# Simultaneous extensions of strictly convex functions

21/03/2017: I have decided to accept Denis Serre's answer, even though it does not exactly answer my question, however I like its simplicity and I'd say it is close enough to the desired claim. Of course, I would still love to see the answer to my original question.

We are given two subspaces $M$ and $N$ of $\mathbb{R}^d$ that have possibly non-empty intersection and let $C_M$ and $C_N$ be compact, convex sets in these subspaces both containing 0.

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 and so, the eigenvalues of their Hessians are bounded below by some (common) $\delta>0$ on $C_M$ and $C_N$, respectively.

The union function $f\cup g$ admits many extensions to a $C^2$-function defined on $\mathbb{R}^d$, some of them may written down explicitly by using projections onto these subspaces, however by using projections we always produce extra zeroes as eigenvalues of the extension. Hence my question:

Is it possible to extend $f\cup g$ to a $C^2$-function $h$ such that $$\inf_{x\in C_M\cup C_N}\min \sigma [D^2h(x)] \geqslant \delta?$$ Or at least $\geqslant \delta-\varepsilon$ for given $\varepsilon \in (0, \delta)$?

(Here, $D^2$ denotes Hessian and $\sigma$ is the set of all eigenvalues. Without loss of generality the canonical basis in $\mathbb{R}^d$ is contained in $M\cup N$.) Note that we do not require $h$ to be convex.

The problem is that obviously $C_M\cup C_N$ is not convex as for compact, convex there are suitable versions of Whitney's extension theorem that would allow for such conclusions, however union of two subspaces is a particularly nice set in $\mathbb{R}^d$.

Here is an elementary proof that the answer is positive, at least if you relax the condition that the extension be $C^2$.

Lemma. Given $x_0\in C_M$, there exists a linear form $\lambda$ such that $f(x_0)+\lambda(x-x_0)\ge0$ over the graphs of both $f$ and $g$.

To see this, consider the tangent space $\{(x,f(x_0)+df_{x_0}(x-x_0))\,|\,x\in C_M\}$ to the graph of $f$. It is located below this graph (touching at $x_0$). If it meats the graph of $G$ at some point $(y,p)\in C_N$, then $y\in M\cap N$ and therefore $f(y)=g(y)$. But then $f(y)=p=f(x_0)+df_{x_0}(y-x_0)$, which implies $y=x_0$. There are therefore two cases. Either $x_0\in C_M\cap C_N$, and because $dg_{x_0}=df_{x_0}$ over $M\cap N$, we may choose $\lambda$ an compatible extension of $dg_{x_0}$ and $df_{x_0}$ to $R^d$. Or $x_0\not\in C_N$, and then you can choose $\lambda$ by applying Hahn-Banach.

Let me now remark that if $x_0$ and $\lambda$ are as in the lemma, then by assumption, the function $$\phi_{x_0,\lambda}(x):=\frac\delta2|x-x_0|^2+\lambda(x-x_0)+f(x_0)$$ satisfies $\phi_{x_0,\lambda}|_{C_M}\le f$ and $\phi_{x_0,\lambda}|_{C_N}\le g$.

Likewise, we consider those pairs $(y_0,\mu)$ such that the functions $$\psi_{y_0,\mu}(x):=\frac\delta2|x-y_0|^2+\mu(x-y_0)+g(y_0)$$ satisfy $\psi_{y_0,\mu}|_{C_M}\le f$ and $\psi_{y_0,\mu}|_{C_N}\le g$.

Finally, we form the function $$S(x)=\max\left\{\max_{x_0,\lambda}\phi_{x_0,\lambda}(x),\max_{y_0,\mu}\psi_{y_0,\mu}(x)\right\}.$$ This is a convex function, which coincides with $f$ over $C_M$ and with $g$ over $C_N$. In addition, it satisfies $D^2S\ge\delta I_d$ in the sense of distributions. In other words, for every $z_0$, there exists a linear form $\zeta$ such that $$S(x)\le S(z_0)+\zeta(x-z_0)+\frac\delta2|x-z_0|^2,\qquad\forall\,x.$$

Notice that $S$ is the largest function satisfying all the queries. Except the $C^2$ regularity of course ; exactly as an ordinary convex enveloppe fails to be $C^1$.

