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Let $h:\mathbb R^{>0}\to \mathbb R^{\ge 0}$ be a smooth function, satisfying $h(1)=0$, and suppose that $h(x)$ is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$.

Let $s>0$ be a parameter, and define $ F(s)=\min_{xy=s,x,y>0} g(x,y), $ where $g(x,y):=h(x)+ h(y)$.

Question: Can the minimum be obtained at two essentially different points?

That is, suppose that $F(s)=g(x,y)=g(\tilde x,\tilde y)$, for some $x,y,\tilde x,\tilde y>0$ satisfying $xy=\tilde x \tilde y=s$. Is it true that

$$ (x,y)=(\tilde x,\tilde y) \, \, \, \text{ or } \,\, (x,y)=(\tilde y,\tilde x)?$$

By symmetry, we can assume W.L.O.G that $x \le \sqrt{s}$.

It is not hard to see that the minimum must be obtained at a point where $x, y \le 1$ (if $s \le 1$) or $x,y \ge 1$ (if $s \ge 1$). Thus, if $s \le 1$, then we have $x,y=\frac{s}{x} \le 1$, which implies $s \le x \le \sqrt{s}$.

Edit:

I tried to produce counter-examples by using $g$ which are invariant under some automorphism of the hyperbola $xy=s$. (Then the set of minimizers is closed under the operation of this automorphism). I couldn't find such an automorphism which preserve the special additive structure of $g$.

Here is a partial analysis of the question for local minima:

Set $\psi(x)=h(x)+h(\frac{s}{x})$. Then

$$\psi'(x)=h'(x)-h'(\frac{s}{x})\frac{s}{x^2}, \tag{1}$$

and

$$\psi''(x)=h''(x)+h''(\frac{s}{x})\frac{s^2}{x^4}+2h'(\frac{s}{x})\frac{s}{x^3}. \tag{2}$$

Now, suppose $x$ is a local minimum of $\psi$. Then, equations $(1),(2)$ imply that

$$ h'(x)=h'(\frac{s}{x})\frac{s}{x^2} \, \, , \, \, h''(x)+h''(\frac{s}{x})\frac{s^2}{x^4}+2\frac{h'(x)}{x} \ge 0\tag{3}. $$

Subquestion: Suppose that $x,y$ satisfy $(3)$. Does $x=y$ or $x=\frac{s}{y} $ hold?

mathworker21
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Asaf Shachar
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  • do you absolutely need smoothness? $h(x) = -(x-2)^2+1$ for $1 \le x \le 2$ and $h(x) = \frac{-1}{4}(x-4)^2+2$ has $(x,y) = (2,2),(1,4)$ both minimum for $s=4$. – mathworker21 Apr 23 '20 at 16:51
  • Well, I think that I prefer $h$ to be at least $C^1$, but counter-examples with continuous $h$ are also very interesting, at least to understand the limitations and scope of what can be expected here. Did you mean to take $h(x) = -(x-2)^2+1$ for any $0<x \le 2$? You wrote $1 \le x$, but your two branches do not agree at $x=1$. I think it is interesting to understand if the fact that your $h$ fails to be differentiable at one of the minimum points $x=2$ is a coincidence or not. I suggest that you will make your comment into a partial answer. – Asaf Shachar Apr 23 '20 at 17:45
  • $-(x-2)^2+1$ for $1 \le x \le 2$ and $\frac{-1}{4}(x-4)^2+2$ for $x \ge 2$. If I understand your question correctly, I don't have to care about $x < 1$. – mathworker21 Apr 23 '20 at 18:11
  • And I think the example can be made smooth, but I don't think I can write it down algebraically easily. – mathworker21 Apr 23 '20 at 18:12
  • Oh yes, you are right of course about the part $x<1$ being irrelevant. But then your extended $h$ won't be differentiable at $x=1$ as well, if we want it to stay positive when $x<1$. Thus it won't be differentiable at both minima points. I find what you are saying-about the possibility of smoothing the example-to be quite reasonable but I am not 100% sure. Again, I think what you found deserves to become a (partial) answer. Thanks for the effort and the clarifications. – Asaf Shachar Apr 23 '20 at 18:21
  • Ah, good point about not being able to be differentiable at $x=1$ as well. I'll think a bit more about smoothing it out. If I can't say something concrete, I'll write a partial answer. I am similarly not completely sure whether it can be smoothed out – mathworker21 Apr 23 '20 at 18:59
  • Final (technical) comment: you can take $h(x)=\frac{-1}{4}(x-4)^2+2$ only up to $x \le 4$. After that, this $h$ starts decreasing, so you need to replace it by something which continue to increase with $x$. But this formal change won't affect the minimum points you have described. – Asaf Shachar Apr 24 '20 at 05:40

1 Answers1

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Yes, it's possible. Define $$h(x)=\begin{cases} (x-1)^2 & x\in (0,2] \\ 2-(\frac{4}{x}-1)^2 & x\in [2+\epsilon,3] \end{cases} $$ for some small $\epsilon$. We'll deal with the values on other domains in a bit, but let's first see what this gets us. For $s=4$, we have $g(2,2)=2$ and $g(x,y) = h(x)+h(4/x) = 2-(\frac{4}{x}-1)^2+(\frac{4}{x}-1)^2 = 2$ for $x>2+\epsilon$. So in fact $g(x,y)$ is constant along the segment of the hyperbola $xy=s=4$ thus defined.

To finish the job, just smoothly interpolate $h$ on the interval $[2,2+\epsilon]$ in such a way that it is greater than $2-(\frac{4}{x}-1)^2$, and similarly finish it off with a smooth segment on $[3,\infty]$ which is increasing and greater than $2-(\frac{4}{x}-1)^2$. This ensures that the minimum is on the segment we defined in the first place, and here there are infinitely many distinct minima.

If you prefer isolated minima, you can add small (so as to keep the resulting function increasing), smooth bump functions to $h$ as defined above in all but a discrete set of points.

Yly
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