***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](https://en.wikipedia.org/wiki/Whitney_extension_theorem) that would allow for such conclusions, however union of two subspaces is a particularly nice set in $\mathbb{R}^d$. 



Simultaneous extensions of strictly convex functions
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 $C^1$.


