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Let $f(x,y),g(x,y),h(x,y)$ be smooth convex functions on $(x,y)\in \mathbb{R}\times \mathbb{R}_{>0}$. Given a positive real number $b>0$, we define a new function $F(x,y)$

$$F(x,y)=\begin{cases} f(x,y) & \text{if }x/y\geq b \\ g(x,y) & \text{if }-b<x/y<b \\ h(x,y) & \text{if }x/y\leq -b \end{cases}$$ Roughly speaking, $F$ is glued from $f,g,$ and $h$ along lines $x/y=\pm b$.

If we furher assume that $F(x,y)$ defined in this way is continous, then is $F(x,y)$ also convex in the region $(x,y)\in \mathbb{R}\times \mathbb{R}_{>0}$?

I have found some discussions here Proving convexity of a function is a local property However, the situation is quite different since now the intersections of different regions in the definition of $F(x,y)$ have no interior points.

Kim
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    Note that this does not even work in one dimension, e.g. glueing $(x+1)^2$ for $x \le 0$ and $(x-1)^2$ for $x \ge 0$ together. – Martin R May 19 '25 at 13:41

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Take $b=1$, $f(x,y)=h(x,y)=y^2$ and $g(x,y)=x^2$.

All these functions are smooth and convex, and $F$ is continuous but not convex.

Christophe Boilley
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  • Thanks for your counterexample! Actually, I'm trying to understand the following theorem: a function defined on a convex domain is a convex function if it is subdifferentiable at every point. For a convex function, differentiable implies subdifferentiable, so it seems that the problem appears when we checking subdifferentiability of $F$ along $x/y=\pm b$? – Kim May 19 '25 at 14:20
  • Yes, that seems a good condition to check. – Christophe Boilley May 19 '25 at 15:24