• (1)I think in both cases you can choose linear form $\lambda$ using H-B due to compatibility. (2)Why is it true that the supporting linear form $\lambda$ for $f$ at $x_0$ also supports $g$ at $x_0$? I think $x_0$ must be minima of $f$ according to your construction. (3) I do not see why the extension $S$ is $C^2$ if the maximum is taken over $x_0,\lambda$ at the same time in the inner maximums...Sorry if I misunderstood. – Henry.L Mar 20 '17 at 22:57
• I don't understand your query (1). About (2), it comes from the compatibility assumption that $f$ and $g$ coincide over $C_M\cap C_N$. About (3), I did not claim that my construction is $C^2$. Exactly as an ordinary convex enveloppe is usually not $C^1$. I said that my construction satisfies $D^2S\ge\delta I_d$ in the distributional sense. – Denis Serre Mar 21 '17 at 6:42
• Thank you Denis, this looks promising. I am now travelling so will return to you in the evening. For my purposes unfortunately this won't be sufficient if I loose $C^2$, however what if we assume that $f,g$ have Hessians bounded below by $\delta$ in the distributional sense (without being $C^2$)? Does your constriction carry over? – Tomasz Kania Mar 21 '17 at 7:57
• @Tania. Yes. We need only that $D^2,D^2g\ge\delta I_d$ in the distributional sense. This is equivalent to saying that $f(x)-\frac\delta2|x|^2$ (the same with $g$) is concave. Therefore $f(x)\le f(x_0)+df_{x_0}(x-x_0)+\frac\delta2|x-x_0|^2$. – Denis Serre Mar 21 '17 at 10:18
• @DenisSerre Okay, I see why (2,3) works, thanks for explanation. – Henry.L Mar 21 '17 at 12:00

Let me know if I missed anything, but I think what you asked is actually a result by C.Fefferman in reponse to "finiteness principle"[Fefferman1] of interpolation if we do following observations.

Claim: As long as we proved some $\omega$-continuity for the Hessian $D^2h$ i.e. we have an extension $h\in C^{2,\omega}(\mathbb{R}^d)$, then the spectrum of Hessian will not oscillate too much in a small neighborhood in $\mathbb{R}^d$ of $C_M\cup C_N$ in form of $$\{ x\in\mathcal{R}^d:dist(x,C_M\cup C_N)\leq\epsilon \},\epsilon>0,dist(x,S):=inf_{y\in > S}d_{\mathbb{R}^d}(x,y)$$

This is true because following observations. The Hessian $D^2h$ is symmetric under your assumption, due to Hoffman-Wielandt Theorem, as long as the entries $\frac{\partial^2h}{\partial x_i^{2}},\frac{\partial^2h}{\partial x_i x_j}$ do not oscillate too much in the neighborhood, the spectrum of Hessian will not oscillate too much. Thus the positivity of $D^2h$ is preserved in such a neighborhood since $h$ coincide with $f\cup g$ on $C_M\cup C_N$ while $f\cup g$ is known to possess positivity in their Hessian.

But the entries will not oscillate too much in a small neighborhood of $C_M\cup C_N$ if $h\in C^{2,\omega}(\mathbb{R}^d)\subset C^2(\mathbb{R}^d)$ ($h$'s second derivative has $\omega$-continuity).

So it suffices to find such an $h\in C^{2,\omega}(\mathbb{R}^d)$. The norm defined on this space is $$\left\Vert F\right\Vert _{C^{m,\omega}(\mathbb{R}^{n})}=max\left\{ \left\Vert F\right\Vert _{C^{m}(\mathbb{R}^{n})},max_{|\beta|=m}sup_{x,y\in\mathbb{R}^{n},0<|x-y|\leq1}\frac{|\partial^{\beta}F(x)-\partial^{\beta}F(y)|}{\omega(|x-y|)}\right\}$$ according to [Fefferman2],(there is a typo in my comment so I deleted it.) and the usual Sobolev norm $$\left\Vert F\right\Vert _{C^{m}(\mathbb{R}^{n})}=max_{|\beta|\leq m}sup_{x\in\mathbb{R}^{n}}|\partial^{\beta}F(x)|$$ where $m=2$ in our case.

Now yield the Theorem 1 of [Fefferman1] by letting tolerance $\sigma\equiv 0$ and the Whitney $\omega$-convexity is trivially satisfied in this case. I would like to hear your motivation if possible.

[Fefferman1]Fefferman, Charles. "A generalized sharp Whitney theorem for jets." Revista Matematica Iberoamericana 21.2 (2005): 577-688.

[Fefferman2]Fefferman, Charles. "Extension of $C^{m,\ omega}$-Smooth Functions by Linear Operators." Revista Matematica Iberoamericana 25.1 (2009): 1-48.

This answer does not solve the OP's problem, please see our comment below.

• It is not a trivial task to verify the norm if we follow Fefferman's original construction, but in certain special situations like $f\cup g$ is a Sobolev function with $\omega$-continuity and regularities, I did try to approximate it using smoothing kernel quite a while ago. I have dropped you an email. – Henry.L Mar 16 '17 at 12:21
• Because he is Fefferman :) I resent it! Sorry for the delay. – Henry.L Mar 16 '17 at 12:37
• Unfortunately any such extension $h$ is not enough, and therefore your answer does not provide a solution to the problem. I am saying this so that other people may try to attempt the problem from different angles. – Tomasz Kania Mar 17 '17 at 19:07
• @TomekKania is correct in this case an arbitrary $h$ may fail to satisfy a (local) extension. After a discussion, we are still unable to claim how/whether such an extension will meet the requirement. – Henry.L Mar 17 '17 at 19:15
• Added, and will keep thinking over it :-) – Henry.L Mar 17 '17 at 22:54