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While working on a variational problem I have reached to the following question:

Let $f:(0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $f(1)=0$. Suppose that $f(x)$ is strictly increasing on $[1,\infty)$ and strictly decreasing on $(0,1]$. Define $F:(0,1) \to [0,\infty)$ by $$ F(s)=\min_{xy=s,x,y\in(0,\infty)} f(x)+ f(y). $$

Question: For which functions $f$, $F(s)$ has an affine part? Can we characterize such functions?

The motivation is that I am applying Jensen inequality with $F$, and an affine part (in contrast to strict convexity) gives some flexiblity.


The only example that I know of is when $f(x)=(x-1)^2$, and $$ F(s) = \begin{cases} 1-2s, & \text{ if }\, 0 \le s \le \frac{1}{4} \\ 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4}, \end{cases} $$ is affine on $[0,\frac{1}{4}]$.

Is this $f$ the only choice which makes $F$ affine?

For qubic and quartic penalizations this is not the case; if $f(x)=(1-x)^3$, then $$ F(s)=\begin{cases} 1 - 3 s - 2s^{3/2} &\text{ if } 0<s\le1/9, \\ 2 + 6 s - 2(3 + s)s^{1/2} &\text{ if } 1/9\le s<1. \end{cases} $$ and similarly for $f(x)=(x-1)^4$.


Here is an attempted analysis:

Assume that there is a $C^1$ map $s \to (x(s),y(s))$ giving a minimizer to the problem, i.e. for any $s \in (0,1]$ $$ F(s)=f(x(s))+f(y(s)), \, \, \,x(s)y(s)=s. \tag{1} $$ Lagrange's multipliers give $$ f'(x(s))=\lambda(s) y(s),f'(y(s))=\lambda(s) x(s). \tag{2} $$ $F'(s)=\lambda (s)$, so $F''(s)=0$ if and only if $\lambda(s) < 0$ is constant. ($\lambda < 0$ since $f'|_{(0,1)} < 0$ by our assumption.)

I don't see how to proceed from here.

For $f(x)=(x-1)^2$ we have $\lambda(s)=-2$: The minimum (for $s \in [0,\frac{1}{4}]$) is obtained at $x(s)+y(s)=1$, so $ f'(x(s))=2(x(s)-1)=-2y(s). $

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  • $\begingroup$ Looks to me like it takes a "perfect storm" to overcome the non-convexity of $xy=s$. $\endgroup$ Commented Jun 10, 2020 at 15:51
  • $\begingroup$ Thanks, I think you might be right. I tried to solve a "easier " question: When is the optimum $F(s)$ affine? This miracle seems even more surprising than convexity. I have edited the question to include an attempted proof for that subproblem. (It reduces to a coupled ODE+a functional equation. which I don't know how to analyze). $\endgroup$ Commented Jun 29, 2020 at 7:38

1 Answer 1

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Say that $F$ is affine on an interval $I$, with $F'\equiv\lambda$, a constant. Then for $x=x(s)$ and $y=y(s)$, one has $f'(x)=\lambda y$ and $f'(y)=\lambda x$, thus $$xf'(x)\quad (=\lambda xy)\quad=yf'(y).$$ This implies a functional equation $$xf'(x)=\frac1\lambda f'(x)f'(\frac1\lambda f'(x)).$$ Simplifying, one finds that the function $\frac1\lambda f'=:g$ is a functional square root of the identity: $$g\circ g={\rm id}_I.$$ Notice that if $g$ is not the identity itself, then it tends to be a dcreasing function: suppose $g(x)\ne x$, say $z:=g(x)>x$, then $g(z)=x<g(x)$.

Conversely, suppose now that $g$ is a decreasing square root of the identity. Then choose a constant $\lambda$ and define $f$ by $f'=\lambda g$. Suppose in addition that $x\mapsto xg(x)$ is strictly convex (perhaps too strong a hypothesis). Then the level sets of $xf'(x)$ consist in pairs $(x,y)$, which turn out to be the $(x(s),y(s))$ above, where $s:=xy$. Then $f$ answers your query.

Now, to construct a functional square root of identity, one proceeds as follows. Choose arbitrarily a point $a>0$ and $g:[0,a]\rightarrow{\mathbb R}$ a decreasing function such that $g(a)=a$. Let $b:=g(0)$, so that $g([0,a])=[a,b]$. Then extend the definition of $g$ to $(a,b]$ by $g(x):=g^{-1}(x)$. Then $g\circ g$ over $[0,b]$.

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  • $\begingroup$ Thank you, that is a very interesting answer. Just to make sure that I understand the 'forward' direction: Suppose that you start with a decreasing square root of the identity $g$ such that $xg(x)$ is strictly convex, and set $f'=\lambda g$ as you described.Then if $x(s),y(s)$ are minimizers, then $xg(x)=yg(y)$ (by the Lagrange multiplier argument) - so are uniquely determined up to order by the convexity assumption on $xg(x)$. (I guess we could also require $xg(x)$ to be strictly concave, right?... $\endgroup$ Commented Feb 17, 2021 at 13:58
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    $\begingroup$ This is what happens in the example described in the question, where $f(x)=(x-1)^2$, and $g(x)=1-x$. $\endgroup$ Commented Feb 17, 2021 at 13:58

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