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Tomasz Kania
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Simultaneous extensions of strictly convex functions

Suppose that we are given two subspaces $M$ and $N$ of $\mathbb{R}^d$ that have possibly non-empty intersection and let us say that the standard basis of $\mathbb{R}^n$ is contained in $M\cup N$.

Moreover, suppose that we are given two strictly convex $C^2$-functions $f\colon M\to \mathbb{R}$ and $g\colon N\to \mathbb{R}$ that are equal on $M\cap N$. That $f,g$ are strictly convex means that their Hessians are positive definite.

The union function $f\cup g$ admits many extensions to a $C^2$-function defined on $\mathbb{R}^d$.

Let $\hat{g}= g\circ Q$, where $Q$ is any projection onto $N$. Set $\psi = f - \hat{g}|_M$. We need to extend $\psi$ to a $C^2$-function $\hat{\psi}$ on $\mathbb{R}^n$ that vanishes on $N$. Indeed, $\hat{g}+\hat{\psi}$ will extend $f\cup g$. Let $S$ be a subspace of $N$ such that $\mathbb{R}^d=M\oplus S$. Let $P_M$ be a projection onto $M$ with $\ker P_M=S$. We set $\hat{\psi}(x)=\psi(P_Mx)$. Certainly, $\hat{\psi}$ is $C^2$ that extends $\psi$. Moreover, if $x\in N$, then $x=y+s$ with $y\in M$ and $s\in S$. As $s\in S\subseteq N$, we have $x-s=y\in N$, thus $P_Mx\in M\cap N$. Consequently, $$\hat{\psi}(x)=\psi(P_Mx)=f(P_Mx)-g(P_Mx)=0.$$

Is it possible to extend $f\cup g$ to a $C^2$-function $h$ such that

$$\inf_{x\in A}\min \sigma [D^2h(x)] = \inf_{x\in S_M}\min \sigma [D^2f(x)]\wedge \inf_{x\in S_N}\min \sigma [D^2g(x)]?$$

(Here $S_M$ and $S_N$ are the unit spheres of $M$ and $N$ respectively and $A=S_M\cup S_N$, $D^2$ denotes Hessian and $\sigma$ is the set of all eigenvalues.)

The problem is that obviously $A$ is not convex as for compact, convex there are suitable versions of Whitney's extension theorem that would allow for such conclusions.

Tomasz Kania
  • 11.3k
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  • 39
  • 75