Let $\Omega \subset \mathbb{R}^{n}$ be a convex open set and $f:\Omega \rightarrow \mathbb{R}$, a differentiable function in $\Omega$, with continuous derivative in $\Omega$.
Prove that if
$\forall x, y \in \Omega$,$$ (f'(y)-f'(x))^{T}(y-x) \geq 2\mu ||y-x||^{2} $$ Then, $\forall x, y \in \Omega$:
$$f(y)-f(x) \geq (f'(x))^{T}(y-x) + \mu ||y-x||^{2}$$
Let $x, y \in \Omega$, and define $g: [0,1] \to \mathbb{R}$ by $g(t) = f(x + t(y-x))$ (since $\Omega$ is convex, this is well defined). As the composition of differentiable functions, $g$ is differentiable with $g'(t) = f'(x + t(y-x))^\top (y-x)$.
We use the given inequality to establish a lower bound on $g'(t)$. From this inequality, we have:
$f'(x + t(y-x))^\top (x + t(y-x) -x) \geq f'(x)^\top (x + t(y-x) -x) + 2\mu ||x + t(y-x) -x||^2$
Simplifying gives:
$tf'(x + t(y-x))^\top (y-x) \geq tf'(x)^\top (y-x) + 2\mu t^2 ||y - x||^2$
Dividing both sides by $t$:
$f'(x + t(y-x))^\top (y-x) \geq f'(x)^\top (y-x) + 2\mu t ||y - x||^2$
The left hand side is $g'(t)$. Finally, notice that $f(y) - f(x) = g(1) - g(0) = \int_0^1 g'(t) dt$. By the bound derived, we have:
$\int_0^1 g'(t) dt = \int_0^1 f'(x + t(y-x))^\top (y-x) dt \geq \int_0^1 f'(x)^\top (y-x) + 2\mu t ||y-x||^2 dt$
Simplifying the right hand side yields:
$f(y) - f(x) \geq f'(x)^\top (y-x) + \mu ||y-x||^2$