Is the optimum of this constraint optimization problem smooth everywhere except at one jump?

Question

Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(1)=0$, and suppose that $\,|f(x)|$ is strictly increasing when $x \ge 1$, and strictly decreasing when $x \le 1$.

For any $s>0$, define $$ F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y). $$

If I am not mistaken, the map $s \to F(s)$ is continuous.

Question: Does there exist an $s^* >0$ such that $F|_{(0,s^*)},F|_{(s^*,\infty)}$ are smooth?

I ask if $F(s)$ is piecewise smooth, with at most one "jump" point. Can there be more than one?

Here is an example which shows that we must allow for at least one point of non-regularity:

Linear penalization: Take $f(x)=x-1$. In that case $$F(s) = \begin{cases} 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\ 1-2s, & \text{ if }\, s \le \frac{1}{4} \end{cases} $$ $F$ is $C^1$ but not $C^2$.

(for a more involved example, see here).

Of course, $F(s)$ can be smooth, e.g. when $f(x)=\log x$. In that case $ F(s)=2f^2(\sqrt s)=\frac{1}{2}(\log s)^2.$

Answer 1

It seems that this answer here settles the question. There can be more than one transition point. (In fact, there can be more than one transition point in $(0,1)$ and also more than one in $(1,\infty)$.

Answer 2

This is not quite an answer, but if $xf(x)f'(x)$ is strictly increasing, then $F(s)$ is smooth.

To simplify the notation, let me write $g(x)=f^2(x)$. You want to minimize $g(x)+g(y)$ subject to the constraint $xy=s$. Using Lagrange multipliers, this reduces to solving $$g'(x) = \lambda y ,\quad g'(y) = \lambda x,\quad xy=s. $$ This implies $x g'(x) = y g'(y)$, which has only the trivial solution $x=y=\sqrt{s}$ if $x g'(x)$ is strictly increasing.

This also differentiates between your two examples; in the first one $x f(x)f'(x) = x(1-x)$, which is not monotone, and in the second, $x f(x)f'(x) = \log x$ is monotone.

I guess you can draw more conclusions from this equality $x g'(x) = y g'(y)$ regarding the cases in which there exist solutions that are not $x=y=\sqrt{s}$, which is the source of a possible failure of smoothness.