*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 $\mathbb{R}^d$.