# Is the optimum of this problem convex in the constraint parameter?

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

Suppose also that $$\lim_{x \to \infty} |f(x)| = \infty$$. For any $$s \in (0,1)$$, define $$F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y)$$

(The minimum exists since $$|f|$$ diverges at infinity.)

Question: Does there exist a convex function $$g(s)$$ such that $$F=g^p$$ for some $$p \ge 1$$? I do not require $$g$$ to be positive.

Here are two examples where this happens:

Linear penalization: $$f(x)=x-1$$. In that case $$F(s) = \begin{cases} 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\ 1-2s, & \text{ if }\, s \le \frac{1}{4} \end{cases}$$ is convex, since $$F'(s)$$ is non-decreasing.

Logarithmic penalization: $$f(x)=\log x$$. In that case $$F(s)=2f^2(\sqrt s)=\frac{1}{2}(\log s)^2$$ is not convex.

However, we have $$F(s)=g^2(s)$$ where $$g(s)=-\frac{1}{\sqrt 2}\log s$$ which is convex.

Is there a general phenomena lying behind these two examples?

• Yes, thank you very much. This was a typo (fixed now). Commented Apr 14, 2020 at 14:12

The answer is no. E.g., let $$f(x):=|x-1|^{3/2}$$. Then $$F(s)=\begin{cases} F_1(s) &\text{ if } 0 where $$F_1(s):=1 - 3 s - 2s^{3/2},$$ $$F_2(s):=2 + 6 s - 2(3 + s)s^{1/2}.$$ One may note here that $$F_1(1/9)=16/27=F_2(1/9)$$ and $$F'_1(1/9)=-4=F'_2(1/9)$$.

From the definition of $$F$$, it is clear that $$F>0$$ on $$(0,1)$$. So, letting $$a:=1/p\in(0,1]$$, we see that the desired goal was that $$h:=F^a$$ be either convex or (if $$p$$ is even) concave. (If $$h$$ is concave and $$p$$ is even, we can take $$g:=-h$$. Then $$g$$ will be convex and we will also have $$g^p=h^p=F$$.) So, letting $$h_j:=F_j^a$$ for $$j=1,2$$, we see that we must have one of the following cases:

(i) $$h_1$$ is convex on $$(0,1/9]$$ and $$h_2$$ is convex on $$[1/9,1)$$;

(ii) $$h_1$$ is concave on $$(0,1/9]$$ and $$h_2$$ is concave on $$[1/9,1)$$.

However, $$h_1''(1/9)$$ equals $$3a-4$$ in sign and hence is $$<0$$ for $$a\in(0,1]$$, whereas $$h_2''(1/9)$$ equals $$2+3a$$ in sign and hence is $$>0$$ for $$a\in(0,1]$$. So, neither one of the cases (i) or (ii) can take place.

Here is the graph of $$F''$$:

• On a second thought, I think that the answer could be remedied as follows: Looking at $\text{sgn}(h_1^{''})=\text{sgn}\big((a-1)F_1^{'}+F_1F_1^{''}\big)$ when $s \to 0$ we must have $h_1^{''}(s)<0$ for sufficiently small $s$. Indeed, since $F_1$ and $F_1^{'}$ tend to finite values at zero, but $F_1''$ tend to $-\infty$, the sum $(a-1)F_1^{'}+F_1F_1^{''}$ must be negative (here we also use the fact that $F_1$ is positive-and in fact tends to $1$ at zero). Do you agree with my analysis? Commented Apr 15, 2020 at 6:24
• @AsafShachar : My calculation was correct. The mistake is actually in your calculation of the expression for $h''$, which must have $F'^2$ in place of $F'$. Commented Apr 15, 2020 at 12:11
• Thank you. Indeed, this was a silly mistake of mine. This is a great answer. I have one more question if you will: Can you say why did you delete your previous example with $f(x)=(x-1)^2$ instead of $f(x)=|x-1|^{\frac{3}{2}}$? In particular, I wonder whether the fact that $f(x)=|x-1|^{\frac{3}{2}}$ is non-smooth was essential in this non-convexity phenomenon, or is it possible to produce counter-examples with smooth $f$. Also, I wonder if you used some program for calculating the minimum $F(s)$ or just calculus...Thanks. Commented Apr 19, 2020 at 14:53
• @AsafShachar : I replaced $f(x):=|x-1|^2$ by $f(x):=|x-1|^{3/2}$ to simplify the resulting expressions. Concerning the smoothness condition on $f$: I'd understand that it can be added hoping that it can help avoid technical complications in a proof of a positive answer. However, I don't see a good point in insisting that this condition be satisfied in a counterexample. Commented Apr 19, 2020 at 18:24
• Previous comment continued: Yet, if you want to insist on the smoothness of $f$, there is a choice: either (i) go back to $f(x):=|x-1|^2$ (found in previous edits) and then deal with the more complicated expressions or (ii) approximate $f(x):=|x-1|^{3/2}$ near $x=0$ (say uniformly) by a smooth function so that the resulting strict inequalities hold. Commented Apr 19, 2020 at 18:25