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Let's take a function $f=sech(x)$ as an example. It is strongly convex in a limited range of $x$.

It is further assumed that:

  1. $f$ is a "black box", $x$ as input ;
  2. output is the value of the function $f(x)$ ;
  3. as well as the gradient $G(x)$ and hessian $H(x)$ ;

How to convert $f(x)$ to a convex one on an infinite interval, i.e. $x=[-\inf;\inf]$, knowing only these 3 output signals.

Clear["Derivative"]
ClearAll["Global`*"]


Plot[{Sech[x], -x^2 + 1}, {x, -1.7, 1.7}, PlotRange -> Full,
PlotPoints -> 100]

The figure shows an example. What is (blue) and what should be (orange).

enter image description here

ayr
  • 793

1 Answers1

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Unless I am missing some subtlety, the problem is very straightforward. If the Hessian is computed in the range for $x$ where $f$ is strongly convex, then $\nabla^2 f( x )$ (I prefer the nabla notation for the Hessian and gradient) is definite positive.

Therefore, the second order Taylor expansion will do the job because it will be strongly convex $$ \tilde f( y ) := f( x ) + \nabla f( x )^T( y - x ) + \frac12 ( y - x )^T\nabla^2f( x )( y - x ). $$ Notice that we have

  1. $\tilde f( x ) = f( x )$
  2. $\nabla \tilde f( x ) = \nabla f( x )$
  3. $\nabla^2\tilde f( x ) = \nabla^2f( x )$

I hope I have not just solved someone's homework. It seems that there is a lot of basic things about convexity that have not been properly learned, if that is indeed the case. So, I leave a couple of important exercises here.

Exercise 1: prove that the Hessian of a strongly convex function is definite positive

Exercise 2: prove that if $\nabla^2 f( x )$ is definite positive, then $\tilde f$ defined above is strongly convex ;-)

Elias
  • 596