Timeline for Smooth convex extensibility of combination of two line segments
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2014 at 0:54 | comment | added | Pietro Majer | (in other words: take, in your last picture, the straight segments not necessarily parallel, but let each of them join a couple of points with the same slope wrto $y$). | |
| Apr 9, 2014 at 0:47 | comment | added | Pietro Majer | A slight variant of your function: define, on the vertical edges, $f(0,y)$ and $f(1,y)$ to be strictly convex, with $\partial_y f(0,0)=\partial_y f(1,0)$ and $\partial_y f(0,1)=\partial_y f(1,1)$. Then join by segments $(0,u,f(0,u))$ and $(1,v,f(1,v))$ whenever $\partial_y f(0,u)=\partial_y f(1,v)$. This should give the graph of a smooth convex function. | |
| Apr 8, 2014 at 23:33 | comment | added | Pietro Majer | I like your suggested construction , that maybe could be formalized as a solution of a suitable PDE with boundary conditions? A doubt on the picture: any $C^1$ function $f$ on the square, such that $f(0,t)=f(t,0)=0$ for $0\le t\le 1$, also verifies $Df(0,0)=0$. If it is convex, this implies that $(0,0)$ is a global minimum, so $f \ge0$ . | |
| Apr 8, 2014 at 22:40 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added image of the idea.
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| Apr 8, 2014 at 10:40 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
catenary ==> parabola
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| Apr 8, 2014 at 1:40 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |