuition about what to do...
