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This is a refined version of my earlier question Convex extensibility of combination of two lines.

Is there a smooth function $f:[0,1]\times [0,1]\rightarrow\mathbb R$ such that for all $x\in [0,1]$, $$ f(x,1)=x,\qquad f(x,0)=0, $$ and $f$ is convex or concave?

If yes, is there a "nice" example?

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A cheap convex solution on $\mathbb{R}^2$ is $$f_0(x,y):= \big(x+3y-2\big)_+ -y \, ,$$
which also verifies $f_0(x,y)=-y$ for all $(x,y)$ in the rectangle $$[-1/4, 5/4]\times [-1/4,1/4]=\big([0,1]\times\{0 \}\big)\;{\bf +}\; [-1/4,1/4]^2 \subset\{x+3y-2\le0\} \, , $$ and $f_0(x,y)=x+2(y-1)$ in the rectangle $$[-1/4, 5/4]\times [3/4,5/4]=\big([0,1]\times\{1 \}\big)\big)\;{\bf +}\; [-1/4,1/4]^2 \subset\{x+3y-2\ge0\} \, .$$ As a consequence, if $\phi$ is a symmetric $C^\infty$ convolution kernel with support in $[-1/4,1/4]^2$, the function $f:=f_0*\phi$ is a $C^\infty$ convex function on $\mathbb{R}^2$ satisfying $f(x,0)=0$ and $f(x,1)=x$.

(We can take e.g. $\phi(x,y):=\psi(x)\psi(y)$ with $\psi\in C^\infty(\mathbb{R})$, $\psi\ge0$, $\psi(-t)=\psi(t)$, $\operatorname{supp}(\psi)\subset[-1/4,1/4]$, $\int_\mathbb{R}\psi(t)dt=1$ ).

For a concave $C^\infty$ solution, the same construction works with $$f_0(x,y):=2y-\big(3y-x-1\big)_+\; $$ and in fact it can be adapted to a more general situation, as the main point of it is just, that convolution with a non-negative mollifier with compact support preserves convexity, and also fixes any affine function, if the mollifier is symmetric (meaning $\phi(x)=-\phi(x)$ for all $x$).

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  • $\begingroup$ Is $\phi$ just a bump function like $e^{-1/x^2}$? $\endgroup$ – Bjørn Kjos-Hanssen Apr 8 '14 at 19:51
  • $\begingroup$ Yes, say $\phi(x)=c\exp(-1/(1-|4x|^2))$ if $|x|<1/4$, and $0$ otherwise. $\endgroup$ – Pietro Majer Apr 8 '14 at 21:06
  • $\begingroup$ Is $\phi$ a function of 2 variables? $\endgroup$ – Bjørn Kjos-Hanssen Apr 9 '14 at 18:35
  • $\begingroup$ Yep. You can also take it if the form $\phi(x,y)=\psi(x)\psi(y)$ as above. $\endgroup$ – Pietro Majer Apr 9 '14 at 19:05
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Here is a partly baked idea: Hang a sheet--a light cloth--from the segment $(0,0,0)-(1,0,0)$ along the bottom of your square, and from the segment $(0,1,0)-(1,1,1)$ slanting above the top of your square, and let it sag under gravity below the $xy$-plane, pinned to these two segments. There are very nice cloth simulation algorithms implemented, e.g., in Blender below. It seems you could approximate the shape of the sheet by hanging a catenary between the points $(x,0,0)$ and $(x,1,x)$. By increasing the lengths of the catenaries as a function of $x$, it seems you should be able to make the surface concave throughout. Likely the catenaries could be replaced by parabolas.


      Cloth simulation
Addendum. I defer to Pietro's more precise analysis, but just to hint toward the idea I suggested, here is a (substantially imperfect) rendition of the sagging sheet that (nearly) matches the boundary conditions:
      SheetSagging

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    $\begingroup$ 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 $f \ge0$ . $\endgroup$ – Pietro Majer Apr 8 '14 at 23:33
  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Apr 9 '14 at 0:47
  • $\begingroup$ (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 $y$). $\endgroup$ – Pietro Majer Apr 9 '14 at 0:54

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