When is the difference of two convex functions convex?

Assume that $X$ is a finite dimensional Banach space. I know that in general if two functions $f:X \mapsto \mathbb{R}$, $g:X \mapsto \mathbb{R}$ are convex then the function $(f-g):X \mapsto \mathbb{R}$ given by $f(x)-g(x)$ is not necessarily convex.

I'd like to ask if are there conditions we can impose on $f$ and $g$ so that the difference is still convex, for instance if $f \geq g$ for every $x$ then can we say it's convex?

Are there any results about convexity of difference of convex functions?

Thanks in advance

For any real-valued function $h$, $\alpha\in[0,1]$ and $x,y$ in the (convex) domain, let $$ D(h,\alpha,x,y)=\alpha h(x)+(1-\alpha)h(y)-h[\alpha x+(1-\alpha)y]. $$ Convexity for $h$ means $D(h,\alpha,x,y)\geq 0$ for all $\alpha,x,y$.

For your situation, one sufficient condition for $f-g$ to be convex is that $$ D(f,\alpha,x,y)\geq D(g,\alpha,x,y)\tag{i} $$ for all $\alpha,x,y$. (I think of this as $f$ being "more convex" then $g$.) Note that (i) doesn't impose convexity directly on $f$ and $g$. For example, $f(x)=-x^2$ and $g(x)=-2x^2$ satisfy (i) so that $f-g$ is convex but $f$ and $g$ are individually concave.

When the domain is $\mathbb{R}$ and $f$ and $g$ are both twice differentiable, it is also sufficient to have $$ f''\geq g''\tag{ii}. $$

$f\ge g$ is not sufficient: for example, take $f(x)=\sqrt{x^2+1}$ and $g(x)=|x|$ (with $X=\mathbb R$); for another, take $f(x)=|x|$ and $g(x)=\max\ \{0,|x|-1\}$.