Let $a$ be a strictly positive $C^\infty$ smooth function on the unit interval. Does there exist a strictly positive $C^\infty$ smooth function $f$ on $I$ such that $$ f’’(x) \leq 0\quad \text{and} \quad (af)’’(x) \leq 0$$ for all $x \in I$?
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2$\begingroup$ Why convex, not concave, in the title? $\endgroup$– Dieter KadelkaCommented Nov 24, 2022 at 17:05
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2$\begingroup$ Isn’t convex/concave just a matter of notation? I will change the title $\endgroup$– AliCommented Nov 24, 2022 at 17:10
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1 Answer
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Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$.
Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $a(1)=1$ and $a(x)=a(-x)\forall x$.
Suppose that we have a convex function $f:[-2,2]\to\mathbb{R}$ such that $f''\leq0$, $(af)''\leq0$. We can assume $f'(0)\leq0$ (if not change $f(x)$ by $f(-x)$). So $f$ is decreasing in $[0,2]$ and we have $f(0)\geq f(1)$. Now apply Jensen's inequality to $af$ to obtain a contradiction:
$$f(1)=(af)(1)\geq\frac{1}{2}((af)(0)+(af)(2))\geq\frac{1}{2}(af)(0)=\frac{3}{2}f(0)\geq\frac{3}{2}f(1)>f(1)$$
