Let $H\colon U\to\mathbb{R}$ with $U\subset\mathbb{R}^n$ be smooth and suppose $H$ has a unique minimum $u_\infty\in U$. Moreoever, suppose that for some $\lambda>0$, we have that $H$ is strong convex, that is, $\nabla^2 H\geq \lambda\textbf{1}$ uniformly on $U$.
It can be shown that $$ H[u]-H[u_\infty]\leq\frac{1}{2\lambda}D_H[u],\tag{1} $$ where $$ D_H[u(t)]=-\frac{d}{dt}H[u(t)]=\lVert\nabla H[u(t)]\rVert^2\geq 0. $$
Not, it is said that from $(1)$, we can deduce by Gronwall's inequality, that $$ e^{2\lambda t}(H[u(t)]-H[u_\infty]) $$ is non-increasing in time $t\geq 0$.
I am not sure if I understand the application of Gronwall's lemma.
Here is what I tried and think:
As far as I see, we have by (1) and the definition of $D_H[u(t)]$ that $$ \frac{d}{dt}H[u(t)]\leq -2\lambda(H[u(t)]-H[u_\infty]) $$
Moreoever, since $\frac{d}{dt}H[u_\infty]=\underbrace{\nabla H[u_\infty]}_{=0}\cdot\frac{d}{dt}u_\infty(t)=0$, we also can write
$$ \frac{d}{dt}(H[u(t)]+H[u_\infty])\leq -2\lambda(H[u(t)]-H[u_\infty]). $$
If I set $g(t):=H[u(t)]-H[u_\infty(t)]$ and am not mistaken, Gronwall's inequality implies $$ g(t)\leq e^{-2\lambda t}g(0) $$ or, equivalently, $$ e^{2\lambda t}g(t)\leq g(0) $$
Isn't that the claim?!
PS. I am referring to this on page 6
Well, since you have that \begin{align} \frac{d}{d t} H[u(t)] \le -2\lambda H[u(t)]+2\lambda H[u_\infty] \end{align} then it follows that \begin{align} e^{2\lambda t}\frac{d}{d t} H[u(t)] \le -2\lambda e^{2\lambda t} H[u(t)]+2\lambda e^{2\lambda t} H[u_\infty] \end{align} or equivalently \begin{align} \frac{d}{d t}\left(e^{2\lambda t} \left(H[u(t)]-H[u_\infty]\right)\right)= e^{2\lambda t}\frac{d}{d t} H[u(t)]+2\lambda e^{2\lambda t} \left(H[u(t)]- H[u_\infty]\right)\le 0. \end{align}
Well, this means that \begin{align} \frac{d}{d t}F(t)\le 0 \end{align} where \begin{align} F(t) = e^{2\lambda t} \left(H[u(t)]-H[u_\infty]\right). \end{align} Hence $F(t)$ must be non increasing since its derivative is non positive.
Although the proof of the inequality does not strictly require the use of Gronwall's lemma, it does rely on the method of integrating factor, which serves as the foundation for the proof of Gronwall's lemma.