Does this relation imply convexity? I'm trying to figure out wheter the following condition inplies convexity or not.
Let $\cal{X}$ be an inner product space with inner product $\langle \cdot, \cdot \rangle$ and a norm $\|\cdot\|$ (not necessarily the one induced by $\langle \cdot , \cdot\rangle$) and $X \subset \cal{X}$ a closed convex set.
Let be $f:X \to \mathbb{R}$, $G: X \to X$ functions and $L>0$ be a constant such that for every $x, x' \in X$ the following condition holds
$$ f(x') + \langle G(x'), x - x'\rangle \leq f(x) \leq f(x') + \langle G(x'), x - x'\rangle + \frac{L}{2}\|x - x'\|^2 $$
Then can one say that $f$ is convex on the set $X$?
I know there's some similar results for the case where $f$ is differentiable and the norm is induced by the inner product, but here the function $G$ not necessarily is the gradient of $f$ and $f$ differentiable may not even have sense since he set is closed. Also the norm is general... So does anyone know if $f$ is convex in this more general context?

Consider the function
$$
g(x,x')=f(x')+\langle G(x'),x-x'\rangle.
$$
The function $g(\cdot,x')$ is linear for all $x'$, hence, convex. The left inequality means that for all $x\in X$
$$
g(x,x')\le f(x),\quad \forall x'\in X \qquad\text{and}\qquad g(x,x)=f(x),
$$
which gives immediately that
$$
f(x)=\max_{x'\in X}g(x,x').
$$
Therefore, the function $f$ is convex as a maximum of convex functions.
