# When is the optimum of an optimization problem affine in the constraint parameter?

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 $$F(s)=\begin{cases} 1 - 3 s - 2s^{3/2} &\text{ if } 0 and similarly for $$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 $$f'(x(s))=2(x(s)-1)=-2y(s).$$

• Looks to me like it takes a "perfect storm" to overcome the non-convexity of $xy=s$. Commented Jun 10, 2020 at 15:51
• 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). Commented Jun 29, 2020 at 7:38

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 $$g(z)=x.
Conversely, suppose now that $$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]$$.
• 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?... Commented Feb 17, 2021 at 13:58
• This is what happens in the example described in the question, where $f(x)=(x-1)^2$, and $g(x)=1-x$. Commented Feb 17, 2021 at 13:58