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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 $\delta_1,\delta_2$?

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  • 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...

enter image description here

Rodrigo de Azevedo
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tituf
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  • What does it mean to minimize $\operatorname{osc}(f, g)$? Does it mean to minimize its integral across the entire domain $\mathbb{R}$? Or something else? – Brian Tung Mar 13 '23 at 22:13
  • @BrianTung $osc(f-g)$ is just a constant defined as the $sup$ minus the $inf$ of the function $f(x)-g(x)$ over $x\in\mathbb R$. I wrote the definition more explicitly. – tituf Mar 13 '23 at 22:17
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    Intuitively, I would anticipate that the minimizer is the convex envelope of the original function. See for instance the image here. – Semiclassical Mar 13 '23 at 22:33
  • @Semiclassical is there a method to compute it? I am mainly interested in estimating $\textrm{osc}(f-g)$ when the parameter $b$ is large – tituf Mar 13 '23 at 22:43
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    The value of $b$ shouldn't make a difference to the optimal oscillation. If you replace $g(x) $ with $g(x) - bx$ you get another arbitrary convex function. This makes the problem equivalent to the $b = 0$ problem. – Ben Martin Mar 14 '23 at 06:32
  • @BenMartin I have to think for a moment about what you are saying. If $b=0$ I think the problem is easy because the two minima are equal ($f(m_1)=f(m_2)$) and I can draw a horizontal line connecting $(m_1,f(m_1))$ and $(m_2,f(m_2))$ to convexify the function. So when $b=0$ it should be $\textrm{osc}(f-g)=\frac{a}{12}$. – tituf Mar 14 '23 at 09:53
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    Finding the convex envelope of the curve amounts to finding the convex hull of the area above said curve. There's a number of convex hull algorithms, but typically these are for point sets not curves. One exception to this is Lucet's LLT algorithm (original paper here, SciLab implementation here.) – Semiclassical Mar 14 '23 at 15:28
  • @BenMartin I think you are right, the minimum of $\textrm{osc(f-g)}$ over convex functions $g$ does not depend on the parameter $b$. Therefore it suffices to compute it for $b=0$. If you want to write a complete answer, I'll vote it – tituf Mar 14 '23 at 17:59
  • @Semiclassical If I understand correctly, defining the Legendre transform $f^(y)=\sup_x(xy-f(x))$ we get that $f^*$ is the convex envelope of $f$ (for any continuous function $f$). So this is the theoretical answer to my question. Then one should figure out how to compute the Legendre transform – tituf Mar 14 '23 at 18:02
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    Yes: the double Legendre transform returns the convex envelope. For more discussion see the paper here. – Semiclassical Mar 14 '23 at 19:47
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    Also, one way to make this particular problem easier to analyze is to write the function of interest as $f(x)=\frac{a}{12}(x^2-1)^2+b x-\frac{a}{12}$. Then the convex envelope in the middle is obtained by finding a linear function $f_c(x)$ which is tangent to the curve at two points, i.e., $f(x)-f_c(x)$ has two double roots. Is there an obvious choice for $f_c(x)$? – Semiclassical Mar 14 '23 at 20:11
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    @Semiclassical very nice, it suffices to take $f_c(x)=b x - \frac{a}{12}$ and this give the convex envelope. So again, as pointed out by BenMartin, the minimal oscilation does not depend on $b$ – tituf Mar 15 '23 at 10:49
  • @tituf Please consider answering your own question when you can find some time. – Rodrigo de Azevedo Mar 18 '23 at 07:27

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