Existence of a strictly convex function interpolating given gradients and values

Question

I'm wondering where to find a proof and reference for the following facts, which I feel sure must be true.

(1) Suppose we are given a finite set of points in $\mathbb{R}^{d+1}$. For each point, we are given a closed halfspace tangent to it and strictly containing all others. Then there exists a strictly convex set contained in the intersection of the halfspaces with all the given points on the boundary.

(If I drop "strictly", then the convex hull of the points works.)

(2) Suppose we are given a finite set of pairs $(x,y)$ with $x \in \mathbb{R}^d$ and $y \in \mathbb{R}$. For each point, we are given a linear function through it and strictly below the others. Then there exists a strictly convex function passing through the points whose gradient at each given point matches the given linear function's.

(If I drop "strictly", then "convex closure" of the pairs works.)

Answer 1

Both facts have short proofs. We first prove (1) and then get a stronger statement for (2) - so that the result is $C^{\infty}$ smooth.

Proof of (1). By assumptions, for each point $x$ of the set there is a half-space $H_x$ containing $x$ in its boundary and containing all other points in its interior. So there is also a radius $R$ ball $B_x^R$ contained in $H_x$, tangent to $\partial H_x$ at $x$ and containing all the other points of the set (just take $R$ large enough). Now, the intersection of all balls $B_x^R$ is the desired strictly convex set.

Proof of (2). Let's show how to construct such a strictly convex function in the $C^{\infty}$ class. So suppose that our collection of points is $(x_i,y_i)$ and the planes are defined by $y=L_i(x)$ (so that $y_i=L_i(x)$). It is clear than the for $\varepsilon>0$ small enough the function $$F(x)=\max_i(L_i(x)+\varepsilon|x-x_i|^2)$$ solves the problem. However $F_{\varepsilon}$ is not smooth. To find a smooth function let's choose a symmetric density $\rho_{\varepsilon}(x)\ge 0$ satisfying

1) $\rho_{\varepsilon }(x)=\rho_{\varepsilon }(-x)$, 2) $\int \rho_{\varepsilon }=1$, and 3) $\rho_{\varepsilon }(x)=0$ for $|x|>\varepsilon$.

The following claim is easy.

Claim. The convolution $\widetilde F_{\varepsilon}=F_{\varepsilon}*\rho_{\varepsilon }$ has the property that $\widetilde F_{\varepsilon}-F_{\varepsilon}=const(\varepsilon)>0$ at points that are on distance more than $\varepsilon$ from the locus of non-smoothness of $F_{\varepsilon}$.

So it is easy to see that for $\varepsilon$ small enough the function $\widetilde F_{\varepsilon}-const(\varepsilon)$ will do the job.

Answer 2

Since the topic of smoothness came up in the other answer: Here is a reference for the case where one asks for the function to be convex and $C^{1,1}$, i.e. Lipschitz continuous derivative.

In general, the problem of constructing a function from some function values and (possibly higher) derivatives is called the "Whitney extension problem". So if you want to read more about this you can look for "convex Whitney extension".