A function is convex if and only if its epigraph $\big\{(\mathbf x,z):z\ge f(\mathbf x)\big\}$ is a convex set. So if your function is nonconvex, take the function corresponding to the convex hull of its epigraph. Other equivalent ways of describing this:
- Let $\hat f$ be the largest convex function dominated by $f$. $\hat f(\mathbf x) = \max\big\{g(\mathbf x):\text{$g$ is convex and $g(\mathbf y)\le f(\mathbf y)$ for all $\mathbf y$}\big\}.$
- Let $\hat f$ be the maximum of all affine functions dominated by $f$. $\hat f(\mathbf x) = \max\big\{\mathbf a\cdot\mathbf x + b:\text{$\mathbf a\cdot\mathbf y+b\le f(\mathbf y)$ for all $\mathbf y$}\big\}.$
- Let $\hat f(\mathbf x)$ be the smallest $\sum\alpha_i f(\mathbf x_i)$ over all possible convex combinations $\sum\alpha_i\mathbf x_i=\mathbf x$. $\hat f(\mathbf x) = \textstyle\min\big\{\sum\alpha_i f(\mathbf x_i):\alpha_i\ge0,\sum\alpha_i=1,\sum\alpha_i\mathbf x_i = \mathbf x\big\}.$
(I'm being sloppy and assuming that the $\max$/$\min$ are attained and we don't need $\sup$/$\inf$ instead; I might be wrong.)
I haven't given any thought to your specific case of $f(\mathbf x)=\lVert\mathbf x\rVert_2\lVert\mathbf x\rVert_1$ yet. If I think of something, I'll edit it into this answer. Surely the symmetry of the function should help a lot here.