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?

Yes, it is convex, and only the first inequality is needed.
Consider any  $a, b \in X$ and $0 < t < 1$.  For convenience write $c = ta + (1-t) b$.
Taking $x = a$ and $x'=c$, we have $x - x' = (1-t) (a - b)$ so
$$f(c) +  (1-t) \langle G(c),a-b\rangle \le f(a) \tag{1}$$
Similarly, taking $x=b$ and $x'=c$, we have
$$f(c) +  t \langle G(c),b-a \rangle \le f(b)\tag{2}$$
Add $t$ times (1) and  $1-t$ times (2) to obtain
$$f(c) \le t f(a) + (1-t) f(b)$$
Thus it is convex.
