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?


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 $x$).

  • $\begingroup$ 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

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:

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.