Majorization in a Neighborhood

Consider the cubic $f(x)=2x-x^2+\frac16 x^3$. It is an increasing function on the real line, with a root at $x=0$, and a saddle point at $x=2$.

We know that cubics do not have quadratic majorizers, but we can try to find a quadratic $$g(x,y)=f(y)+f'(y)(x-y)+\frac12 a(x-y)^2$$ such that $g(x,y)\geq f(x)$ for all $x\geq 2$ and such that $$y-\frac{f'(y)}{a}\geq 2$$ if $y\geq 2$.

The condition $g(x,y)\geq f(x)$ for all $x\geq 2$ is $$a\geq f''(y)+\frac13 f'''(y)(2-y),$$ which is in our case $$a\geq \frac23(y-2).$$ We have $$y-\frac{f'(y)}{a}\geq 2$$ for $$a\geq\frac12(y-2).$$

Minimize $f:x\rightarrow x^4$ on $[-1,+1]$, where $f''(x)\leq 12$, and thus if $-1\leq x\leq +1$ and $-1\leq y\leq +1$ $$f(x)\leq g(x,y)=y^4+4y^3(x-y)+6(x-y)^2.$$ The majorizer is minimized at $x=y-\frac13 y^3$. Thus the majorization algorithm is $$x^{(k+1)}=x^{(k)}-\frac13 (x^{(k)})^3,$$ which converges monotonically to zero if started with $-1\leq x^{(0)}\leq +1$. Because $$\lim_{k\rightarrow\infty}\frac{x^{(k+1)}}{x^{(k)}}=1$$ convergence is sublinear.