Upperbound for max degree of k-tree completion Definitions: For a graph $G$, a $k$-tree completion of $G$ is a $k$-tree obtained by adding edges to $G$ (if $G$ has a $k$-tree completion, $G$ is said to be a partial $k$-tree). The least integer $k$ such that $G$ has a $k$-tree completion is called the treewidth of $G$. The maximum degree of a graph $G$ is denoted by $\Delta(G)$.
Question: Suppose that $G$ has a $k$-tree completion.
Is there an upper bound $U$ in terms of $k$ and $\Delta(G)$ such that
there exist a $k$-tree completion $G'$ of $G$ with $\Delta(G')\leq U\,?$
(note: To be explicit, $U$ is a function of $k$ and $\Delta(G)$ )
Context: I have a graph parameter $x(G)$ such that $x(G)\leq \Delta(G)+p$ for every chordal graph $G$ where $p$ is a constant. Using this result, I am trying to give an upper bound for $x(G)$ in general graphs in terms of treewidth of $G$ and $\Delta(G)$.
It is known that a graph of treewidth $k$ and maximum degree $\Delta$ has tree partition width at most $O(k\Delta)$. See Wood, arXiv:math/0602507.
From a tree partition of width $O(k\Delta)$ a triangulation of width $O(k\Delta)$ and maximum degree $O(k\Delta^2)$ follows pretty directly (start with the tree partition, make every bag into a clique, make every vertex $v$ in $U$ adjacent to all vertices in all bags appearing together with neighbors of $v$).
So you get a (minimal) triangulation with the bounds above.
Note: I would not be surprised if you can get a better degree bound.. the additional $\Delta$ term feels unnecessary.